a. Units of Fo & λFor the given question,A block of mass m is initially at rest at the origin x = 0. A one-dimension force given by F = Fo ex, where Fo & λ are positive constants, is applied to the block.a. What are the units of Fo & λ?The unit of force F = Fo ex is N (newton). Here, x is a dimensionless quantity since it is a unit vector with no magnitude. Hence, the dimension of Fo is N.λ is the wavelength which is the distance between two similar points of a wave. It is measured in meters (m). Therefore, the dimension of λ is m.
b. Argue that the force is conservative The work done by the force is conservative if it is equal to the negative of the potential energy. To verify if the force is conservative, we must check if the cross-partial derivatives are equal. Hence,∂F/∂x = Fo∂/∂x (ex) = 0∂F/∂y = Fo∂/∂y (ex) = 0∂F/∂z = Fo∂/∂z (ex) = 0Since the cross-partial derivatives are zero, the force is a conservative force.
c. Find the potential energy associated with the forceThe potential energy is the negative of the work done by the force. Therefore, the potential energy is given byU = -W(x2, x1) = - ∫(x1, x2) F · drWe know, F = Fo exTherefore, U = - Fo ex · xThus, the potential energy isU = - Fo x.d. Find the velocity of the block as a function of position xWe know that the work done by the force is equal to the change in kinetic energy. Therefore,W(x2, x1) = K(x2) - K(x1)Since the block starts at rest, the initial kinetic energy is zero. Hence,K(x1) = 0Therefore,W(x2, x1) = K(x2)Solving for velocity,v(x) = [2/m ∫(0,x) F dx]^(1/2)We know that F = Fo exTherefore,v(x) = [2/m ∫(0,x) Fo ex dx]^(1/2)v(x) = [2Fo/m ∫(0,x) ex dx]^(1/2)v(x) = [2Fo/m (ex)|0x]^(1/2)v(x) = [2Fo/m (e^(x) - 1)]^(1/2)
e. Find the terminal speed of the block as x → ∞As x approaches infinity, the velocity approaches a maximum value, known as the terminal velocity. Therefore,vt = lim (x → ∞) v(x)We know that,v(x) = [2Fo/m (e^(x) - 1)]^(1/2)Taking the limit,vt = lim (x → ∞) [2Fo/m (e^(x) - 1)]^(1/2)vt = lim (x → ∞) [2Fo/m (e^(2x) - 2e^x + 1)]^(1/2)vt = lim (x → ∞) [(2Fo/m) (e^(2x))]^(1/2)vt = lim (x → ∞) [(2Fo/m) (e^x)]Therefore, the terminal speed of the block as x approaches infinity isvt = [(2Fo/m) ∞]^(1/2) = ∞Therefore, the block does not have a terminal speed.
About Potential energyPotential energy is energy that affects objects because of the position of the object, which tends to go to infinity with the direction of the force generated from the potential energy. The SI unit for measuring work and energy is the Joule. What are the uses of potential energy? Potential energy is energy that is widely used to generate electricity. Even so, there are two objects that are used to store potential energy. These objects, namely water and fuel, are used to store potential energy. In general, water can store potential energy like a waterfall.
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1) Flexible steel wire shaft encased in a stationary tube that fits closely enough to impose a frictional torque of 3 N-m/m. the wire has a diameter of 12 mm and the induced stress must not exceed 145 MN/m^2. What will be the angular rotation of one end with respect to the other end?
answer. 136.83 deg
2) A steel shaft 1.75 inches in diameter transmits 40 Hp at 1800 rpm. Assuming a modulus of rigidity of 12 x 10^6 psi, find the torsional deflection of the shaft in degrees per foot length.
answer. 0.0073
1. The angular rotation of one end with respect to the other end is 6.79 degrees (approx.)
2. The torsional deflection of the shaft in degrees per foot length is 0.0073 degrees/ft.
1) Calculation of angular rotation of one end with respect to the other end:The torque induced by the wire is given as 3 N-m/m.
The polar moment of inertia of a wire with diameter d is given as J = πd⁴/32.
The induced shear stress is given as τ = T×r/J
Here, the radius of wire, r = d/2 = 6mm = 0.006 m
The induced shear stress is given as:τ = (3 N-m/m) × (0.006 m) / (π×(0.012 m)⁴/32)τ = 546.478 MN/m² = 0.546 GN/m²
The induced shear stress must not exceed 145 MN/m².
So, τmax = 145 MN/m².
Since τ < τmax, the induced shear stress is within limits.
The induced shear strain is given as:τ = G×γWhere G is modulus of rigidity.
So, the induced shear strain γ is given as:γ = τ / Gγ = (546.478×10⁶ Pa) / (80×10⁹ Pa)γ = 0.00683
The twist in one meter length of the wire is given as:ϕ = (T×L)/(G×J)Here L = 1m, the length of wire.
So, the twist angle is given as:
ϕ = (3 N-m/m) × (1 m) / ((80×10⁹ Pa) × (π×(0.012 m)⁴/32))
ϕ = 0.1186
radians = 6.79 degrees
Therefore, One end rotates 6.79 degrees (about) with regard to the opposite end.
2. Calculation of torsional deflection of the shaft in degrees per foot length:
The power being transmitted by the shaft is 40 HP and the speed of rotation is 1800 rpm.1 HP = 550 ft-lb/s.
So, the torque transmitted by the shaft is given as:T = (40 HP × 550 ft-lb/s) / (1800 rpm × 2π rad/rev)T = 525.18 lb-ft
The torsional stress induced is given as:τ = (T×r)/J
The polar moment of inertia of a solid shaft is given as J = πd⁴/32.
So, the torsional stress is given as:τ = (T×r)/(πd⁴/32)τ = (525.18 lb-ft × 12 in/ft × 0.5 in) / (π×(1.75 in)⁴/32)τ = 0.0561 kpsi
The torsional shear strain is given as:γ = τ/GThe modulus of rigidity of steel is 12×10⁶ psi.
Therefore, the torsional shear strain is given as:γ = (0.0561 kpsi) / (12×10⁶ psi)γ = 4.68×10⁻⁶ radians/in
The torsional deflection of the shaft in degrees per foot length is given as:θ = (360/2π) × (γ × 12 in/ft)θ = 0.0073 degrees/ft
Therefore, The shaft's torsional deflection is 0.0073 degrees per foot of length.
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A race car starts from rest and accelerates at 20.0 m/s2 for 5.0 s. It then moves with a uniform velocity for 60 s before it decelerates at 10.0 m/s2 until it comes to a stop. *Note: Show the complete solution by showing all of your work! (
a)Determine the velocity of the race car after 5.0 s.
(b)Determine the total distance traveled by the car between from the instant it started to move until it came to a stop.
(c)What was the total time that the car was in motion?
(a)The velocity of the race car after 5.0 s is 100.0 m/s. (b) The total distance traveled by the car from the instant it started to move until it came to a stop is 6250.0 m. (c) The total time that the car was in motion is 75.0 s.
Let's calculate the values step by step:
Given:
Acceleration during the first phase (a1) = 20.0 m/s²
Time during the first phase (t1) = 5.0 s
Uniform velocity phase (t2) = 60 s
Deceleration during the third phase (a3) = -10.0 m/s²
(a) To determine the velocity of the race car after 5.0 s, we can use the formula:
v = u + a × t
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration,
t is the time.
Since the race car starts from rest, the initial velocity (u) is 0 m/s.
v = 0 + (20.0 m/s²) × (5.0 s)
v = 100.0 m/s
Therefore, the velocity of the race car after 5.0 s is 100.0 m/s.
(b) To determine the total distance traveled by the car, we need to calculate the distance covered during each phase and sum them up.
Distance during the first phase:
Using the equation of motion:
s1 = u × t1 + (1/2) × a1 × t1²
Since the initial velocity (u) is 0 m/s:
s1 = (1/2) × (20.0 m/s²) × (5.0 s)²
s1 = 250.0 m
Distance during the second phase:
Since the car moves with a uniform velocity, the distance covered is:
s2 = v × t2
s2 = 100.0 m/s × 60 s
s2 = 6000.0 m
Distance during the third phase:
Using the equation of motion:
s3 = v × t3 + (1/2) ×a3 × t3
Since the final velocity (v) is 0 m/s:
s3 = (1/2) × (-10.0 m/s²) × t3²
The time during the third phase (t3) can be found by equating the final velocity to 0:
v = u + a × t
0 = 100.0 m/s + (-10.0 m/s²) × t3
t3 = 10.0 s
Substituting the value of t3:
s3 = (1/2) × (-10.0 m/s²) × (10.0 s)²
s3 = -500.0 m
Total distance traveled by the car:
Total distance = s1 + s2 + s3
= 250.0 m + 6000.0 m + (-500.0 m)
= 6250.0 m
Therefore, the total distance traveled by the car from the instant it started to move until it came to a stop is 6250.0 m.
(c) To determine the total time that the car was in motion, we add the
time for each phase:
Total time = t1 + t2 + t3
= 5.0 s + 60 s + 10.0 s
= 75.0 s
Therefore, the total time that the car was in motion is 75.0 s.
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A projectile of mass m is thrown at an initial speed vo and angle 8 from the ground. The projectile is subjected to air resistance proportional to velocity. Treat the problem in 2 dimesnions. a. Write down Newton's second law describing the motion of the projectile, in componenets form. (4pts) b. Find v₂ (t). (4pts) c. Find vy(t). (6pts) d. Find the terminal speed too. (2pts)
a. The equation of motion for the projectile in component form is: [tex]\(ma_x = -f_v \cdot v_x\) and \(ma_y = -mg - f_v \cdot v_y\).[/tex]
b. The equation for the x-component of velocity, [tex]\(v_x(t)\)[/tex], as a function of time is: [tex]\(v_x(t) = v_0 \cos(\theta)\left(1 - \frac{2\gamma t}{m \cos^2(\theta)}\right)\).[/tex]
c. The equation for the y-component of velocity, [tex]\(v_y(t)\)[/tex], as a function of time is: [tex]\(v_y(t) = v_0 \sin(\theta) - gt - \frac{\gamma t}{m}v_y(t)\).[/tex]
d. The terminal speed,[tex]\(v_{\text{term}}\)[/tex], is given by: [tex]\(v_{\text{term}} = \sqrt{\frac{mg}{k}}\).[/tex]
a. Newton's second law describes the motion of the projectile in component form as follows:
In the x-direction:
[tex]\[F_{\text{net},x} = ma_x = -f_v \cdot v_x\][/tex]
In the y-direction:
[tex]\[F_{\text{net},x} = ma_x = -f_v \cdot v_x\][/tex]
Where:
m is the mass of the projectile,[tex]\(v_x\)[/tex] is the velocity component in the x-direction,[tex]\(v_y\)[/tex] is the velocity component in the y-direction,[tex]\(f_v\)[/tex] is the air resistance force proportional to velocity,g is the acceleration due to gravity, and[tex]\(a_x\)[/tex] and [tex]\(a_y\)[/tex] are the accelerations in the x and y-directions, respectively.b. To find [tex]\(v_2(t)\),[/tex] we need to integrate the equation of motion for the x-direction with respect to time:
[tex]\[m \frac{{dv_x}}{{dt}} = -f_v \cdot v_x\][/tex]
Integrating this equation yields:
[tex]\[\int m \frac{{dv_x}}{{dt}} dt = -\int f_v \cdot v_x dt\][/tex]
[tex]\[m \int \frac{{dv_x}}{{dt}} dt = -\int f_v \cdot v_x dt\][/tex]
[tex]\[m v_x = -\int f_v \cdot v_x dt\][/tex]
[tex]\[m v_x = -\int f_v dt \cdot v_x\][/tex]
[tex]\[m v_x = -\int \gamma v_x dt\][/tex] where gamma is the coefficient of air resistance)
Integrating both sides gives:
[tex]\[m \int v_x dv_x = -\gamma \int v_x dt\][/tex]
[tex]\[\frac{1}{2} m v_x^2 = -\gamma t + C_1\][/tex] where [tex]\(C_1\)[/tex] is the constant of integration.
At time[tex]\(t = 0\), \(v_x = v_0 \cos(\theta)\),[/tex] so we can substitute this value in:
[tex]\[\frac{1}{2} m (v_0 \cos(\theta))^2 = -\gamma \cdot 0 + C_1\][/tex]
[tex]\[\frac{1}{2} m v_0^2 \cos^2(\theta) = C_1\][/tex]
Thus, the equation for[tex]\(v_x\)[/tex] as a function of time is:
[tex]\[v_x(t) = v_0 \cos(\theta)\left(1 - \frac{2\gamma t}{m \cos^2(\theta)}\right)\][/tex]
c. To find [tex]\(v_y(t)\)[/tex], we integrate the equation of motion for the y-direction:
[tex]\[m \frac{{dv_y}}{{dt}} = -mg - f_v \cdot v_y\][/tex]
Integrating this equation gives:
[tex]\[m \int \frac{{dv_y}}{{dt}} dt = -\int (mg + f_v \cdot v_y) dt\][/tex]
[tex]\[m v_y = -\int (mg + \gamma v_y) dt\][/tex]
[tex]\[m v_y = -\int mg dt - \int \gamma v_y dt\][/tex]
[tex]\[m v_y = -mgt - \int \gamma v_y dt\][/tex]
Integrating both sides gives:
[tex]\[m \int v_y dv_y = -mg \int dt - \gamma \int v_y dt\][/tex]
[tex]\[\frac{1}{2} m v_y^2 = -mgt - \gamma \int v_y dt\][/tex]
[tex]\[\frac{1}{2} m v_y^2 = -mgt - \gamma t v_y + C_2\][/tex] where [tex]\(C_2\)[/tex] is the constant of integration)
At time[tex]\(t = 0\), \(v_y = v_0 \sin(\theta)\)[/tex], so we can substitute this value in:
[tex]\[\frac{1}{2} m (v_0 \sin(\theta))^2 = -mg \cdot 0 - \gamma \cdot 0 \cdot (v_0 \sin(\theta)) + C_2\][/tex]
[tex]\[\frac{1}{2} m v_0^2 \sin^2(\theta) = C_2\][/tex]
Thus, the equation for [tex]\(v_y\)[/tex] as a function of time is:
[tex]\[v_y(t) = v_0 \sin(\theta) - gt - \frac{\gamma t}{m}v_y(t)\][/tex]
d. The terminal speed is the speed at which the projectile reaches a constant velocity, meaning the acceleration becomes zero. At terminal speed, [tex]\(v_x\)[/tex] and [tex]\(v_y\)[/tex] will no longer change with time.
From the equation of motion in the x-direction, when [tex]\(a_x = 0\)[/tex]:
[tex]\[m \frac{{dv_x}}{{dt}} = -f_v \cdot v_x\][/tex]
[tex]\[0 = -f_v \cdot v_x\][/tex]
Since [tex]\(v_x\)[/tex] cannot be zero (otherwise the projectile won't be moving horizontally), we can conclude that [tex]\(f_v\)[/tex] must be zero at terminal speed.
From the equation of motion in the y-direction, when [tex]\(a_y = 0\)[/tex]:
[tex]\[m \frac{{dv_y}}{{dt}} = -mg - f_v \cdot v_y\][/tex]
[tex]\[0 = -mg - f_v \cdot v_y\][/tex]
[tex]\[f_v \cdot v_y = -mg\][/tex]
Since [tex]\(f_v\)[/tex] is proportional to v, we can write:
[tex]\[f_v = k \cdot v_y\][/tex]
Substituting this into the equation, we have:
[tex]\[k \cdot v_y \cdot v_y = -mg\][/tex]
[tex]\[v_y^2 = -\frac{mg}{k}\][/tex]
The terminal speed [tex]\(v_{\text{term}}\)[/tex] is the absolute value of [tex]\(v_y\)[/tex] at terminal velocity:
[tex]\[v_{\text{term}} = \sqrt{\frac{mg}{k}}\][/tex]
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QUESTION 3 What must the mass of a speed skater be if they are moving with a linear velocity of 3.40 m/s and a total linear momentum of 220.0 kgm/s? Note 1: The units are not required in the answer in this instance. Note 2: If rounding is required, please express your answer as a number rounded to 2 decimal places. QUESTION 4 Calculate the linear velocity of a speed skater of mass 69.8 kg moving with a linear momentum of 322.47 kgm/s. Note 1: The units are not required in the answer in this instance. Note 2: If rounding is required, please express your answer as a number rounded to 2 decimal places.
The mass of the speed skater in the first question is approximately 64.71 kg, and in QUESTION 4, the linear velocity of the speed skater in the second question is approximately 4.62 m/s.
To find the mass of the speed skater in the first question, use the formula for linear momentum:
momentum = mass × velocity.
Rearranging the formula,
mass = momentum / velocity.
Plugging in the given values,
mass = 220.0 kgm/s / 3.40 m/s ≈ 64.71 kg.
QUESTION 4: In the second question, need to calculate the linear velocity. Again, using the formula for linear momentum, rearrange it to:
velocity = momentum / mass.
Plugging in the given values,
velocity = 322.47 kgm/s / 69.8 kg ≈ 4.62 m/s.
Therefore, the mass of the speed skater in the first question is approximately 64.71 kg, and the linear velocity of the speed skater in the second question is approximately 4.62 m/s.
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Unpolarized light of intensity 30 W/cm2W/cm2 is incident on a linear polarizer set at the polarizing angle θ1θ1 = 28 ∘∘. The emerging light then passes through a second polarizer that is set at the polarizing angle θ2θ2 = 152 ∘∘. Note that both polarizing angles are measured from the vertical.
What is the intensity I2I2 of the light after passing through both polarizers? I2 =
4.69
W/cm2W/cm2
Suppose the second polarizer is rotated so that θ2θ2 becomes 118 ∘∘. What is the intensity of the transmitted light I2 now?
The intensity of the transmitted light I₂ now is 12.31 W/cm². Unpolarized light of intensity 30 W/cm² is incident on a linear polarizer set at the polarizing angle θ₁ = 28°.
The emerging light then passes through a second polarizer that is set at the polarizing angle θ₂ = 152°.
Both polarizing angles are measured from the vertical.
The intensity I₂ of the light after passing through both polarizers is 4.69 W/cm².
So, 30 = I₁Cos²28°I₁ = 38.83 W/cm²
Intensity of light after passing through the first polarizer is 38.83 Cos²28° = 24.62 W/cm²
Then, 24.62 = I₂Cos²30°I₂ = 21.32 W/cm²
Suppose the second polarizer is rotated so that θ₂ becomes 118°.
Angle between the polarizers is changed by 34° (i.e. 152° − 118°).
Hence, Intensity of the transmitted light I₂ = I₁/2 [Cos²34°]I₂ = 12.31 W/cm².
Therefore, the intensity of the transmitted light I₂ now is 12.31 W/cm².
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Shiprock has a latitude of 36.78o. On the equinoxes (the start of spring and autumn), this is also the angle of the sunlight falling on Shiprock. In the applet, set the slider to this angle. What is the relative intensity?
At a latitude of 36.78° on the equinoxes, the relative intensity of sunlight falling on Shiprock can be determined using the given angle.
The relative intensity of sunlight refers to the amount of solar radiation received at a specific location and angle compared to the maximum intensity received when the Sun is directly overhead (at a 90° angle). In this case, Shiprock's latitude of 36.78° is also the angle of sunlight falling on it during the equinoxes (the start of spring and autumn), as mentioned.
When the Sun's rays are perpendicular to the Earth's surface (at a 90° angle), the intensity of sunlight is at its maximum. As the angle of incidence decreases, the intensity of sunlight decreases. To determine the relative intensity, it is necessary to compare the angle of incidence at Shiprock (36.78°) to the angle of maximum intensity (90°).
The relative intensity can be calculated using the formula: relative intensity = cos(angle of incidence). Plugging in the given angle (36.78°) into the cosine function, we can determine the relative intensity of the sunlight falling on Shiprock during the equinoxes.
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Stopping distance of a car. 5 of 16 Review | Constants Part A If the coefficient of kinetic friction between tires and dry pavement is 0.73, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 30.1 m/s?
B. In Haiti, public transportation is often by tap, small pickup trucks with seats along the sides of the pickup bed and railings to which passengers can hang on. Typically they carry two dozen or more passengers plus an assortment of chickens, goats, luggage, etc. Putting this much into the back of a pickup truck puts quite a large load on the truck springs.
A truck has springs for each wheel, but for simplicity assume that the individual springs can be treated as one spring with a spring constant that includes the effect of all the springs. Also for simplicity, think that all four springs compress equally when weight is added to the truck and that the equilibrium length of the springs is the length they have when they support the load of an empty truck.
Part B Question
A 61 kg driver gets into an empty tap to start the day's work. The springs compress 1.8×10−2 mm. What is the effective spring constant of the spring system in the tap?
Enter the spring constant numerically in newtons per meter using two significant figures.
The effective spring constant of the spring system in the tap is approximately 2.19 × 10^5 N/m.
To find the effective spring constant, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from equilibrium. The formula for Hooke's Law is F = -kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, we know the mass of the driver (61 kg) and the displacement of the springs (1.8 × 10^-2 mm, which is converted to meters). We can use the equation F = mg to find the force exerted by the weight of the driver, where g is the acceleration due to gravity (approximately 9.8 m/s^2). Since the force exerted by the springs is equal and opposite to the weight, we can equate the two forces: -kx = mg.
Rearranging the equation, we can solve for the spring constant: k = -mg/x. Substituting the given values, we get k = -(61 kg × 9.8 m/s^2) / (1.8 × 10^-2 m).
Calculating the values, we find that the effective spring constant of the spring system in the tap is approximately 2.19 × 10^5 N/m.
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An object of height 0.75 cm is placed 1.50 cm away from a converging lens with a focal length of 1.00 cm. The final image is cm tall. The final image is cm from the lens. The magnification of the lens is . Is the final image inverted or upright? Is final image enlarged or diminished? Is the final image real or virtual? When entering calculated values, enter them using proper significant figures, include any negative signs needed before the value, and do NOT include units.
The final image is inverted, enlarged, real, with a height of -1.50 cm, and located at a distance of 3 cm from the lens.
We can use the lens formula:
[tex]\(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)[/tex]
f is the focal length of the lens,
[tex]\(d_o\)[/tex]is the object distance from the lens, and
[tex]\(d_i\)[/tex] is the image distance from the lens.
Object height ([tex]\(h_o\)[/tex]) = 0.75 cm
Object distance ([tex]\(d_o\)[/tex]) = 1.50 cm
Focal length [tex](\(f\))[/tex] = 1.00 cm
We can calculate the image distance [tex](\(d_i\))[/tex] using the lens formula:
[tex]\(\frac{1}{1.00} = \frac{1}{1.50} + \frac{1}{d_i}\)[/tex]
Solving this equation:
[tex]\(d_i = \frac{1}{\frac{1}{1.00} - \frac{1}{1.50}}\)[/tex]
[tex]\(d_i = \frac{1}{\frac{1}{1.00} - \frac{2}{3}}\)[/tex]
[tex]\(d_i = \frac{1}{\frac{3 - 2}{3}}\)[/tex]
[tex]\(d_i = \frac{1}{\frac{1}{3}}\)[/tex]
[tex]\(d_i = 3\)[/tex]
Therefore, the image distance ([tex]\(d_i\)[/tex]) is 3 cm.
The magnification M of the lens is given by:
[tex]\(M = -\frac{d_i}{d_o}\)[/tex]
[tex]\(M = -\frac{3}{1.50}\)[/tex]
[tex]\(M = -2\)[/tex]
Therefore, the magnification [tex](\(M\)[/tex]) of the lens is -2.
The height of the final image ([tex]\(h_i\)[/tex]) can be calculated using the magnification formula:
[tex]\(M = \frac{h_i}{h_o}\)[/tex]
Rearranging the formula:
[tex]\(h_i = M \times h_o\)[/tex]
[tex]\(h_i = -2 \times 0.75\)[/tex]
[tex]\(h_i = -1.50\)[/tex]
The height of the final image ([tex]\(h_i\)[/tex]) is -1.50 cm.
From the negative magnification and height, we can conclude that the final image is inverted.
Since the magnification is greater than 1, the final image is enlarged.
The final image is real because it is formed on the opposite side of the lens from the object.
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How much force must be applied by a soccer player to give a stationary 0.45 kg ball an acceleration of 20 m/s ^2
and why? A 250gm ice-cube is thrown in an ice-rink and it travels 40 meters. The same cube is thrown with the same horizontal velocity on a football field and it travels 5 meters. Why is that? What is contributing to the differences in the distance traveled by the ice-cube and the soccer ball on the two surfaces?
The force required to accelerate a stationary 0.45 kg ball by [tex]20 m/s^2[/tex] is 9 N. The difference in distance traveled by the ice cube on the ice rink and the football field can be attributed to the varying levels of friction between the cube and the surfaces, with the lower friction on the ice allowing for a greater distance traveled compared to the higher friction on the grass.
To determine the force required to give the ball an acceleration of 20 [tex]m/s^2[/tex], we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a): F = m * a. Plugging in the values, we have F = 0.45 kg * 20[tex]m/s^2[/tex] = 9 N.
Therefore, a force of 9 Newtons must be applied by the soccer player to give the ball the desired acceleration.
The difference in the distances traveled by the ice cube on the ice rink and the football field can be attributed to the frictional forces acting on the cube. On the ice rink, the friction between the ice cube and the ice surface is significantly lower compared to the football field, resulting in less resistance to the cube's motion.
This lower friction allows the cube to slide and travel a greater distance. On the football field, the higher friction between the cube and the grass surface impedes its motion, causing it to come to a stop after covering a shorter distance. In essence, the frictional forces between the cube and the surfaces play a crucial role in determining the distances traveled.
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Two motorcycles are traveling due east with different velocities. However, 4.09 seconds later, they have the same velocity. During this 4.09-second interval, motorcycle A has an average acceleration of 3.03 m/s^2 due east, while motorcycle B has an average acceleration of 18.8 m/s^2 due east. (a) By how much did the speeds differ at the beginning of the 4.09-second interval, and (b) which motorcycle was moving faster?
(a) The speeds of the motorcycles differed by 12.4 m/s at the beginning of the 4.09-second interval.
(b) Motorcycle B was moving faster.
(a) The difference in speeds at the beginning of the 4.09-second interval can be determined by multiplying the average acceleration of motorcycle A (3.03 m/s²) by the time interval (4.09 s). Thus, the difference in speeds is:
Δv = (3.03 m/s²) × (4.09 s) = 12.4 m/s.
Therefore, the speeds of the motorcycles differed by 12.4 m/s at the beginning of the 4.09-second interval.
(b) Since motorcycle B had a higher average acceleration (18.8 m/s²) compared to motorcycle A, it means that motorcycle B experienced a larger change in velocity over the 4.09-second interval. This indicates that motorcycle B was moving faster during that time period. Therefore, motorcycle B was moving faster than motorcycle A.
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Calculate the force of Gravity between the following objects.
a.)The Earth has a mass of 6.0 x 10 ^24 kg and the moon has a mass of 1.34 x 10^22 kg. They are separated by a distance of 3.84 x 10 ^8 m.
b.)The Earth has a mass of 6.0 x 10^24 kg and the sun has a mass of 2.00 x 10^30 kg. The distance from the sun to the Earth is 1.49 x 10^11 meters.
The force of gravity between the Earth and the moon is approximately 1.982 x [tex]10^{20[/tex] Newtons. The force of gravity between the Earth and the sun is approximately 3.52 x [tex]10^{22[/tex] Newtons.
a) To calculate the force of gravity between the Earth and the moon, we can use the formula for gravitational force:
[tex]F = (G * m1 * m2) / r^2[/tex]
where F is the force of gravity, G is the gravitational constant (approximately 6.67 x 10^-11 N[tex](m/kg)^2[/tex]), m1 and m2 are the masses of the objects, and r is the distance between their centers.
Plugging in the values:
m1 = 6.0 x [tex]10^{24[/tex]kg (mass of the Earth)
m2 = 1.34 x [tex]10^{22[/tex] kg (mass of the moon)
r = 3.84 x [tex]10^8[/tex] m (distance between the Earth and the moon)
F = (6.67 x [tex]10^{-11[/tex]N[tex](m/kg)^2[/tex]) * (6.0 x [tex]10^{24[/tex] kg) * (1.34 x [tex]10^{22[/tex]kg) / [tex](3.84 * 10^8 m)^2[/tex]
Calculating this expression, we find that the force of gravity between the Earth and the moon is approximately 1.982 x [tex]10^{20[/tex] Newtons.
b) Similarly, to calculate the force of gravity between the Earth and the sun, we can use the same formula:
m1 = 6.0 x [tex]10^{24[/tex] kg (mass of the Earth)
m2 = 2.00 x [tex]10^{30[/tex] kg (mass of the sun)
r = 1.49 x [tex]10^{11[/tex] m (distance between the Earth and the sun)
F = (6.67 x [tex]10^{-11} N(m/kg)^2[/tex]) * (6.0 x [tex]10^{24[/tex] kg) * (2.00 x [tex]10^{30[/tex] kg) / [tex](1.49 * 10^{11} m)^2[/tex]
Calculating this expression, we find that the force of gravity between the Earth and the sun is approximately 3.52 x [tex]10^{22[/tex] Newtons.
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A uniform rod AB is 1 m long and weighs 17N. It is suspended by strings AC and BD as shown. A block P weighing 85N is attached at E, 0.5m from A. The magnitude of the tension force of the string BD is N. Round off only on the final answer expressed in 3 decimal places. E A B P Add your answer
A uniform rod AB is 1 m long and weighs 17N. It is suspended by strings AC and BD as shown. A block P weighing 85N is attached at E, 0.5m from A. Thus, the magnitude of the tension force of the string BD is 98.971N (approx.) to 3 decimal places.
A uniform rod AB is 1 m long and weighs 17N. A block P weighing 85N is attached at E, 0.5m from A.The length of the rod AB is 1m. The distance of the block P from end A is 0.5m.
The weight of the rod, W1= 17N. The weight of the block, W2= 85N.
The forces acting on the rod are the weight, W1, tension, T1 in the string AC, tension, T2 in the string BD, and the reaction, R1, at A.
The forces acting on the block are the weight, W2, and the tension, T2, in the string BD.
Taking moments about A:
Sum of anticlockwise moments = Sum of clockwise moments
Taking moments about A:
Sum of anticlockwise moments = T2 × AB = T2 × 1
Sum of clockwise moments = (W1 × AE) + (W2 × EP) = (17 × AE) + (85 × 0.5).
Therefore,T2 = (17 × AE + 42.5) N.
For equilibrium in the vertical direction: Taking upward forces as positive,T1 + T2 = W1 + W2
For equilibrium in the horizontal direction:Taking forces towards the right as positive,R1 = 0.
The magnitude of the tension force of the string BD is 98.971N (approx) to 3 decimal places.
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Is it correct to say that a radio wave is a low-frequency light wave? Is a radio wave also a sound wave? Justify. Also, explain the nature of light and the electromagnetic spectrum. Elaborate your answer.
No, it is not correct to say that a radio wave is a low-frequency light wave. A radio wave and a light wave are both forms of electromagnetic radiation, but they differ in frequency and wavelength. Additionally, a radio wave is not a sound wave as they belong to different types of waves.
A radio wave is a type of electromagnetic wave with a long wavelength and low frequency. It is used for long-distance communication, such as radio broadcasting or cellphone signals. Light waves, on the other hand, encompass a broader range of frequencies and wavelengths, including visible light, which is the range of electromagnetic radiation that is visible to the human eye.
Sound waves, on the other hand, are mechanical waves that require a medium (such as air, water, or solids) to travel through. They are created by vibrations and can be detected by the human ear. Unlike radio waves and light waves, which are forms of electromagnetic radiation, sound waves cannot propagate through a vacuum.
The nature of light is best described by the theory of electromagnetic radiation, which states that light is composed of particles called photons that exhibit both wave-like and particle-like properties. The electromagnetic spectrum encompasses the entire range of electromagnetic waves, including radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Each type of wave has different properties, such as wavelength, frequency, and energy, and they are used in various applications ranging from communication to medical imaging.
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Only two forces act on an object (mass=3.19 kg). Find
(a) the magnitude and (b) the
direction (relative to the x axis) of the acceleration of
the object.
(a) The magnitude of the acceleration of the object is 9.81 m/s².
(b) The direction of the acceleration is vertically downward (opposite to the positive y-axis).
The magnitude of the acceleration can be calculated using Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). In this case, there are two forces acting on the object, so the net force can be found by summing up these forces.
Since we know the mass of the object (3.19 kg), we can calculate the net force. However, the question does not provide information about the forces acting on the object. Therefore, we cannot determine the net force or the acceleration directly.
However, if we assume that only two forces act on the object, we can deduce that the net force is the vector sum of these two forces. In the absence of any other information, we can consider the gravitational force (weight) as one of the forces acting on the object.
The weight of an object can be calculated by multiplying its mass by the acceleration due to gravity (9.81 m/s²). As the object is on Earth, the gravitational force acts vertically downward, opposite to the positive y-axis. Therefore, the direction of the acceleration is also vertically downward.
In summary, the magnitude of the acceleration is 9.81 m/s², and its direction is vertically downward (opposite to the positive y-axis).
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Which of the following is not an advantage of nuclear power plants, when compared to fossil fuel plants? 1 The fuel for nuclear power plants has a higher specific energy than fossil fuels. 2 Nuclear power plants use renewable fuel. 3 Nuclear power plants do not produce greenhouse gases. 4 Nuclear power plants can be used to establish a 'baseline' power generation
The option that is not an advantage of nuclear power plants when compared to fossil fuel plants is: 2) Nuclear power plants use renewable fuel.
While options 1, 3, and 4 provide advantages of nuclear power plants, option 2 is not accurate. Nuclear power plants do not use renewable fuel. The fuel used in nuclear power plants is uranium or plutonium, which are non-renewable resources. These fuels are obtained through mining and have finite reserves on Earth. Once the fuel is used up, it cannot be replenished.
Advantages of nuclear power plants include:
1) The fuel for nuclear power plants has a higher specific energy than fossil fuels. This means that a small amount of nuclear fuel can produce a large amount of energy.
3) Nuclear power plants do not produce greenhouse gases. Unlike fossil fuel plants, which release carbon dioxide and other pollutants, nuclear power plants generate electricity without contributing to air pollution and climate change.
4) Nuclear power plants can be used to establish a 'baseline' power generation. Nuclear reactors can provide a constant and reliable source of electricity, operating continuously without depending on external factors like weather conditions or fuel availability.
Therefore, the correct option is 2) Nuclear power plants do not use renewable fuel.
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Suppose a car is mowing along a flat piece of road. Moreover, let us suppose that we know the coefficient of friction within the axles and wheel bearines of the car to be μ=0.015. If the car let's off the accelerator (gas pedal) and starts rolling and you measure that it takes the car Boo meters to roll to a stop (without using the breaks). how fast was the car moving the moment the driver removed her/his foot from the pedal? Give your answer in units of m/s, however do not include the units explicitly in your answer. If you include units, the answer will be counted wrong.
A spinning table has radius 2.50 m and moment of inertia 1900 kg×m^2 about a vertical axle through its center, and it can turn with negligible friction. Two persons, one directly in from of the other. (consider that the persons are standing at opposite ends of a line that passes through the center of rotation) apply each a force of 8.0 N tangentially to the edge of the table for 10.0 s. A. If the table is initially at rest, what is its angular speed after this 10.0 s interval? \{10 points\} rad/s^2
B. How much work is done on the table by EACH person?
A spinning table with radius 2.50 m and moment of inertia 1900 kg×m^2 remains at rest after two people apply forces of 8.0 N tangentially to the edge for 10.0 s. Each person does zero work, and the table's angular speed remains zero rad/s.
A. To find the angular speed of the table after the 10.0 s interval, we can use the principle of angular momentum conservation. Initially, the table is at rest, so its initial angular momentum is zero (L₀ = 0).
The angular momentum of an object is given by the formula:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
The total angular momentum after the 10.0 s interval is the sum of the angular momenta contributed by each person:
L = L₁ + L₂
Since the forces applied are tangential to the edge of the table, the torque exerted by each person's force is equal to the force multiplied by the radius:
τ = Fr
where F is the force and r is the radius.
The change in angular momentum is equal to the torque multiplied by the time interval:
ΔL = τΔt
Since the table is initially at rest, the change in angular momentum is equal to the final angular momentum:
L = τΔt
Substituting the values into the equation, we get:
I₁ω - I₂ω = F₁r₁Δt + F₂r₂Δt
where I₁ and I₂ are the moments of inertia of the table with respect to the first and second person, respectively, ω is the final angular speed, F₁ and F₂ are the forces applied by the first and second person, r₁ and r₂ are the distances from the axis of rotation to the points where the forces are applied, and Δt is the time interval.
Since both persons apply the same force (8.0 N) and the same radius (2.50 m), we can simplify the equation:
I₁ω - I₂ω = 8.0 N * 2.50 m * 10.0 s
The moment of inertia of the table (I) is given as 1900 kg×m^2, so we have:
1900 [tex]kg*m^2[/tex] * ω - 1900 kg×m^2 * ω = 8.0 N * 2.50 m * 10.0 s
0 = 200 N * m * s
Therefore, the angular speed of the table after the 10.0 s interval is zero rad/s.
B. The work done by each person can be calculated using the work-energy theorem, which states that the work done is equal to the change in kinetic energy.
The change in kinetic energy (ΔK) is equal to the work done (W). The work done by each person is given by:
W = ΔK = 1/2 * I * ω²
Substituting the given values, we have:
W = 1/2 * 1900 [tex]kg*m^2\\[/tex] * (0 rad/s)²
W = 0 Joules
Therefore, each person does zero work on the table.
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E= mc2, according to Einstein, i.e. energy is mass times the speed of light (squared).
If E= mc2 is a true equation, the units must be equal on both sides of the equation. Show that the units are consistent. Use CGS units.
We show that the units on both sides of the equation are consistent, we can conclude that the units in E = mc^2 are consistent in CGS units.
To show that the units are consistent in the equation E = mc^2, we can use CGS (centimeter-gram-second) units. Let's break down the units on each side of the equation:
E: Energy (ergs) in CGS units.
m: Mass (grams) in CGS units.
c: Speed of light (centimeters per second) in CGS units.
Now let's analyze the units on each side of the equation:
Left side of the equation (E):
Energy (E) is measured in ergs in CGS units.
Right side of the equation (mc^2):
Mass (m) is measured in grams in CGS units.
The speed of light (c) is measured in centimeters per second in CGS units.
To determine the units of mc^2, we multiply the units of mass (grams) by the square of the units of speed (centimeters per second). This gives us:
mc^2 = (grams) × (centimeters per second)^2
Expanding the units further:
mc^2 = grams × (centimeters/second)^2
= grams × centimeters^2/second^2
Now, comparing the units on each side of the equation:
Left side (E) = ergs
Right side (mc^2) = grams × centimeters^2/second^2
Since the units on both sides of the equation are consistent, we can conclude that the units in E = mc^2 are consistent in CGS units.
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If it is necessary to test the relieving capacity of a safety valve, a(n) __________:
A. accumulation test is put on the boiler
B. hydrostatic test is put on the boiler
C. boiler must be secured and off-line
D. bench test must be done
The brightest star in the night sky, Sirius, has a radius of about 1,189,900 km. The spherical surface behaves as a black body radiator. The surface temperature is about 8500 K. What is the rate at which energy is radiated from the star (W)?
The rate at which energy is radiated from the star is p=[tex](5.67 × 10^(-8) W·m^(-2)·K^(-4)) * (4 * 3.1416 * (1,189,900,000 m)^2) * (8500 K)^4[/tex]
The rate at which energy is radiated from a black body can be calculated using the Stefan-Boltzmann law, which states that the power radiated per unit area (P) is proportional to the fourth power of the temperature (T) of the object and its surface area (A). The Stefan-Boltzmann constant (σ) is used in this calculation. The formula is given by:
P = σ * A * [tex]T^4[/tex]
P is the power radiated (in watts)
σ is the Stefan-Boltzmann constant (5.67 × [tex]10^(-8) W·m^(-2)·K^(-4))[/tex]
A is the surface area of the star [tex](4πr^2)[/tex]
T is the temperature of the star (in Kelvin)
r is the radius of the star
Substituting the values:
r = 1,189,900 km = 1,189,900,000 m
T = 8500 K
First, calculate the surface area (A):
A = [tex]4πr^2[/tex]
A = 4 * 3.1416 * [tex](1,189,900,000 m)^2[/tex]
Next, substitute the values into the formula:
P =[tex](5.67 × 10^(-8) W·m^(-2)·K^(-4)) * (4 * 3.1416 * (1,189,900,000 m)^2) * (8500 K)^4[/tex]
Calculating this expression will give you the rate at which energy is radiated from the star.
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1000 J of heat are added to 400 g of potatoes which have a
specific heat capacity of 3430 J kg-1. What is the change in
temperature of the potatoes (in C)?
When 1000 J of heat are added to 400 g of potatoes with a specific heat capacity of 3430 J/kg°C. Hence, the change in temperature of the potatoes is approximately 0.733°C.
The change in temperature (ΔT) of an object can be determined using the equation \[ Q = mcΔT \].
where Q is the heat energy added, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
Q = 1000 J
m = 400 g = 0.4 kg
c = 3430 J/kg°C
Substituting these values into the equation:
\[ 1000 = (0.4)(3430)ΔT \]
Simplifying:
[ ΔT = \frac{1000}{0.4 \times 3430}
[ ΔT ≈ 0.733 \, °C \]
Therefore, the change in temperature of the potatoes is approximately 0.733°C.
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You measure the length of the waterfall along the map, and determine that it is 0.4 mm in length. How many meters does this represent?
a) 4.0 meters
b) 9.6 meters
c) 16.7 meters
d) 40.0 meters
To convert the length of the waterfall from millimeters (mm) to meters (m), we need to divide the length in millimeters by 1,000 since there are 1,000 millimeters in a meter.
0.4 mm / 1,000 = 0.0004 meters. Therefore, the length of the waterfall represents 0.0004 meters. Among the given options:
a) 4.0 meters is not the correct answer because 0.0004 meters is much smaller than 4.0 meters. b) 9.6 meters is not the correct answer because 0.0004 meters is much smaller than 9.6 meters. c) 16.7 meters is not the correct answer because 0.0004 meters is much smaller than 16.7 meters. d) 40.0 meters is not the correct answer because 0.0004 meters is much smaller than 40.0 meters.
None of the provided options match the converted value of 0.0004 meters.
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The outward electric flux through a spherical surface
is 4.5×104N⋅m2/C×104N⋅m2/C.
What is the net charge, in coulombs, enclosed by the
surface?
qenc =
To determine the net charge enclosed by the surface, we can use Gauss's Law, which states that the electric flux through a closed surface is proportional to the net charge enclosed by that surface.
The formula for electric flux is given as:
Electric Flux = (Net Charge Enclosed) / (ε₀)
Given that the electric flux is 4.5 ×[tex]10^4[/tex] N·m²/C, and the electric constant (ε₀) is approximately 8.85 ×[tex]10^(-12)[/tex] N·m²/C², we can rearrange the equation to solve for the net charge:
Net Charge Enclosed = Electric Flux × ε₀
Net Charge Enclosed = 4.5 × [tex]10^4[/tex] N·m²/C × 8.85 × [tex]10^(-12)[/tex] N·m²/C²
After performing the multiplication, we find that the net charge enclosed by the surface is approximately 3.9825 × [tex]10^(-7)[/tex] C.
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Problem 4. In physics, the torque is defined by τ=r×F, where r is the position vector (a vector from the point about which the torque is being measured to the point where the force is applied), and F is the force vector, for a rotation. Suppose there is a bolt connecting the main and rear frame of a mountain bike. You apply 40 N of force at a position of 0.2 m away from the center of the bolt with wrench. Suppose the angle between the force and the wrench is 90°. 1. Draw a diagram to represent the vectors. 2. What is the direction of the torque vector? Is the bolt being loosened or tightened? 3. What is the magnitude of the torque vector?
The magnitude of the torque vector is 8 Nm. The direction of the torque vector can be determined as counterclockwise.
1. A diagram to represent the vectors: The given diagram shows the position vector r (from the point about which the torque is being measured to the point where the force is applied) and force vector F.
2. The direction of the torque vector: To determine the direction of the torque vector, the right-hand rule is used. The right-hand rule is given as follows: if the fingers of the right hand are curled around the axis of rotation in the direction of rotation, then the thumb points in the direction of the torque vector.
Hence, from the diagram, the direction of the torque vector can be determined as counterclockwise.
Therefore, the bolt is being loosened.
3. The magnitude of the torque vector: The formula to find torque is τ=r×F. Given that r = 0.2 m, F = 40 N, and the angle between r and F is 90°.
Therefore, τ=r×F=sin(90°)×r×F=1×0.2×40= 8 Nm.
Hence, the magnitude of the torque vector is 8 Nm.
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The large red L's on a surface map represent centers of low pressure, also known as ____ storms.
a. high-latitude anti-cyclonic
b. mid-latitude cyclonic
c. high-latitude cyclonic
d. mid-latitude anti-cyclonic
Answer:
The large red L's on a surface map represent centers of low pressure, also known as mid-latitude cyclonic storms.
Explanation:
These storms are characterized by rotating winds that move counterclockwise in the Northern Hemisphere and clockwise in the Southern Hemisphere. The low pressure at the center of the storm causes air to rise, leading to cloud formation and precipitation. Mid-latitude cyclonic storms are also known as extratropical cyclones and are common in the middle latitudes (around 30-60 degrees) of both hemispheres.
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Answer the following questions using the
knowledge you gained from the
hygrothermographs.
Q2- (0.25pt) Why is the maximum temperature higher
in summer than in winter? Relate your answer to the
hours of daylight and height of the sun at zenith.
Q3- (0.25pt) At what time does the minimum
temperature occur in June?
Q4- (0.25pt) At what time does the minimum
temperature occur in January?
Q5- (0.25pt) Why does the minimum temperature
occur at a different time in summer than in winter?
Q6- (0.5pt) In general, when the temperature falls the
relative humidity (increases or decreases) and when
the temperature rises the relative humidity (increases
or decreases).
Summer Chart Temperature :
Date Max temp Time of max Mini temp Time of mini Daily range
June 17 80 4 pm 60 6am 20
June 18 82 6 pm 60 5am 22
June 19 79 7 pm 59 5am 20
Summer chart relative humidity :
Date Max temp Time of max Mini temp Time of mini Daily range
June 17 66 7am 29 5pm 37
June 18 78 8am 3 7pm 40
June 19 74 4am 44 4pm 30
Winter Chart Temperature
Date Max temp Time of max Mini temp Time of mini Daily range
Jan 30 20 12pm 10 6am 10
Jan 31 18 3pm 8 9am 10
Feb 1 18 4pm] 2 8am 16
Winter chart relative humidity :
Date Max temp Time of max Mini temp Time of mini Daily range
Jan 30 96 3am 50 3pm 46
Jan 31 80 7am 50 4pm 30
Feb 1 90 12am 58 4pm 32
The maximum temperature is higher in summer than in winter because in summer there are more hours of daylight and the sun is at a higher height at zenith.
During summer, the sun is directly overhead and the days are longer, so the maximum temperature will be higher compared to the winter when the sun is at an angle and days are shorter.
Q3: The minimum temperature generally occurs in the early morning hours before sunrise in June at 4am.
This is because during the night, the Earth's surface cools down by radiating heat away from the surface. As the sun begins to rise, the Earth's surface starts to warm up again.
Q4: In January, the minimum temperature occurs at 6 am. The minimum temperature in January usually occurs during the early morning hours before sunrise.
This is because at night, there is less incoming solar radiation, which means that the earth's surface cools down and continues to radiate away heat. As a result, the lowest temperature of the day is usually reached just before sunrise, after which temperatures begin to rise again.
Q5: The minimum temperature occurs at a different time in summer than in winter because the amount of solar radiation changes from summer to winter.
In summer, the sun is up longer and at a higher angle, which causes the minimum temperature to occur earlier in the morning. In winter, the sun is up less, and at a lower angle, which causes the minimum temperature to occur later in the morning.
Q6: When the temperature falls, the relative humidity increases, and when the temperature rises, the relative humidity decreases.
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The following well is to be equipped with a surface choke operating in critical flow. Determine the well's producing capacity and the choke size required for the following conditions
o No choke
o Choke at wellhead
o Choke at separator
Well depth = 10,000ft
Tubing Size = 1.995 inch
Pr = 3000psia
fw =0
C=0.0023STB/ day-psi
Flowline length =5,000ft
Flowline Size =2 inch
GLR=500scf/STB
P sep =100psig
n=0.85
For the given well, the pressure gradient in the well and tubing are expresses by the following two equations
Tubing: P wh =0.9Pwf−0.95Q−100
Flowline: P sep = P wh q −0.35Q+2.5
To determine the well's producing capacity and the required choke size, we need to analyze three scenarios: no choke, choke at the wellhead, and choke at the separator.
In the case of no choke, the well is unrestricted, and the pressure at the wellhead (Pwh) is equal to the flowing bottomhole pressure (Pwf). We can use the Tubing equation to calculate the producing capacity:
Pwh = 0.9Pwf - 0.95Q - 100
For the choke at the wellhead, we need to consider the critical flow condition. This means that the pressure at the wellhead is determined by the flow rate (Q) and the choke size (nozzle diameter). By rearranging the Tubing equation, we can solve for the required choke size:
Nozzle diameter = (0.9Pwf - Pwh - 100) / 0.95
For the choke at the separator, we use the Flowline equation to determine the well's producing capacity. Rearranging the equation, we find:
Pwh = (Psep + 0.35Q - 2.5) / q
Now, we can substitute the values for the given conditions (well depth, tubing size, Pr, fw, C, flowline length, flowline size, GLR, Psep, and n) into these equations to calculate the producing capacity and the required choke size for each scenario.
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Three displacement vectors of a croquet ball are 5 hown in the figure, where ∣
A
∣=12.0 units, ∣
B
∣=20.0 units, and ∣
C
∣=15.0 unr (a) Find the resultants in unit-vector notation.
R
= units
The resultant in unit-vector notation is `R = (-10√6 - 6√3) i + (5√2 - 15√6/4) j` units. three displacement vectors of a croquet ball are shown in the figure, where `|A| = 12.0 units, |B| = 20.0 units, and |C| = 15.0 units`.
To find the resultants in unit-vector notation, we can use the parallelogram law of vector addition, which states that "if two vectors are represented by two adjacent sides or a parallelogram then the diagonal of the parallelogram will be equal to the resultant of two vectors".
Here, the vector has a length of 12.0 units and is directed at an angle of 30°, vector has a length of 20.0 units and is directed at an angle of 180°, and vector has a length of 15.0 units and is directed at an angle of 285°.
Now, we can calculate the resultant by finding the vector sum of vector , vector , and vector .
Let's assume the vector sum is `R` in the form of unit-vector notation.
We can find it as follows:R = A + B + C.
We can write the given vectors in the form of the unit-vector notation as follows:A = 12(cos 30° i + sin 30° j)B = 20(cos 180° i + sin 180° j)C = 15(cos 285° i + sin 285° j).
We know that `cos 180° = -1`, `cos 30° = √3/2`, `cos 285° = √2 - √6/4`, `sin 180° = 0`, `sin 30° = 1/2`, and `sin 285° = -√2 - √6/4`.
Substituting the given values in the above expressions, we getA = 12(√3/2 i + 1/2 j)B = 20(-i)C = 15(√2 - √6/4 i - √2 - √6/4 j).
Now, we can substitute these values in the above expression of `R` and simplify it.R = 12(√3/2 i + 1/2 j) + 20(-i) + 15(√2 - √6/4 i - √2 - √6/4 j)R = (-20√6/4 - 12√3) i + (-20 + 15√2 - 15√6/4) j.
Simplifying further,R = (-10√6 - 6√3) i + (5√2 - 15√6/4) j.
Therefore, the resultant in unit-vector notation is `R = (-10√6 - 6√3) i + (5√2 - 15√6/4) j` units.
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A small water-heating coil is submerged in a container with 200g of water and 100g of ice, initially in thermal equilibrium. The heating coil is plugged to a 120V AC outlet for 5 minutes. The resistance of the coil is 720. a. (5 pts) what is the initial temperature of the water-ice mixture? b. (10 pts) what is the average power dissipated in the coil? C. (5 pts) how much er.ergy does the coil supply to the water-ice mixture in 5 minutes? d. (15 pts) what is the final temperature of the mixture (assume all of the ice melts and the final temp. is more than 0°C)
a. The initial temperature of the water-ice mixture was - (16675 °C). b. The average power dissipated in the coil is 20 W. c) The coil supply 6000J energy to the water-ice mixture in 5 minutes. d) The final temperature of the mixture is + 16675 °C.
a. To find the initial temperature of the water-ice mixture, we need to consider the thermal equilibrium between the water and ice.
At this point, they are both at the same temperature, which we will denote as T_initial. Since the water and ice are in thermal equilibrium, we can use the principle of energy conservation:
Energy gained by the water = Energy lost by the ice
The energy gained by the water can be calculated using the specific heat capacity of water (c_water), mass of water (m_water), and the change in temperature (T_final - T_initial):
Energy gained by the water = c_water * m_water * (T_final - T_initial)
The energy lost by the ice can be calculated using the heat of fusion (Q_fusion) and the mass of ice (m_ice):
Energy lost by the ice = Q_fusion * m_ice
Since the system is in thermal equilibrium, the energy gained by the water is equal to the energy lost by the ice:
c_water * m_water * (T_final - T_initial) = Q_fusion * m_ice
Substituting the given values, we have:
(4182 J/(kg·°C)) * (0.2 kg) * (T_final - T_initial) = (333500 J/kg) * (0.1 kg)
Solving for T_initial, we find:
T_initial ≈ T_final - (16675 °C)
b. The average power dissipated in the coil can be calculated using the formula:
Power = ([tex]voltage^{2}[/tex]) / Resistance
Substituting the given values, we have:
Power = [tex]120 V^{2}[/tex] / 720 Ω
Simplifying the expression:
Power ≈ 20 W
c. The energy supplied by the coil to the water-ice mixture can be calculated using the formula:
Energy = Power * Time
Substituting the given values, we have:
Energy = 20 W * (5 min * 60 s/min)
Simplifying the expression:
Energy ≈ 6000 J
d. To find the final temperature of the mixture, we need to consider the heat absorbed by the ice during its phase change from solid to liquid and the heat gained by the water. The heat absorbed by the ice can be calculated using the formula:
Heat absorbed by ice = Q_fusion * m_ice
The heat gained by the water can be calculated using the specific heat capacity of water (c_water), mass of water (m_water), and the change in temperature (T_final - T_initial):
Heat gained by water = c_water * m_water * (T_final - T_initial)
Since the ice melts completely, the heat absorbed by the ice is equal to the heat gained by the water:
Q_fusion * m_ice = c_water * m_water * (T_final - T_initial)
Substituting the given values, we have:
(333500 J/kg) * (0.1 kg) = (4182 J/(kg·°C)) * (0.2 kg) * (T_final - T_initial)
Solving for T_final, we find:
T_final ≈ T_initial + (16675 °C)
Therefore, the final temperature of the mixture is approximately T_initial + 16675 °C.
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A circular loop of wire has an area of 0.27 m2 . It is tilted by 44 ∘ with respect to a uniform 0.35 T magnetic field. Part A What is the magnetic flux through the loop?
6.8×10^−2 T⋅m2T⋅m2
0.14 T⋅m2T⋅m2
0.37 T⋅m2T⋅m2
0.80 T⋅m2T⋅m2
1.5 T⋅m2
The magnetic-flux through the loop is approximately 0.1 T⋅m^2.
To calculate the magnetic flux through the loop, we can use the formula:
Φ = B * A * cos(θ)
Where:
Φ is the magnetic flux
B is the magnetic field strength
A is the area of the loop
θ is the angle between the magnetic-field and the normal to the loop
Given:
Area of the loop (A) = 0.27 m^2
Magnetic field strength (B) = 0.35 T
Angle (θ) = 44°
Plugging in the values into the formula:
Φ = (0.35 T) * (0.27 m^2) * cos(44°)
Calculating:
Φ ≈ 0.35 T * 0.27 m^2 * cos(44°)
Φ ≈ 0.0975 T⋅m^2
Rounded to one decimal place, the magnetic flux through the loop is approximately 0.1 T⋅m^2.
Therefore, the correct option is 0.1 T⋅m^2.
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