The force acting on the particle at t = 5s is determined to be 245 kg m/s. This value is obtained by differentiating the given linear momentum equation with respect to time and substituting t = 5 into the resulting expression.
To find the force on a particle, we need to differentiate the linear momentum with respect to time: p = 2t^2 kg m/s + 3t^3 kg m/s^2
Taking the derivative of p with respect to time (t), we get:
dp/dt = d/dt (2t^2 kg m/s + 3t^3 kg m/s^2)
= 4t kg m/s + 9t^2 kg m/s^2
Now, to find the force, we use Newton's second law of motion, which states that the force (F) acting on an object is equal to the rate of change of momentum (dp/dt) with respect to time:
F = dp/dt
= 4t kg m/s + 9t^2 kg m/s^2
To find the force at t = 5s, we substitute t = 5 into the equation:
F(5) = 4(5) kg m/s + 9(5)^2 kg m/s^2
= 20 kg m/s + 9(25) kg m/s^2
= 20 kg m/s + 225 kg m/s^2
= 245 kg m/s
Therefore, the force acting on the particle at t = 5s is 245 kg m/s.
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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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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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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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A current-carrying gold wire has a diameter of Part A 0.80 mm. The electric field in the wire is 0.46 V/m. Use the resistivity at room femperature for gold rho=2.44×10
−8
Ω+m. What is the current carried by the wire? For related problemsolving tips and strategies, you Express your answer with the appropriate units. may want to view a Video Tutor Solution of Electric field potential difference. and resistance in a wire What is the potential difference between two points in the wire 6.0 m apart? Express your answer with the appropriate units. Part C What is the resistance of a 6.0 m length of this wire? Express your answer with the appropriate units.
The current, potential difference, and resistance: Current (I) = 0.46 V/m / R, Potential difference = E * 6.0 m, Resistance (R) = 2.44 × 10^(-8) Ω•m * (6.0 m / A). We need to use Ohm's Law.
To calculate the current carried by the gold wire, we need to use Ohm's Law, which states that the current (I) is equal to the electric field (E) divided by the resistance (R). The resistance of the wire can be determined using its resistivity (ρ), length (L), and cross-sectional area (A).
Given:
Diameter of the wire = 0.80 mm = 0.80 × 10^(-3) m
Electric field in the wire = 0.46 V/m
Resistivity of gold (ρ) = 2.44 × 10^(-8) Ω•m
First, let's calculate the radius of the wire:
Radius (r) = diameter / 2 = 0.80 × 10^(-3) m / 2 = 0.40 × 10^(-3) m
Next, we can calculate the cross-sectional area of the wire:
A = πr^2 = π(0.40 × 10^(-3) m)^2
Now we can find the resistance of the wire:
R = ρ * (L / A) = 2.44 × 10^(-8) Ω•m * (6.0 m / A)
To find the current, we can use Ohm's Law:
I = E / R = 0.46 V/m / R
To calculate the potential difference between two points in the wire 6.0 m apart, we can multiply the electric field by the distance:
Potential difference = E * 6.0 m
Now we can solve for the current, potential difference, and resistance:
Current (I) = 0.46 V/m / R
Potential difference = E * 6.0 m
Resistance (R) = 2.44 × 10^(-8) Ω•m * (6.0 m / A)
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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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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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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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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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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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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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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.
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
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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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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Which if the following are true statements regarding electrical safety (circle all that apply)?
A) Current will go through the path of most resistance B) Current going through your body can cause burns C) Current going through your body can interfere with your nervous system’s control of your muscles D) it is safe to stand in the middle of a field during a lightning storm E) A lightning rod attracts lightning
Current going through your body can cause burns and Current going through your body can interfere with your nervous system's control of your muscles are true statements regarding electrical safety. Both statements B and C are true regarding electrical safety.
Statement B highlights the danger of electric current passing through the human body. When current flows through the body, it encounters resistance in the form of tissues and organs, which can lead to the generation of heat. This heat can cause burns, ranging from mild to severe, depending on the magnitude and duration of the current.
Statement C addresses the potential interference of electric current with the nervous system's control over muscles. The passage of current through the body can disrupt the normal functioning of nerves, leading to muscle contractions, spasms, and loss of control over muscle movement. Electric shocks can potentially cause paralysis or result in difficulty in breathing or heart function.
Statements A, D, and E are false or misleading in terms of electrical safety.
Statement A, "Current will go through the path of most resistance," is incorrect. Electric current follows the path of least resistance, meaning it takes the easiest path available rather than the one with the most resistance.
Statement D, "It is safe to stand in the middle of a field during a lightning storm," is incorrect. Standing in an open field during a lightning storm is extremely dangerous as it increases the risk of being struck by lightning. It is advisable to seek shelter indoors or in a vehicle during such conditions.
Statement E, "A lightning rod attracts lightning," is also misleading. A lightning rod does not attract lightning but rather provides a preferred path for lightning to follow, directing it safely into the ground to prevent damage to structures and potential harm to people.
Therefore, Statement B and C are correct.
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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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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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If the fusion of two unknown atoms results in a particle with 0.0000023 kg less mass than the sum of the masses of the two unknown atoms, how much energy is released? Report your answer to zero decimal places.
The answer rounded to zero decimal places, the energy released is approximately 206148 joules. The mass defect of a particle resulting from the fusion of two unknown atoms is 0.0000023 kg. To find out how much energy is released in this process, we can use Einstein's famous equation E = mc², where E is energy, m is mass and c is the speed of light.
The energy released is given by the mass defect multiplied by the speed of light squared.
Therefore,E = (0.0000023 kg)(299,792,458 m/s)²⇒E = (0.0000023 kg)(89875517873681764 m²/s²)⇒E = 206148.408 joules
Rounding the answer to zero decimal places, the energy released is approximately 206148 joules.
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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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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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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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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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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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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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what is the tensile strength of the aluminum foil sample
The tensile strength of the aluminum foil sample refers to the maximum stress or force per unit area that the sample can withstand before it breaks.
To determine the tensile strength of the aluminum foil sample, a tensile test is typically conducted. In this test, a sample of the aluminum foil is subjected to a gradually increasing tensile force until it reaches its breaking point. The tensile strength is then calculated by dividing the maximum force applied to the sample by its cross-sectional area.
Tensile strength is measured in units of force per unit area, such as pascals (Pa) or megapascals (MPa). The actual value of the tensile strength of an aluminum foil sample can vary depending on various factors, including the thickness of the foil, the purity of the aluminum, and any additional treatments or coatings applied to the foil.
To obtain the specific tensile strength of a particular aluminum foil sample, it would be necessary to perform a tensile test on that specific sample and measure the force at which it breaks. This would provide the maximum stress or force per unit area, indicating the tensile strength of the sample.
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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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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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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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A conductor with resistance R carries a constant positive current , and hence dissipates a power P = Ri? This causes the conductor to heat up above the ambient temperature. Let T denote the temperature of the conductor above the ambient temperature at time t. T satisfies the equation
a dT/dt + bT =P
where a > 0, b>0 are thermal coefficients and P is the power dissipated in the conductor. The resistance R of the conductor changes with temperature according to:
R = Ro (1 + cT)
where the constant c is called the resistance temperature coefficient of the conductor and Ro > 0 is the resistance of the conductor at ambient temperature. Consider a metal wire, for which c>0. If the current i is smaller than a critical value crie the temperature T converges to a steady-state value as t . If the current is larger than this critical value of current, then the temperature T goes to const goes to o. (In practice, the temperature increases until the conductor is destroyed). This phenomenon is called thermal runaway.
Assume a = 1J/°C, b = 0.4 W/°C , Ro = 1.9 Ω and c = 0.010/°C
Find the critical value ferit. above which thermal runaway occurs.
Enter your answer, in Amperes (A), to 2 decimal places in the box below.
Enter the numerical value only without units
The critical value of current (ferit) above which thermal runaway occurs is 40 Amperes (A).
To find the critical value of current (ferit) above which thermal runaway occurs, we need to determine the steady-state temperature and the condition under which the temperature diverges to infinity.
For steady-state temperature, the equation becomes:
bT = P
Substituting P = Ri, we get:
bT = Ri
Solving for T, we have:
T = Ri / b
Now, we can substitute the expression for resistance R in terms of temperature:
T = (iRo(1 + cT)) / b
Rearranging the equation, we have:
bT = iRo(1 + cT)
bT - iRo(cT) = iRo
T(b - ic) = iRo / b
T = (iRo / b) / (b - ic)
To ensure convergence to a steady-state temperature, we need the denominator to be non-zero. Therefore:
b - ic ≠ 0
Solving for i, we have:
i ≠ b / c
The critical value of current (ferit) is the value of i that satisfies the condition above. Therefore, the critical value of current is:
ferit = b / c
Plugging in the given values:
ferit = 0.4 W/°C / (0.010/°C) = 40 A
Therefore, 40 Amperes (A) is the critical current (ferit) threshold above which thermal runaway occurs.
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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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