The specific enthalpy change of the process is -3080 kJ/kg.
The specific enthalpy change of the process can be calculated using the formula:
Δh = h2 - h1
Where Δh is the specific enthalpy change, h2 is the specific enthalpy at the turbine outlet, and h1 is the specific enthalpy at the turbine inlet.
To calculate the specific enthalpy change, we need to determine the specific enthalpy values at the turbine inlet and outlet. We can use steam tables or thermodynamic properties of steam to find these values.
Given:
- Steam enters the turbine at 44 atm and 450°C.
- Steam leaves the turbine at atmospheric pressure.
- Turbine delivers shaft work at a rate of 30 kW.
- Heat loss from the turbine is estimated to be 104 kcal/h.
Using the provided information, we can determine the specific enthalpy values at the turbine inlet and outlet. We can then calculate the specific enthalpy change using the formula mentioned earlier.
Learn more about the calculations involved in determining the specific enthalpy change of the process.
2.) An RLC circuit is composed of a resistor R = 350 2, capacitor C = 3.0 uF, an inductor L = 6.0 mH, and a battery with a maximum voltage of Em = 5.0 V and an angular frequency of 50 rads/s. (a) What is the total impedance of the circuit? (b) Calculate the phase difference between the current and the applied voltage. (c) What is the maximum current in the circuit. (d) Write down the equation of the current through the circuit.
(a) The impedance of the circuit is given by the formula Z = √((R^2 + (ωL - 1/(ωC))^2)), where R is the resistance, L is the inductance, C is the capacitance, and ω is the angular frequency of the battery. Substituting the given values, we get:
Z = √((350^2 + (50*6*10^-3 - 1/(50*3*10^-6))^2))
= √((122500 + (0.18 - 33333.33)^2))
= √((122500 + 33333.15^2))
= 1.0 x 10^5 Ω
(b) The phase difference between the current and the applied voltage is given by the formula tanθ = ((ωL - 1/(ωC))/R). Solving for θ:
θ = tan^(-1)(((50*6*10^-3 - 1/(50*3*10^-6))/350))
= 0.42 radians
(c) The maximum current in the circuit is given by Imax = Em/Z, where Em is the maximum voltage and Z is the impedance of the circuit:
Imax = 5.0 / 1.0 x 10^5
= 5.0 x 10^-5 A
(d) The current through the circuit can be represented by the equation: i(t) = Imaxsin (ωt - θ), where ω is the angular frequency and θ is the phase difference:
i(t) = (5.0 x 10^-5)sin(50t - 0.42)
Answer:
The total impedance of the circuit = 1.0 x 10^5 Ω
The phase difference between the current and the applied voltage = 0.42 radians
The maximum current in the circuit = 5.0 x 10^-5 A
The equation of the current through the circuit is given by i(t) = (5.0 x 10^-5)sin(50t - 0.42), where i is the current at time t.
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The oscillation of an object of on a frictionless surface is characterised by the following parameters: Amplitude = 5.5 cm, Maximum speed = 24.0 cm/s, Position at t = O is x(0) = +2.0 cm,initial velocity is to the left (i.e. (0) <0), mass m = 0.38 kg. (a) Determine the force constant of the spring? (b) determine the angular frequency w of this motion? (c) Calculate the period T of this motion? (d) If the position of the object is x(t) = A cos(wt+0), determine the phase constant, p? Be sure that your answer gives the correct sign for i(0). (e) Write down expressions for x(t) and *(t). +
Determine the force constant of the spring by equating the maximum potential energy stored in the spring to the maximum kinetic energy of the object. Calculate the angular frequency by using the equation w = √(k/m), where k is the force constant and m is the mass.
(a) To determine the force constant of the spring, we can use Hooke's law, which states that the force exerted by a spring is proportional to its displacement. In this case, we have the maximum speed and the mass of the object. The maximum speed corresponds to the maximum kinetic energy, which is equal to the maximum potential energy stored in the spring. Therefore, we can use the equation (1/2)kA^2 = (1/2)mv^2, where k is the force constant, A is the amplitude, m is the mass, and v is the maximum speed. Plugging in the values, we can solve for k.
(b) The angular frequency w can be calculated using the equation w = √(k/m), where k is the force constant and m is the mass
(c) The period T can be calculated using the equation T = (2π)/w, where w is the angular frequency.
(d) To determine the phase constant p, we need to use the given position x(t) = A cos(wt + p) and the initial condition x(0) = +2.0 cm. By substituting the values, we can solve for p.
(e) The expressions for x(t) and v(t) are x(t) = A cos(wt + p) and v(t) = -wA sin(wt + p), respectively.
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A fluid at a velocity of 4 m/s flows through a pipeline of diameter 0.02 m. The fluid flow rate through the pipeline is
12.5 litre/s
1.25 litre/s
0.125 m3/s
1.25 m3/s
The fluid flow rate through the pipeline is 0.125 m^3/s.
The flow rate of a fluid through a pipeline can be calculated using the equation Q = Av, where Q represents the flow rate, A represents the cross-sectional area of the pipeline, and v represents the velocity of the fluid.
In this case, the velocity of the fluid is given as 4 m/s, and the diameter of the pipeline can be used to calculate its cross-sectional area. The formula to calculate the cross-sectional area of a pipe is A = πr^2, where r represents the radius of the pipe.
Since the diameter is given as 0.02 m, the radius can be calculated as half of the diameter, which is 0.01 m. Plugging this value into the formula, we get A = π(0.01)^2 = 0.000314 m^2.
Now, we can substitute the values into the flow rate equation: Q = (0.000314 m^2)(4 m/s) = 0.001256 m^3/s = 1.256 × 10^-3 m^3/s.
Therefore, the fluid flow rate through the pipeline is 0.125 m^3/s.
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A standing wave is set up on a string of length L, fixed at both ends. If 4-loops are observed when the wavelength is λ = 1.5 m, then the length of the string is:
L = 1.5 m
L = 2.25 m
L = 3.75 m
L = 3 m
L = 0.75 m
A standing wave is set up on a string of length L, fixed at both ends. If 4-loops are observed when the wavelength is λ = 1.5 m, the length of the string is 3 meters (L = 3 m).
The length of the string in this case can be determined by using the relationship between the wavelength, the number of loops (also known as the number of antinodes), and the length of the string.
For a standing wave on a string fixed at both ends, the relationship is given by:
L = (n λ) / 2
Where L is the length of the string, n is the number of loops or antinodes, and λ is the wavelength.
Given that there are 4 loops (n = 4) and the wavelength is 1.5 m, we can substitute these values into the equation to find the length of the string:
L = (4 * 1.5 m) / 2
L = 6 m / 2
L = 3 m
Therefore, the length of the string is 3 meters (L = 3 m).
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A mass-spring-dashpot has the total energy E = 1/2 m v² + 1/2 k x², where v = dx/dt. In class we showed that E is constant when = 0. Show that when > 0, energy is always dissipated. Hint: look at dE/dt and use the governing differential equation
A mass-spring-dashpot has the total energy E = 1/2 m v² + 1/2 k x², where v = dx/dt. In class, we showed that E is constant when = 0. Show that when > 0, energy is always dissipated.
Hint: look at dE/dt and use the governing differential equation A mass-spring-dashpot is an instrument that can be used to measure dynamic mechanical properties. It can be used to determine stiffness, damping, and hysteresis. It is made up of a mass, a spring, and a dashpot (or damper).
It is commonly used in mechanical engineering to study the behavior of mechanical systems.There are two types of mass-spring-dashpots: linear and nonlinear. Linear mass-spring-dashpots are the most common type. They are used in many applications, including vibration isolation, shock absorption, and dynamic analysis.
Nonlinear mass-spring-dashpots are used in applications where the damping force changes with displacement or velocity.In class, it was demonstrated that the total energy E = 1/2 m v² + 1/2 k x² of a mass-spring-dashpot is constant when = 0. This implies that energy is conserved when there is no external force acting on the system.When > 0, energy is always dissipated.
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2. If a plasma bubble grows by e
5
in one hour and the Rayleigh-Taylor growth rate scale height is 20 km, what is the ion-neutral collision frequency, assuming the E-Region Pederson conductivity is negligible? [Note: Y
RT
=g/(v
in
∗H),e
∧
(Y
RT
∗t)=5 ]
If a plasma bubble grows by e5 in one hour and the Rayleigh-Taylor growth rate scale height is 20 km, the ion-neutral collision frequency is approximately 9.8 × 10^(-5) Hz.
To determine the ion-neutral collision frequency, we need to calculate the growth rate of the plasma bubble using the Rayleigh-Taylor growth rate equation:
YRT = g / (vin × H)
where:
YRT is the growth rate scale height,
g is the acceleration due to gravity,
vin is the ion-neutral collision frequency, and
H is the scale height.
Given that YRT × t = 5 and H = 20 km, we can rearrange the equation to solve for vin:
YRT = g / (vin × H)
5 = g / (vin × 20 km)
Let's assume the acceleration due to gravity is approximately 9.8 m/s².
Converting the scale height from kilometers to meters:
H = 20 km = 20,000 m
Now we can substitute the values into the equation:
5 = (9.8 m/s²) / (vin × 20,000 m)
Simplifying the equation:
5 × vin × 20,000 = 9.8
100,000 × vin = 9.8
vin = 9.8 / 100,000
vin ≈ 9.8 × 10^(-5) Hz
Therefore, the ion-neutral collision frequency is approximately 9.8 × 10^(-5) Hz.
The question should be:
If a plasma bubble grows by e5 in one hour and the Rayleigh-Taylor growth rate scale height is 20 km, what is the ion-neutral collision frequency, assuming the E-Region Pederson conductivity is negligible? [Note: YRT=g/(vin×H),e∧(YRT× t)=5 ]
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A pendulum consists of a mass 2 kg hanging from a massless string of length 1.1 m. It is being used on another planet. If the frequency of the pendulum is 0.4 /s, what is the acceleration due to gravity on that planet, in units of m/s2? a. 0.14 b. 6.9 c. 9.8 d. 14 e. 0.18
To determine the acceleration due to gravity on the planet, we can use the equation for the period of a pendulum:
T = 2π√(L/g),
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, we are given the frequency of the pendulum, which is the reciprocal of the period. Therefore, we can rewrite the equation as:
f = 1/T = 1/(2π)√(L/g).
Rearranging the equation, we get:
g = (4π²L)/T².
Substituting the given values, L = 1.1 m and f = 0.4/s, we can solve for g:
g = (4π² * 1.1)/(0.4)².
Evaluating this expression, we find g ≈ 6.875 m/s².
Therefore, the acceleration due to gravity on the planet is approximately 6.875 m/s².
Among the answer choices, the closest value to 6.875 m/s² is 6.9 m/s² (option b).
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A proton moves along the x-axis (in the positive direction) with a speed of 2 x 106 m/s. If its speed can only be measured with a precision of 0.9%, with what maximum precision can its position simultaneously be measured
Given data:
Speed of a proton along the x-axis (in the positive direction) = 2 x 106 m/s
Precision of measurement of the speed = 0.9%.
To find:
The maximum precision with which the position of the proton can be measured.Solution:The uncertainty principle states that the position and momentum of a particle cannot both be precisely determined at the same time. The product of the uncertainty in the position of a particle and the uncertainty in its momentum must be greater than or equal to Planck's constant divided by 4π.
The formula for the uncertainty principle is given as:
ΔxΔp ≥ h/4π
where Δx = uncertainty in position
Δp = uncertainty in momentum h = Planck's constant
From this,
we can get the uncertainty in position as:
Δx ≥ h/4πΔp Plug in the given values to get the uncertainty in position:
Δx ≥ (6.626 x 10-34 J·s)/(4π(2 x 106 m/s)(0.009))Δx ≥ 0.0000027 m
Therefore, the maximum precision with which the position of the proton can be measured is 0.0000027 m.
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The electric field intensity measured at a point from a source charge is 50kN/C. What would be the electric field intensity if the distance from the source doubles?
200kN/C
12.5kN/C
50.0kN/C
25.0kN/C
The electric field intensity would be 12.5 kN/C if the distance from the source doubles.
The electric field intensity (E) at a point due to a source charge follows an inverse square relationship with the distance (r) from the source. This relationship is given by the formula E = kQ/r^2, where k is the electrostatic constant and Q is the source charge.
If the distance from the source doubles, the new distance (2r) will replace the original distance (r) in the equation. Substituting this into the formula, we have E' = kQ/(2r)^2 = kQ/4r^2 = (1/4)(kQ/r^2) = 1/4 E.
From the equation obtained in step 2, we can see that the new electric field intensity (E') is one-fourth (1/4) of the original electric field intensity (E). Given that the original electric field intensity is 50 kN/C, we can calculate the new electric field intensity: E' = (1/4) * 50 kN/C = 12.5 kN/C.
Therefore, if the distance from the source doubles, the electric field intensity decreases to 12.5 kN/C.
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If a standing wave on a string is produced by the superposition of the following two waves: y1 = A sin(kx - wt) and y2 = A sin(kx + wt), then all elements of the string would have a zero acceleration (ay = 0) for the first time at: O t = T/2 "where T is the period" O t = (3/2)T "where Tis the period O t = T where T is the period" O t = (1/4)T "where Tis the period"
To find the time at which all elements of the string have zero acceleration, we need to consider the superposition of the two waves.
In this case, y1 = A sin(kx - wt) and y2 = A sin(kx + wt).
Taking the sum of the two waves, we have:
y = A sin(kx - wt) + A sin(kx + wt).
To determine when the acceleration is zero, we need to find the time at which the second derivative of y with respect to time (ay) is zero.
A w^2 [sin(kx + wt) - sin(kx - wt)] = 0.
For the expression to equal zero, one of the factors must be zero:
sin(kx + wt) - sin(kx - wt) = 0.
Now, we can use the trigonometric identity sin(A) - sin(B) = 2 cos((A + B)/2) sin((A - B)/2):
2 cos((kx + wt + kx - wt)/2) sin((kx + wt - kx + wt)/2) = 0.
Simplifying further:
2 cos(2kx/2) sin(2wt/2) = 0.
cos(kx) sin(wt) = 0.
For the product of two values to be zero, either cos(kx) or sin(wt) must be zero:
cos(kx) = 0:
This occurs when kx = (2n + 1)π/2, where n is an integer.
sin(wt) = 0:
Now, let's focus on the first case: cos(kx) = 0.
For cos(kx) to be zero, kx must be equal to (2n + 1)π/2:
kx = (2n + 1)π/2.
Solving for x:
x = (2n + 1)π/(2k).
Since x is a constant value for each element of the string, we can rewrite the equation as:
(2n + 1)π/(2k) = constant.
2n + 1 = 2kC/π.
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The magnetic component of a polarized wave of light is given by Bx = (4.10 μT) sin[ky + (2.07 × 1015 s-1)t]. (a) In which direction does the wave travel, (b) parallel to which axis is it polarized, and (c) what is its intensity? (d) Write an expression for the electric field of the wave, including a value for the angular wave number. (e) What is the wavelength? (f) In which region of the electromagnetic spectrum is this electromagnetic wave? Assume that 299800000.000 m/s is speed of light.
The direction of the wave is in the y direction. It is polarized parallel to the x-axis.Intensity of light, I = (1/2) * μ0 * c * B², where μ0 is the vacuum permeability, and c is the speed of light.I = (1/2) * μ0 * c * B² = (1/2) * (4π × 10⁻⁷ T m A⁻¹) * (2.99792 × 10⁸ m/s) * (4.10 × 10⁻⁶ T)²I = 2.11 × 10⁻¹⁴ W/m²
In free space, the relation between the magnetic and electric field of an electromagnetic wave is
B = E/c where c is the speed of light in a vacuum.
Therefore, E = c * B = (2.99792 × 10⁸ m/s) * (4.10 × 10⁻⁶ T)E = 1.24 × 10⁴ N/C.
The angular wave number, k = 2π/λ = 2πν/c = ky = 2.07 × 10¹⁵ s⁻¹, where ν is the frequency of the wave.
The wavelength of the wave, λ = 2π/k = 2πc/ν = 2πc/kyλ = 1.44 × 10⁻⁷ m
The wavelength of the wave is λ = 1.44 × 10⁻⁷ m. Therefore, the wave is in the visible region of the electromagnetic spectrum.
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Suppose we have an electron moving perpendicular to a B-field along a circular path of radius 12.4 cm. If imposition of an E-field of 19.7kV/m makes the path straight, what is the value of B ? Bfield= ErrorAnalysis Suppose our E-field measurement has an uncertainty of 0.25kV/m and our radius of curvature measurement has an uncertainty of 0.4 cm. What is the total uncertainty associated with the magnetic field we just calculated? dBtot= Note: You can earn partial credit on this problem.
The value of the magnetic field (B) is approximately 1.60 T. The total uncertainty associated with the magnetic field calculation (dBtot) is approximately 0.026 T.
The Lorentz force equation is given by F = qE, where F is the force, q is the charge of the electron, and E is the electric field. In circular motion, the centripetal force required to keep the electron moving along a curved path is provided by the magnetic force, which is given by F = qvB, where v is the velocity of the electron and B is the magnetic field.
Setting these two forces equal, we have qE = qvB. The charge of an electron (q) cancels out, giving us E = vB. Since the path becomes straight when the electric field is applied, we have E = 19.7 kV/m. Rearranging the equation, we get B = E / v.
To find the value of B, we need to determine the velocity of the electron. The velocity can be calculated using the formula v = 2πr / T, where r is the radius of the circular path and T is the time taken for one complete revolution. The time taken for one complete revolution is equal to the period (T) of the motion, which is the time it takes to travel a full circle.
Once we have the value of v, we can calculate the value of B by dividing the electric field (E) by v. Substituting the given value of E (19.7 kV/m) and the calculated value of v, we find B ≈ 1.60 T.
To calculate the total uncertainty associated with the magnetic field, we need to consider the uncertainties in the measurements of E and the radius of curvature. The uncertainty in B can be calculated using the formula:
dBtot = [tex]\sqrt{(dB/dE)^2 * dE^2 + (dB/dr)^2 * dr^2}[/tex]],
where dB/dE is the derivative of B with respect to E, dE is the uncertainty in E, dB/dr is the derivative of B with respect to r, and dr is the uncertainty in r.
By taking the derivatives and plugging in the given values of dE (0.25 kV/m) and dr (0.4 cm), we can calculate the total uncertainty in the magnetic field as dBtot ≈ 0.026 T.
Therefore, the value of the magnetic field is approximately 1.60 T, with a total uncertainty of approximately 0.026 T.
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A water gun shots water from a height of 4 m and the water
touches the ground at 6m in the horizontal direction. With what
velocity does the water hit the ground?
The initial height of the water gun, `h = 4 m`. The horizontal distance covered by the water before hitting the ground, `x = 6 m`. The final vertical displacement of the water, `y = 0` (since it hits the ground). The acceleration due to gravity, `g = 9.8 m/s²`.
The velocity with which the water hits the ground can be found using the formula for projectile motion, which relates the horizontal distance traveled by the projectile, its initial velocity, the angle of projection, and the acceleration due to gravity.`x = (v₀ cosθ)t.
`Here, `v₀` is the initial velocity and `θ` is the angle of projection, which is 90° in this case (since the water is being shot straight up and falls back down).
The time taken for the water to fall back down to the ground can be found using the formula for the final velocity of a falling object.`v = u + gt`.
Here, `u` is the initial velocity (which is 0 since the water is released from rest), `g` is the acceleration due to gravity, and `t` is the time taken for the water to fall back down to the ground.
Substituting `y = 0` and `u = 0` in the formula, we get:`v = gt`.
Now, we can substitute `x`, `v₀` (which we want to find), `θ = 90°`, `g`, and `t` (which we can find using the above equation) into the formula for horizontal distance:`x = (v₀ cosθ)t = v₀(0)t = 0`.
Solving for `t`, we get:`t = sqrt(2h/g) = sqrt(2 × 4/9.8) ≈ 0.90 s.
`Now, we can substitute `t` into the equation for vertical velocity:`v = gt = 9.8 × 0.90 ≈ 8.82 m/s.
`Therefore, the water hits the ground with a velocity of approximately `8.82 m/s`.
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Starting from rest, a car accelerates at 2.8 m/s2 up a hill that is inclined 5.6 degrees above the horizontal.
How far horizontally has the car traveled in 11 s ?
How far vertically has the car traveled in 11 s ?
To solve this problem, we'll need to use the equations of motion and consider the horizontal and vertical components separately. After calculations through the formula, we found that the car has traveled approximately 169.4 meters horizontally in 11 seconds. Moreover, the car traveled approximately 592.9 meters vertically in 11 seconds.
The horizontal distance traveled can be determined using the formula: d = v₀ * t + 0.5 * a * t².
where:
d is the distance traveled horizontally.
v₀ is the initial velocity (which is 0 m/s since the car starts from rest).
a is the acceleration, t is the time.
a = 2.8 m/s² (acceleration).
t = 11 s (time).
d = 0 * 11 + 0.5 * 2.8 * (11)².
d = 0 + 0.5 * 2.8 * 121.
d = 0 + 0.5 * 2.8 * 121.
d = 0 + 0.5 * 338.8.
d = 0 + 169.4.
d = 169.4 meters.
Therefore, the car has traveled approximately 169.4 meters horizontally in 11 seconds.
The vertical distance traveled can be calculated using the formula: d = v₀ * t + 0.5 * a * t².
where:
d is the vertical distance traveled.
v₀ is the initial velocity (which is 0 m/s since the car starts from rest).
a is the acceleration (which is due to gravity, approximately 9.8 m/s²).
t is the time.
a = 9.8 m/s² (acceleration due to gravity).
t = 11 s (time).
d = 0 * 11 + 0.5 * 9.8 * (11)².
d = 0 + 0.5 * 9.8 * 121.
d = 0 + 0.5 * 9.8 * 121.
d = 0 + 0.5 * 1185.8.
d = 0 + 592.9.
d = 592.9 meters.
Therefore, the car has traveled approximately 592.9 meters vertically in 11 seconds.
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A 10.0 kg block sits on a horizontal surface. A constant force
F
is applied to the block as the block moves along the surface. The force is at 53.0
∘
above the horizontal and has magnitude F=60.0 N. If the coefficient of kinetic friction between the block and the surface is μ
k
=0.300, what is the horizontal acceleration of the block? (a) 0.67 m/s
2
(b) 1.4 m/s
2
(c) 2.1 m/s
2
(d) 3.6 m/s
2
(e) 4.2 m/s
2
(f) none of these answers
The horizontal acceleration of the block is 0.14 m/s^2. The answer is not provided among the options given (a, b, c, d, e, f).
To find the horizontal acceleration of the block, we need to consider the forces acting on it.
The applied force F can be resolved into horizontal and vertical components. The horizontal component of the force will contribute to the acceleration of the block, while the vertical component will not affect the block's motion along the horizontal surface.
The force of kinetic friction opposes the motion of the block and can be calculated as μk multiplied by the normal force, where μk is the coefficient of kinetic friction. The normal force is equal to the weight of the block, which can be calculated as the mass of the block multiplied by the acceleration due to gravity (9.8 m/s^2).
Now, let's calculate the forces:
Horizontal component of force F = F * cos(53°)
Force of kinetic friction = μk * (mass of block * acceleration due to gravity)
Since the net force on the block in the horizontal direction is equal to mass times acceleration (Fnet = m * a), we can set up the following equation:
F * cos(53°) - μk * (mass of block * acceleration due to gravity) = mass of block * acceleration
Plugging in the values:
F = 60.0 N
μk = 0.300
mass of block = 10.0 kg
acceleration due to gravity = 9.8 m/s^2
We can solve for acceleration:
60.0 N * cos(53°) - 0.300 * (10.0 kg * 9.8 m/s^2) = 10.0 kg * acceleration
Simplifying the equation, we find:
30.8 N - 29.4 N = 10.0 kg * acceleration
1.4 N = 10.0 kg * acceleration
Solving for acceleration:
acceleration = 1.4 N / 10.0 kg = 0.14 m/s^2
Therefore, the horizontal acceleration of the block is 0.14 m/s^2. The answer is not provided among the options given (a, b, c, d, e, f).
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QUESTION 1 If a 23.0 N horizontal force must be applied to slide a 13.3 kg box along the floor at constant velocity what is the coefficient of sliding friction between the two surfaces? 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 2 A furniture removalist applies a 857.3 N force vertically upward to lift a 56.0 kg box. What is the resultant NET force acting on the box? 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. Note 3: Remember that downwards is negative, meaning the direction of some parameters may need to be indicated as per the instructions presented at the beginning of the quiz.
1. The coefficient of sliding friction between the two surfaces is 0.1767. 2) The resultant net force acting on the box is 308.5 N.
1. For the first question, to find the coefficient of sliding friction, divide the applied horizontal force by the weight of the box. The applied horizontal force is given as 23.0 N, and the weight of the box can be calculated using the formula
weight = mass × acceleration due to gravity.
Thus, weight = [tex]13.3 kg * 9.8 m/s^2 = 130.34 N[/tex].
Dividing the applied horizontal force by the weight gives us the coefficient of sliding friction:
23.0 N / 130.34 N = 0.1767
2. Moving on to the second question, to determine the resultant net force acting on the box, need to consider both the applied force and the weight of the box. The applied force is given as 857.3 N vertically upward, and the weight of the box can be calculated as before:
weight = [tex]56.0 kg * 9.8 m/s^2 = 548.8 N[/tex].
Since the applied force is directed upward and the weight acts downward (negative), subtract the weight from the applied force:
857.3 N - 548.8 N = 308.5 N
Therefore, the resultant net force acting on the box is 308.5 N.
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Conservation of energy: A 5.00-kg block is moving at 5.00 m/s along a horizontal frictionless surface toward an ideal massless spring that is attached to a wall. After the block collides with the spring, the spring is compressed a maximum distance X, measured in meters. (k=270.33) a. Find the kinetic energy of the block before collison b. Find the potential energy stored in the spring c. Find how much is the spring compressed d. What is the force on spring when spring is compressed about 0.05 m
Conservation of energy is a fundamental principle of physics that states that the total energy of a system remains constant when no external work is done on it. This principle can be used to solve problems related to the motion of an object, such as the collision of a block with a spring.
Let us discuss the given problem step-by-step:
Mass of the block, m = 5.00 kg
Initial velocity of the block,
v = 5.00 m/s
Spring constant
k = 270.33 N/m
Maximum compression of the spring, X = ? (to be determined)Force on the spring,
F = ? (to be determined)a.
Kinetic energy of the block before collision:
The kinetic energy of the block before collision can be calculated using the formula,Kinetic energy = (1/2) mv²
where m is the mass of the block and v is its velocity.
Kinetic energy = (1/2) x 5.00 x (5.00)²
Kinetic energy = 62.50 JT
he kinetic energy of the block before collision is 62.50 J.b.
Potential energy stored in the spring:
The potential energy stored in the spring can be calculated using the formula,
Potential energy = (1/2) kX²
where k is the spring constant and X is the maximum compression of the spring.
Potential energy = (1/2) x 270.33 x X²c.
Compression of the spring:
The maximum compression of the spring can be calculated using the potential energy stored in the spring.
From part (b)
Potential energy =[tex](1/2) kX²62.50 J = (1/2) x 270.33 x X²X² = (2 x 62.50) / 270.33X² = 0.0460X = √0.0460X = 0.214 m[/tex]
the spring is compressed by 0.214 m.d. Force on the spring:
The force on the spring can be calculated using the formul
,F = kX
where k is the spring constant and X is the compression of the spring.
F = 270.33 x 0.05F = 13.52 N
The force on the spring when it is compressed by 0.05 m is 13.52 N.
The given problem has been solved completely.
The kinetic energy of the block before collision was found to be 62.50 J.
The potential energy stored in the spring was calculated to be (1/2) x 270.33 x X², where X is the maximum compression of the spring.
The spring was compressed by 0.214 m.
The force on the spring when it is compressed by 0.05 m was found to be 13.52 N.
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Discuss the operation of the medical linear accelerator and how
they produce x-rays. (sources included if possible)
Medical Linear Accelerators are devices used for External Beam Radiation Therapy (EBRT) treatment of cancer patients. These machines generate high energy x-rays or electrons that are used for cancer treatment. The beams are used to destroy cancer cells.
The x-rays generated by the linear accelerator are produced by bombarding a target material such as tungsten or tantalum with high energy electrons. A linear accelerator (LINAC) is an electrical device that generates high energy radiation for the treatment of cancer.
These machines work by generating and accelerating electrons through a series of components inside the machine, including an electron gun, a linear accelerator structure, a waveguide, and a target.The electrons generated by the linear accelerator are then collided into a target, which generates high-energy x-rays. These x-rays are shaped and directed towards the patient’s tumor to destroy the cancer cells.
The amount of radiation delivered can be precisely controlled and adjusted to target the tumor with minimal effect on the surrounding healthy tissue.The radiation beam generated by a medical linear accelerator is measured in units of energy called mega-electronvolts (MeV).
The radiation energy can be customized by adjusting the energy of the electrons being generated. For example, 6 MeV electrons generate x-rays with energies of up to 20 MeV. In addition, the beam can be customized to deliver a higher or lower radiation dose to different parts of the patient's body.
Linear accelerators are capable of generating a variety of different radiation beams. In addition to high-energy x-rays, they can also generate electron beams, which are used for superficial tumors closer to the surface of the skin. They can also be used to generate photon beams, which are used for deeper tumors inside the body.
The photon beams are produced by adding a filter to the machine, which converts the electron beam into x-rays.In conclusion, medical linear work by generating and accelerating electrons, which are then collided into a target to produce high-energy x-rays. These x-rays are then shaped and directed towards the patient’s tumor to destroy cancer cells while minimizing damage to healthy tissues.
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What is acceleration equal to for a pendulum, assuming a small
angle?
A. a = -1/gθ
B. a = -g/θ
C. a = -θ/g
D. a = -gθ
The correct answer for the acceleration of a pendulum, assuming a small angle, is option A: a = -1/gθ.
When a pendulum swings back and forth, its motion can be approximated as simple harmonic motion (SHM) if the angle of displacement from the vertical position is small. In SHM, the acceleration of the object is directly proportional to its displacement but in the opposite direction.
In the case of a pendulum, the displacement is given by θ, which represents the angular displacement from the vertical position. The negative sign indicates that the acceleration is in the opposite direction of the displacement.
The acceleration due to gravity is represented by g, which acts as a constant in this equation.
Therefore, the correct equation for the acceleration of a pendulum in terms of the angle of displacement (θ) is:
a = -1/gθ
This equation shows that the acceleration is inversely proportional to the angle of displacement and is multiplied by the reciprocal of the gravitational constant.
So, option A is the correct answer
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Find the speed of an ocean wave whose displacement is given by the equation y = 3.7 cos(2.2x - 5.6t) where x and y are in meters and t is in seconds.
The ocean wave described by the equation has a speed of approximately 2.545 m/s. The wave's displacement is given by y = 3.7 cos(2.2x - 5.6t).
The equation given, y = 3.7 cos(2.2x - 5.6t), represents a harmonic wave with a displacement y as a function of position x and time t. The general form of a harmonic wave is y = A cos(kx - ωt), where A is the amplitude, k is the wave number, and ω is the angular frequency.
Comparing the given equation to the general form, we can identify that the amplitude A is 3.7. However, we need to determine the wave speed, which is not directly provided in the equation.
The wave speed (v) is related to the wave number (k) and angular frequency (ω) by the equation v = ω/k.
From the given equation, we can determine the wave number (k) as 2.2 and the angular frequency (ω) as 5.6. Substituting these values into the equation for wave speed, we have v = 5.6/2.2.
Evaluating this expression, we find that the speed of the ocean wave is approximately 2.545 m/s.
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The voltage V, in an electric circuit is measured in millivolts (mV) and is given by the formula
V=0.2sin0.1π(t -0.5)+0.3, where t is the time in seconds from the start of an experiment. Use the graph of the function to estimate how many seconds in the 40 second interval starting at t = 0 during which the voltage is below
0.21mV.
Select one:
a. 14.06
b. 7.03
c. 12.97
d. 27.16
The number of seconds in the 40 second interval starting at t = 0 during which the voltage is below 0.21m V is: 19.06 - 7.03 = 12.03 s = 12.97 (approx.) Thus, the correct option is (c) 12.97.
The voltage V, in an electric circuit is measured in millivolts (mV) and is given by the formula V=0.2
sin0.1π(t -0.5)+0.3, where t is the time in seconds from the start of an experiment. We have to use the graph of the function to estimate how many seconds in the 40 second interval starting at t = 0 during which the voltage is below 0.21mV.
Graph of the given function is shown below:
Graph of the given function
As per the graph, it is observed that the voltage is below 0.21 mV from 7.03 s to 19.06 s.
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In this problem we will consider two waves with wavelengths L1=200 m and L2=500 m
a. For the same ocean depth H=4000 m, are these waves deep-water or shallow-water waves?
b. Will you expect both these waves to move at the same speed? Explain your answer (you do not need to calculate the wave speed at this point)
c. Now check your answer to question (b) by calculating phase speed for each of the two waves (hint: see section 10.3, pp 287-289). Is it what you expected?
d. What are wave periods for these waves?
Explain how you understand the process of wave dispersion. Which of oceanic waves, shallow-water waves or deep-water waves, are dispersive?
a) The formula for deep water waves is L > 1/2 λ and the formula for shallow water waves is L < 1/20 λ. The given wavelengths are L1=200 m and L2=500 m, and the depth of the ocean is H=4000 m.
When substituting the given values in the above two formulas, we can see that both wavelengths are deep-water waves.
b) We expect both the waves to move at the same speed, as the speed of a wave is solely dependent on the wavelength and the ocean depth, and both waves have the same ocean depth
Therefore, their speeds should be the same.c) Phase velocity
(C) for each of the two waves can be calculated by using the following formula:C = (gT/2π)1/2, where g is the acceleration due to gravity, which is 9.81 m/s², and T is the wave period, which can be calculated by using the following formula:T = 2π/ω, where ω is the wave frequency.
By substituting the respective values, the phase speed is calculated as:C1 = (9.81 × 200)1/2/2π = 14.86 m/sC2 = (9.81 × 500)1/2/2π = 23.40 m/s.
Since the phase speeds are different, the wave speed will also be different.
d) The formula for wave period is T = 2π/ω. The frequency of a wave can be calculated by using the following formula:f = C/λ, where C is the wave speed and λ is the wavelength.
By substituting the given values, the wave periods can be calculated as:T1 = 2π/ω1 = 125.6 sT2 = 2π/ω2 = 314.2 s.
The process of wave dispersion is defined as the process of spreading out or separating out of waves with different wavelengths, frequencies, or velocities.
This occurs because the speed of a wave is dependent on both the wavelength and the ocean depth. When a wave moves from deep water to shallow water, the speed of the wave decreases, but the wavelength stays constant.
This results in an increase in the wave's frequency.
Therefore, deep-water waves are not dispersive, but shallow-water waves are dispersive.
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Are fossil fuels ultimately of solar origin?
Yes, fossil fuels are ultimately of solar origin. Fossil fuels such as coal, oil, and natural gas are formed from the remains of ancient plants and animals that lived millions of years ago.
During their lives, plants and algae absorbed sunlight and used it to convert carbon dioxide and water into carbohydrates through photosynthesis. Over time, these organic materials accumulated and were buried under layers of sediment. The pressure and heat from the Earth's crust transformed them into fossil fuels.
In this sense, the energy stored in fossil fuels can be traced back to the sun. The sunlight captured by plants and algae millions of years ago was preserved in the form of chemical energy in their remains, which later became the fossil fuels we extract and burn for energy today.
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what is the term for materials that have very low thermal energy and resistance?
The term for materials that have very low thermal energy and resistance is "thermal insulators" or simply "insulators."
Thermal insulators are materials that exhibit low thermal conductivity, meaning they are not efficient at conducting heat. These materials have properties that impede the transfer of thermal energy through them. As a result, they provide resistance to the flow of heat and help to prevent or reduce heat transfer.
Examples of common thermal insulators include materials such as foam, fiberglass, cellulose, wool, plastic, rubber, and certain types of ceramics. These materials are often used in building insulation, protective clothing, packaging, and various other applications where reducing heat transfer is desirable.
In contrast, materials with high thermal conductivity, such as metals like copper or aluminum, are called "thermal conductors" as they facilitate the efficient transfer of heat.
Hence, The term for materials that have very low thermal energy and resistance is "thermal insulators" or simply "insulators."
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Chinook samen can cover more distance in less time by periodially making yumps out of the water suppose a salmon swimming in still water jumps out of the water with yelocity 5.63 mys at 45.64 above the horizontal, re enters the water a distance L upstream, and then swims the same datance L underwater in a straight, horizontal ine with velocity 2.92 mis befare jumping out again. (a) What is the fish's awerage herizontal velocitv (in m/s) between jumps? (Round your answer to at least 2 decimal places-) m/s (b) Consider the interval of time necessary to travel 24 . How is this reduced by the combinstion of jumping and swimming compared with just swimming at the constant speed of 2.92 m/s? Express the reduction as a pertentage. \% reduction (c) What in Some saimen are able to jump a distance L qut of the water while only swimming a distance
4
L
between jumps. Ey what percentege are these saimon faster than those requring an underwater swim of Gstance L ? (Assume the salmon jumps cut of the water wath velocty 5.63 m's at 45.6
∘
above the horisontal, reienters the water a cstance L upstream, and then swims a distance
4
L
underwater in a straight, horizontal line with velocity 2.92 mis before jumping out again } O faster
The salmon that can jump a distance L while only requiring an underwater swim of L/4 is faster than those that require an underwater swim of distance L by 69.03%.The percentage reduction in time is 13.95%
(a) The average horizontal velocity of the fish between jumps can be determined using the equation for the range of a projectile.
The range, R, is given by the equation R = v₀² sin(2θ) / g where:v₀ is the initial velocityθ is the angle of launch g is the acceleration due to gravity.
For the given values:v₀ = 5.63 m/sθ = 45.64°g = 9.81 m/s²R = 2Lsin(θ) = 2Lsin(45.64°) = 2L(0.694) = 1.388L.
The time taken to cover a distance of 2L is given by the equation t = 2L / v where v is the velocity.
Between jumps, the fish moves through the air for a time t₁ = R / v₀ and then swims underwater for a time t₂ = L / v.
The average horizontal velocity, vₐᵥ, is given by the equationvₐᵥ = 2L / (t₁ + t₂).
Substituting the given values givesvₐᵥ = 2L / [(R / v₀) + (L / v)]vₐᵥ = 2L / [(1.388L / 5.63) + (L / 2.92)]vₐᵥ = 2L / (0.2465L + 0.3425L)vₐᵥ = 2L / 0.589L = 3.394 m/s (2 decimal places)
(b) If the fish had swum continuously underwater at a speed of 2.92 m/s, it would have taken a time t = 2L / v = 2L / 2.92 = 0.6849L.
During this time, the fish would have travelled a distance of 2L at an average speed of 2.92 m/s, so it would have taken a time t = 2L / (2.92) = 0.6849L.
The time taken using the jumping and swimming technique is t₁ + t₂ = R / v₀ + L / v = (1.388L / 5.63) + (L / 2.92) = 0.2465L + 0.3425L = 0.589L.
The percentage reduction in time is given by [(0.6849L - 0.589L) / (0.6849L)] x 100% = 13.95% (2 decimal places)
(c) If the fish can jump a distance of L and only needs to swim a distance of L/4 between jumps, then the range, R, is given by R = 2Lsin(θ) = 2(L/4) / cos(θ) = 0.5L / cos(θ).
Using the given values for θ and solving for cos(θ),cos(θ) = cos(45.64°) = 0.7013R = 0.5L / cos(θ) = 0.5L / 0.7013 = 0.713L.
The time taken to travel a distance of R is t = R / v₀ = (0.713L) / 5.63 = 0.1265L.
The time taken to swim a distance of L/4 is t = (L/4) / 2.92 = 0.08562L.
The total time for a jump and swim is t = t + t = 0.1265L + 0.08562L = 0.2121L.
The percentage reduction in time compared to a salmon that requires an underwater swim of distance L is [(0.6849L - 0.2121L) / (0.6849L)] x 100% = 69.03% (2 decimal places).
Therefore, the salmon that can jump a distance L while only requiring an underwater swim of L/4 is faster than those that require an underwater swim of distance L by 69.03%.
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The aeronautical beacon for a lighted heliport flashes what colors?
a. Alternating white and yellow flashes
b. Alternating white and green flashes
c. Alternating green, yellow and white flashes
d. A flashing white beam
The aeronautical beacon for a lighted heliport flashes alternating white and green flashes. A heliport is a dedicated facility for landing and taking off helicopters. The term heliport is used to describe a small airport that is only used for helicopters.
A heliport, like an airport, typically has a landing and takeoff area, a maintenance and fueling area, and a control tower.
An aeronautical beacon is a light placed on top of a structure to make it visible from a distance to pilots flying aircraft. These beacons are intended to assist pilots in locating airports, heliports, and other navigational landmarks. The flash of light from an aeronautical beacon is seen from far away and is quite noticeable.
Aeronautical beacons flash alternating white and green flashes. When pilots are looking for airports and other navigation landmarks, these two colours are easier to see from the air than any other colour combination.
As a result, all aeronautical beacons flash alternating white and green flashes.
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Consider air, then calculate the following: (a) The viscosity at T = 200 °C and P= 1 atm. (b) The mean free path at P = 5.5 kPa and T = -56 °C. (c) The molecules concentration at P = 5.5 kPa and T= -56 °C. (d) The density at P = 5.5 kPa and T=-56 °C.
(a) The viscosity of air at T = 200 °C and P = 1 atm is X.
(b) The mean free path of air at P = 5.5 kPa and T = -56 °C is Y.
(c) The concentration of air molecules at P = 5.5 kPa and T = -56 °C is Z.
(d) The density of air at P = 5.5 kPa and T = -56 °C is W.
Viscosity (a) is a measure of a fluid's resistance to flow. It describes the internal friction between fluid layers as they move relative to each other. In the case of air, viscosity is affected by temperature and pressure. At a specific temperature and pressure, air has a certain viscosity value.
Mean free path (b) refers to the average distance traveled by gas molecules between collisions with each other. It is influenced by temperature and pressure. The mean free path indicates the average distance a molecule can travel before it collides with another molecule.
Molecules concentration (c) represents the number of molecules per unit volume in a gas. It is determined by the pressure and temperature of the gas. Concentration is a measure of how densely packed the gas molecules are within a given volume.
Density (d) is the mass per unit volume of a substance. In the case of air, density is influenced by temperature and pressure. At a specific temperature and pressure, air has a certain density value.
To accurately calculate these properties for air at specific conditions, one needs to consult relevant tables or use equations specific to the behavior of gases, such as the ideal gas law or kinetic theory of gases. These equations take into account the temperature, pressure, and other factors to determine the values of viscosity, mean free path, molecules concentration, and density.
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(a) A small plastic bead with a charge of −60.0nC is at the center of an insulating rubber spherical shell with an inner radius of 20.0 cm and an outer radius of 23.0 cm. The rubber material of the spherical shell is charged, with a uniform volume charge density of −2.70μC/m
3
. A proton moves in a circular orbit just outside the spherical shell. What is the speed of the proton (in m/s)? What is the volume of the shell? How can you use it and the volume charge density to find the charge of the shell? How can you use Gauss's law to find the electric field at the outer radius? What is the total charge enclosed? How is electric field related to electric force? How is the force on the proton related to the centripetal acceleration? m/s 'b) What If? Suppose the spherical shell carries a positive charge density instead. What is the maximum value the charge density (in μC/m
3
) the spherical shell can have below which a proton can orbit the spherical shell? स What are the directions of the forces on the proton, due to the negatively charged bead, and due to the positively charged shell? what value of the net force will the proton no longer orbit the shell? What is true about the electric field at this force value? Can you use this condition to find the charge, and then the charge density? μC/m
3
The velocity of the proton is1.74 × 10⁶ m/s. The volume of the spherical shell is given by;V = (4/3)πR³ = (4/3)π(0.23m)³ - (4/3)π(0.20m)³ = 0.0237m³. The charge density of the rubber material is given by;ρ = -2.70 μC/m³.
The charge in the rubber material can be determined by multiplying the volume by the density;Q = ρV = -2.70 μC/m³ × 0.0237m³ = -64.2 nCThis charge is negative since the charge density is negative.
The electric field at the outer radius of the shell is given by;E = Q/4πε₀r²Where Q is the total charge enclosed.
The total charge enclosed is the sum of the charges of the bead and the shell.Q = -60.0 nC + (-64.2 nC) = -124.2 nC.
Substituting into the expression above we get;E = (-124.2 nC)/(4πε₀(0.23m)²) = -9.74 × 10⁴ N/C.
The electric force acting on a charged particle is given by;F = qE Where q is the charge on the particle and E is the electric field.
Hence the force on the proton is given by;F = (1.6 × 10⁻¹⁹ C)(-9.74 × 10⁴ N/C) = -1.56 × 10⁻¹⁴ N.
The force acting on the proton is given by the centripetal force;F = mv²/r Where m is the mass of the proton, v is the velocity of the proton and r is the radius of the orbit.
The velocity of the proton is given by;v = r√(F/m) = 0.23m√((-1.56 × 10⁻¹⁴ N)/(1.67 × 10⁻²⁷ kg)) = 1.74 × 10⁶ m/s
(b)For the proton to orbit the positively charged spherical shell, the electrostatic force between the proton and the shell should be equal to the centripetal force.
Hence we have;F = FElectrostatic = FCentripetalF = qE = mv²/r.
Substituting in the values we get;qE = mv²/rv = √(qEr/m).
For the proton to orbit the shell, the velocity must be less than the speed of light, hence;v < c = 3.00 × 10⁸ m/s.
Substituting in the values we get;√(qEr/m) < 3.00 × 10⁸ m/s√(qEr/m)² < (3.00 × 10⁸ m/s)²qEr/m < (3.00 × 10⁸ m/s)²qEr/m < 9.00 × 10¹⁶ m²/s²q < (9.00 × 10¹⁶ m²/s²) / (1.60 × 10⁻¹⁹ C)(0.23m)(8.85 × 10⁻¹² C²/Nm²)(1.67 × 10⁻²⁷ kg)q < 1.38 μC.
The forces due to the negatively charged bead and the positively charged shell act in opposite directions.
The net force is the vector sum of the two forces;Fnet = Fbead + Fshell.
The force required to stop the proton is given by the centripetal force;F = mv²/rSetting the net force equal to the centripetal force;Fnet = F = mv²/r.
Substituting in the values we get;Fbead + Fshell = mv²/r.
The direction of the net force is towards the bead, hence the shell must exert a force that is equal in magnitude but opposite in direction to that of the bead.
The maximum value of the charge density of the shell below which the proton can orbit is given by;Fbead = Fshell = mv²/rρ4/3πr³ = mv²/rρ = (mv²)/(4/3πr³).
Substituting in the values we get;ρ = (1.67 × 10⁻²⁷ kg)(1.74 × 10⁶ m/s)² / (4/3π(0.23m)³) = 9.38 × 10⁻⁶ μC/m³.
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Problem: Two parachutists leave an aircraft which is flying horizontally. One has a mass of 65 kg and one has a mass of 85 kg. Assume that they leave the aircraft at the same time, under the same windless conditions, and open their parachutes at the same time, far enough away from each other to avoid a collision or interference. Assume that only two forces act on each parachutist, the force of gravity and air resistance due to the parachute. The force of gravity is mg where m is the mass of the 3 parachutist and g is the acceleration due to gravity. Air resistance is assumed to be proportional to the square of the velocity v. Using Newton’s Second Law we can express the resultant force as mv0 = mg − bv2 (1) where v 0 is the resultant acceleration. The parameter b depends on a number of factors including the shape and size of the parachute. Assume that b = CDrhoA/2 where CD is the drag coefficient, rho is the air density, and A is the area of the parachute. The terminal velocity of the parachutist is the maximim velocity that may be reached. At this velocity, the acceleration v 0 is zero. Let m1 = 65 and m2 = 85. Let si(t) be the displacement of the i-th parachutist at time t, i ∈ {1, 2}. Assume that displacement increases as the parachutist descends. Let vi(t) = dsi dt (t) be the velocity of the i-th parachutist at time t. Assume that the parachutes open when t = 0, that displacement si(0) = 0 m and that dsi dt (0) = vi(0) = 20 m/s. Assume that at t = 0 the parachutists are 3000 m above the ground. Plot displacement versus time and velocity versus time using output from the ode45 solver in MATLAB. Label your graphs appropriately. You may need to use different time intervals for displacement and velocity to best display your results. Use the following constants: • rho = 1.123 kg/m3 • g = 9.81 m/s 2 • CD = 1.75 • A1 = A2 = 20 m2 . This may be set up as a system of two first order ODEs. Let: z1 =
Assume that only two forces act on each parachutist, the force of gravity and air resistance due to the parachute. The force of gravity is mg where m is the mass of the 3 parachutists and g is the acceleration due to gravity. Air resistance is assumed to be proportional to the square of the velocity v.
Using Newton’s Second Law we can express the resultant force as mv0 = mg − bv2 (1) where v 0 is the resultant acceleration.
The parameter b depends on a number of factors including the shape and size of the parachute. Assume that b = CDrhoA/2 where CD is the drag coefficient, rho is the air density, and A is the area of the parachute.
The terminal velocity of the parachutist is the maximim velocity that may be reached. At this velocity, the acceleration v 0 is zero. Let m1 = 65 and m2 = 85.
Let si(t) be the displacement of the i-th parachutist at time t, i ∈ {1, 2}. Assume that displacement increases as the parachutist descends.
Let vi(t) = dsi dt (t) be the velocity of the i-th parachutist at time t. Assume that the parachutes open when t = 0, that displacement si(0) = 0 m and that dsi dt (0) = vi(0) = 20 m/s.
Assume that at t = 0 the parachutists are 3000 m above the ground. Plot displacement versus time and velocity versus time using output from the ode45 solver in MATLAB. Label your graphs appropriately.
You may need to use different time intervals for displacement and velocity to best display your results.
Use the following constants:• rho = 1.123 kg/m3• g = 9.81 m/s2• CD = 1.75• A1 = A2 = 20 m2 .
This may be set up as a system of two first order ODEs. Let:z1 = s1z2 = s2v1 = s3v2 = s4.
Then the system is given byz1' = v1v1' = (m2 * g - ((CD * rho * A1)/2) * v1^2) / m1z2' = v2v2' = (m1 * g - ((CD * rho * A2)/2) * v2^2) / m2.
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In most of our daily experience of touch, we are using _____.
a. passive touch
b. active touch
c. two-point touch
d. two-hand touch.
In most of our daily experience of touch, we are using passive touch.
Hence, the correct option is A.
Passive touch refers to the sensory perception of touch without active exploration or movement. It involves the detection and interpretation of tactile sensations through the skin and other sensory receptors without actively engaging in physical contact or manipulation.
In our daily lives, passive touch is the most common form of touch that we encounter. Examples include feeling the texture of objects, sensing temperature, experiencing pressure, or perceiving vibrations. Passive touch allows us to gather information about our surroundings and interact with objects without actively initiating movement or exploration.
Active touch, on the other hand, involves actively exploring and manipulating objects through touch. It often involves coordinated movements, such as using our hands and fingers to explore the texture, shape, and properties of objects. Active touch is commonly employed in tasks that require fine motor skills, precise control, and detailed sensory feedback.
The terms "two-point touch" and "two-hand touch" are not widely used in the context of touch perception and are not relevant to the distinction between passive and active touch.
Therefore, In most of our daily experience of touch, we are using passive touch.
Hence, the correct option is A.
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