The maximum amount of charge that can be packed onto a green pea before it spontaneously discharges is approximately 1.3 microcoulombs (μC).
When the electric field exceeds a certain threshold, dry air can break down and generate a spark. In this case, the threshold is given as 3.0×[tex]10^6[/tex] N/C. To determine the maximum charge that can be packed onto a green pea before it discharges, we need to calculate the electric field strength at the surface of the pea.
The electric field strength (E) near the surface of a uniformly charged sphere is given by E = k * Q / [tex]r^2[/tex], where k is Coulomb's constant (k ≈ 9.0×[tex]10^9[/tex] [tex]N m^2/C^2[/tex]), Q is the charge on the sphere, and r is the radius of the sphere. In this case, the diameter of the pea is 0.85 cm, so the radius (r) is 0.85 cm / 2 = 0.425 cm = 0.00425 m.
We can rearrange the formula to solve for the charge (Q): Q = E * [tex]r^2[/tex] / k. Plugging in the values, we have Q = (3.0×[tex]10^6[/tex] N/C) * [tex](0.00425 m)^2[/tex] / (9.0×[tex]10^9 N m^2/C^2[/tex]). Calculating this, we find that the maximum charge is approximately 1.3 microcoulombs (μC), rounded to two significant figures. Therefore, any charge exceeding this amount would cause the green pea to spontaneously discharge due to the breakdown of dry air.
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CONSERVATION OF ENERGY. Calculate the final velocity (at the ground, h=0) the chandelier would have if the rope used to hang it suddenly broke, and the chandelier plummets to the ground. (mass of the chandelier 18kg, height from the ground 6ft)
The final velocity of the chandelier when it hits the ground would be approximately 9.8 m/s.
To calculate the final velocity of the chandelier, we can use the principle of conservation of energy. The potential energy of the chandelier at its initial height can be calculated using the formula PE = mgh,
where m is the mass (18 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (6 ft = 1.83 m). The potential energy at the initial height is converted entirely into kinetic energy at ground level, given by KE = 0.5mv^2.
By equating the potential energy and kinetic energy and solving for v, we find that the final velocity of the chandelier is approximately 9.8 m/s.
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An electromagnetic wave has a wavelength of 768 micrometers.
What is the frequency of this electromagnetic wave? Express your answer in GigaHertz and keep three significant digits
The frequency of this electromagnetic wave is approximately 391 GHz.
The frequency of an electromagnetic wave can be calculated using the formula: f = c / λ, where f is the frequency, c is the speed of light (approximately 3.00 × 10^8 meters per second), and λ is the wavelength.
First, let's convert the wavelength from micrometers to meters:
λ = 768 micrometers * (1 × 10^(-6) meters per micrometer) = 0.000768 meters.
Now we can calculate the frequency:
f = (3.00 × 10^8 meters per second) / (0.000768 meters) = 3.91 × 10^11 Hz.
To express the frequency in GigaHertz (GHz), we divide the frequency by 10^9: f = (3.91 × 10^11 Hz) / (10^9) = 391 GHz (rounded to three significant digits).
Therefore, the frequency of this electromagnetic wave is approximately 391 GHz.
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Starting frem rest, a rectangular toy block with mass 275.9 sides in 2.10 s all the way across a table 1.80 min length that Zak has tilted at an angle of 32.5∘ to the horizontal. (o) What is the magnitude of the acceleration of the toy block? m/s2 (b) What is the coeticient of winetic friction between the block and the table? We Your response defers from the correc answer by more than 1096. Doubie check your calculations. (c) What are the macnitude and crection of the frictien force acting on the block? magnitude direction (d) What is the speed of the tilock when is is at the end of the table, having sid a distance of 1.80 m ? an Yeur response chers from the correct answer by mere than 10%, Double check your calculatians: nVs
The magnitude of the acceleration of the toy block is 0.015 m/s2. the coefficient of kinetic friction between the block and the table is `0.001792`. The direction of the frictional force is towards the rest of the block.The speed of the block when it is at the end of the table is 0.25 m/s.
a) The formula for acceleration is given by the formula,`a = (vf - vi)/t`Where `vf` is the final velocity, `vi` is the initial velocity, and `t` is the time taken. In this case, `vf = d/t`.
Here, `d` is the distance travelled by the block and `t` is the time taken.
The initial velocity of the block is zero.
Therefore, the formula for acceleration can be rewritten as `a = 2d/t^2`.
The block has moved a distance equal to the length of the table, which is 1.80 m.
Therefore, the distance travelled by the block is d = 1.80 m.
The time taken by the block to travel the length of the table is t = 2.10 s.
Substituting the values, we get: `a = 2(1.80)/2.10^2 = 0.015 m/s^2`.
Therefore, the magnitude of the acceleration of the toy block is 0.015 m/s2
.b) The formula for kinetic friction is given by the formula `f = µkN`. Where `f` is the force of friction, `µk` is the coefficient of kinetic friction, and `N` is the normal force.
The normal force acting on the block is given by `N = mg cosθ`.
Here, `m` is the mass of the block, `g` is the acceleration due to gravity, and `θ` is the angle of inclination of the table.
Substituting the values, we get: `N = 275.9 × 9.81 × cos 32.5∘ = 2309.7 N`.
The force of friction acting on the block is given by `f = ma`.
Substituting the values, we get: `f = 275.9 × 0.015 = 4.14 N`.
Therefore, the coefficient of kinetic friction between the block and the table is `µk = f/N = 4.14/2309.7 = 0.001792`.
c) The frictional force acting on the block is given by `f = µkN = 0.001792 × 2309.7 = 4.14 N`.
The frictional force acts in a direction opposite to the direction of motion of the block.
Therefore, the direction of the frictional force is towards the rest of the block.
d) The final velocity of the block can be calculated using the formula `v^2 = u^2 + 2as`.
Here, `u` is the initial velocity, `v` is the final velocity, `a` is the acceleration of the block, and `s` is the distance travelled by the block.
The initial velocity of the block is zero.
Therefore, the formula can be simplified as `v = √(2as)`.
Substituting the values, we get: `v = √(2 × 0.015 × 1.80) = 0.25 m/s`.
Therefore, the speed of the block when it is at the end of the table is 0.25 m/s.
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Bubba drops a rock down a well. The rock hits the water at the bottom of the well after 2.06 seconds. (Ignore air resistance.) A) How deep is the well? B) How fast is the rock moving when it reaches the bottom of the well?
Bubba drops a rock down a well.The rock hits the water at the bottom of the well after 2.06 seconds. (Ignore air resistance.)We have to calculate:A) How deep is the well?B) How fast is the rock moving when it reaches the bottom of the well.
The velocity of the rock at any time t is given by:v(t) = g * t where, g = 9.8 m/s² (acceleration due to gravity)When the rock reaches the water at the bottom of the well, the distance it travels is equal to the depth of the well. The distance traveled by the rock is given by:s(t) = (1/2) * g * t²We know the time it takes for the rock to hit the water,
so we can find the depth of the well by plugging in the time into the equation for distance:s(2.06) = (1/2) * 9.8 * (2.06)²s(2.06) = 21.16 mTherefore, the depth of the well is 21.16 meters.To find the velocity of the rock when it hits the water, we can use the formula:v(t) = g * tWe know the acceleration due to gravity, and we know the time it takes for the rock to hit the water:v(2.06) = 9.8 * 2.06v(2.06) = 20.17 m/sTherefore, the velocity of the rock when it hits the water is 20.17 m/s.
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The force exerted by an ideal spring is a conservative force, and the vertical forces do no work, so the total mechanical energy of the system is (A) conserved (C) zero (D) none of the above
the total mechanical energy of the system is (A) conserved.
Conservative forces, such as the force exerted by an ideal spring, have the property that the total mechanical energy of a system is conserved. Mechanical energy is the sum of the kinetic energy and potential energy of the system.
In the case of an ideal spring, as long as there are no non-conservative forces (such as friction or air resistance) acting on the system, the total mechanical energy remains constant. The potential energy of the spring changes as it is compressed or stretched, but this change is compensated by the corresponding change in kinetic energy of the system.
Therefore, the total mechanical energy of the system, which includes the kinetic and potential energies, is conserved.
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How many turns would be needed to make a 1mH inductor with the following specifications: 18 AWG magnet wire (diameter of 1.02 mm ) 2 cm cclil diameter Air filled (ur=1) 140 153 161 175
To calculate the number of turns needed to make a 1mH inductor with the given specifications, we can use the formula:
L = (n^2 x d^2 x ur x A) / (18 x 10^10)
where:
L = inductance in Henrys (1mH in this case)
n = number of turns
d = wire diameter in meters (1.02 mm = 0.00102 m)
ur = relative permeability (1 for air)
A = cross-sectional area of the coil in square meters (π x r^2, where r is the radius of the coil)
First, let's find the cross-sectional area of the coil. The coil diameter is given as 2 cm, which means the radius is 1 cm or 0.01 m. Therefore, the cross-sectional area (A) is π x (0.01)^2.
Next, let's substitute the given values into the formula and solve for n:
1 x 10^(-3) = (n^2 x (0.00102)^2 x 1 x π x (0.01)^2) / (18 x 10^10)
Simplifying the equation:
1 x 10^(-3) = (n^2 x 1.0404 x 10^(-6) x 3.1416 x 10^(-4)) / (18 x 10^10)
Multiplying both sides by (18 x 10^10):
18 x 10^7 = n^2 x 1.0404 x 3.1416 x 10^(-6)
Dividing both sides by (1.0404 x 3.1416 x 10^(-6)):
n^2 = 18 x 10^7 / (1.0404 x 3.1416 x 10^(-6))
Taking the square root of both sides:
n = √(18 x 10^7 / (1.0404 x 3.1416 x 10^(-6)))
Evaluating the expression:
n ≈ 153.78
Therefore, approximately 154 turns would be needed to make a 1mH inductor with the given specifications.
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A skydiver jumps out of an airplane. What happens to her acceleration after her jump? Include air resistance. Mark the two correct answers. When she reaches terminal velocity, her acceleration becomes constant. Before she reaches terminal velocity, her velocity increases. Before she reaches terminal velocity, her acceleration remains constant. Before she reaches terminal velocity, her velocity decreases.
The correct answers are 1. When she reaches terminal velocity, her acceleration becomes constant and 4. Before she reaches terminal velocity, her velocity decreases.
When a skydiver jumps out of an aeroplane, her acceleration initially increases due to the force of gravity. However, as she falls, the opposing force of air resistance gradually becomes stronger. At a certain point, the force of air resistance equals the force of gravity, resulting in a net force of zero. This is known as terminal velocity.
At terminal velocity, the skydiver's acceleration becomes constant because the net force acting on her is zero. This means that her velocity no longer increases or decreases, and she falls at a constant speed. Therefore, statement 1 is correct: when she reaches terminal velocity, her acceleration becomes constant.
Before reaching terminal velocity, the force of gravity is stronger than the force of air resistance. As a result, the skydiver's velocity continues to increase. However, as the force of air resistance increases, it eventually equals the force of gravity. At this point, the net force becomes zero, and the skydiver's velocity stops increasing. In fact, her velocity starts to decrease because the force of air resistance now exceeds the force of gravity. Hence, statement 4 is also correct: before reaching terminal velocity, her velocity decreases.
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This problem checks that you can use the formula that gives the electric field due to a spherical shell of charge. This formula can be calculated using the superposition principle we discussed in class and gives
E
=
4πϵ
0
1
r
2
Q
r
^
outside the shell and zero inside the shell. The distance r is the distance between the center of the shell and the point of interest. Consider a sphere with radius 4 cm having a uniformly distributed surface charge of +25nC. What is the magnitude of the electric field it creates at a point 6 cm from its center, in units of kN/C ?
A sphere with radius 4 cm having a uniformly distributed surface charge of +25nC. Its magnitude will be approximately 1.48 × [tex]10^7[/tex] N/C.
To find the magnitude of the electric field created by the spherical shell at a point outside the shell, we can use the formula you mentioned:
E = (1 / (4πε₀)) * (Q / [tex]r^2[/tex])
where:
E is the electric field,
ε₀ is the permittivity of free space,
Q is the charge of the spherical shell, and
r is the distance between the center of the shell and the point of interest.
Given:
Radius of the sphere, R = 4 cm = 0.04 m
Surface charge density, σ = +25 nC (uniformly distributed over the surface of the sphere)
To calculate the magnitude of the electric field at a point 6 cm from the center of the sphere, we need to find the total charge Q of the sphere. The charge Q can be obtained by multiplying the surface charge density σ by the surface area of the sphere.
Surface area of the sphere, A = 4π[tex]r^2[/tex]
Substituting the given values:
A = 4π[tex](0.04)^2[/tex]
A = 0.0201 π [tex]m^2[/tex]
Total charge, Q = σ * A
Q = (25 × [tex]10^{(-9)[/tex]) * (0.0201 π)
Q ≈ 1.575 × [tex]10^{(-9)[/tex]C
Now we can calculate the electric field using the formula:
E = (1 / (4πε₀)) * (Q / [tex]r^2[/tex])
Substituting the values:
E = (1 / (4π(8.85 × [tex]10^{(-12)[/tex]))) * (1.575 × [tex]10^{(-9)[/tex] / [tex](0.06)^2[/tex])
E ≈ 1.48 × [tex]10^7[/tex] N/C
Therefore, the magnitude of the electric field created by the spherical shell at a point 6 cm from its center is approximately 1.48 × [tex]10^7[/tex] N/C (kilonewtons per coulomb).
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in fast-pitch softball, a pitcher might use a "windmill" motion in which she moves her hand through a circular arc to pitch a ball at 70 mph. The 0.19 kg ball is 50cm from the pivot point at her shoulder and the ball reaches its maximum speed at the lowest point of the circular arc. at the bottom of the circle, just before the ball leaves her hand, what is it centripetal acceleration? what are the magnitude and direction of the force her hand exerts on the ball at this point? please show all work
The centripetal acceleration of the ball is 1960.8 m/s².2.
The magnitude of the force her hand exerts on the ball at this point is 372.352 N. The direction of this force is towards the center of the circle
Given values:
Speed of the ball = 70 mph = 31.2928 m/s
Mass of the ball = 0.19 kg
Radius of the circle = 50 cm = 0.5 m1.
Find the centripetal acceleration of the ball.
The centripetal acceleration is given by the formula:
ac = v²/r
where
ac = centripetal acceleration
v = velocity
r = radius of the circle
Substitute the given values, we get:
ac = (31.2928 m/s)²/(0.5 m)ac = 1960.8 m/s²
Therefore, the centripetal acceleration of the ball is 1960.8 m/s²
Find the magnitude and direction of the force her hand exerts on the ball at this point.
The force her hand exerts on the ball is the centripetal force. It is given by the formula:
F = mac
where
F = force applied
m = mass of the ball
a = centripetal acceleration
Substitute the given values, we get:
F = (0.19 kg)(1960.8 m/s²)F = 372.352 N
The magnitude of the force her hand exerts on the ball at this point is 372.352 N. The direction of this force is towards the center of the circle.
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In photoelectric effect, the stopping potential value is 0.6 V when the light source is kept at a distance of 10 cm. When the source is kept at 20 cm away, the stopping potential will be A) 0.6 V B) 0.3 V C) 1.2 V D) 2.4 V
In the photoelectric effect, the stopping potential value is 0.6 V when the light source is placed at a distance of 10 cm. When the source is moved to a distance of 20 cm, the stopping potential will be 0.3 V. Therefore the correct option is B) 0.3 V.
The stopping potential in the photoelectric effect is determined by the energy of the incident photons and the work function of the material. The stopping potential depends on the intensity of the incident light, which is inversely proportional to the square of the distance from the source.
Given that the stopping potential is 0.6 V when the light source is 10 cm away, we can use the inverse square law to determine the stopping potential when the source is 20 cm away.
Let's denote the stopping potential when the source is 20 cm away as V'.
According to the inverse square law, the intensity of the light at a distance of 20 cm would be (10 cm / 20 cm)^2 = 0.25 times the intensity at 10 cm.
Since the stopping potential is directly proportional to the intensity of the incident light, the stopping potential when the source is 20 cm away would also be 0.25 times the stopping potential at 10 cm.
0.25 * 0.6 V = 0.15 V
Therefore, the stopping potential when the source is kept at 20 cm away would be 0.15 V.
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A ball is thrown downward with an initial velocity of 13 m/s. Using the approximate value of g=10 m/s
2
, calculate the velocity of the ball 1.8 seconds after it is released. The velocity of the ball is m/s.
The velocity of the ball 1.8 seconds after it is released is 5.6 m/s.ExplanationA ball is thrown downward with an initial velocity of 13 m/s. Given,Initial velocity u = 13 m/sThe ball is thrown downward which means it is thrown in the downward direction.
Hence, acceleration will be acting in the downward direction and acceleration due to gravity g = 10 m/s²Time taken t = 1.8 secondsWe have to find the velocity of the ball after 1.8 seconds.Solution:Using the third equation of motion, we have, v = u + atwherev is the final velocity of the ballu is the initial velocity of the balla is the acceleration of the ballt is the time taken by the ballFrom the given values, we can substitute,u = 13 m/sa = 10 m/s²t = 1.8 secondsSubstituting these values in the equation above,
we get:v = u + atv = 13 m/s + 10 m/s² × 1.8 secondsv = 13 m/s + 18 m/sv = 31 m/sTherefore, the velocity of the ball 1.8 seconds after it is released is 31 m/s.Approximately, g = 10 m/s²Using the approximate value of g=10 m/s², we can also use the below formula,v = u + gtwherev is the final velocity of the ballu is the initial velocity of the ballg is the acceleration of the ballt is the time taken by the ballFrom the given values, we can substitute,u = 13 m/sg = 10 m/s²t = 1.8 secondsSubstituting these values in the equation above, we get:v = u + gtv = 13 m/s + 10 m/s² × 1.8 secondsv = 13 m/s + 18 m/sv = 31 m/sTherefore, the velocity of the ball 1.8 seconds after it is released is 31 m/s.
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The phase velocity of surface waves in deep water is approximately given by
k
ω
=
k
g
+
rho
m
T
s
k where T
s
is the surface tension (7.3×10
−2
N/m) and rho
m
=10
3
kg/m
3
is the water mass density. (a) Find the expression for the group velocity. (b) When a stone is thrown into a pond, waves of many wavelength components start propagating. What would be the slowest wavelength? Note: A stone thrown into water gives an impulse-like disturbance, which contains many wavelength components.
(a) To find the expression for the group velocity, we start with the dispersion relation for surface waves in deep water:
ω² = gk + (T_s / ρ_m)k³
where ω is the angular frequency, k is the wave number, g is the acceleration due to gravity, T_s is the surface tension, and ρ_m is the water mass density.
To find the group velocity, we differentiate the dispersion relation with respect to k:
dω/dk = g + (3T_s / ρ_m)k²
The group velocity (v_g) is given by the derivative of the angular frequency with respect to the wave number:
v_g = dω/dk = g + (3T_s / ρ_m)k²
Therefore, the expression for the group velocity is v_g = g + (3T_s / ρ_m)k².
(b) When a stone is thrown into a pond, waves of various wavelength components start propagating. The slowest wavelength (λ_slowest) corresponds to the longest wavelength component present in the disturbance.
In deep water, the slowest wavelength can be determined by finding the wave number (k_slowest) that gives the lowest group velocity. To minimize the group velocity, we can set the derivative of the group velocity with respect to k to zero:
dv_g/dk = 0
2gk_slowest + (6T_s / ρ_m)k_slowest² = 0
Solving this equation, we find:
k_slowest = -3(T_s / 2ρ_mg)
The slowest wavelength (λ_slowest) is related to the wave number by λ_slowest = 2π / k_slowest.
Therefore, the slowest wavelength would be λ_slowest = -2πρ_mg / (3T_s).
It's important to note that the negative sign indicates a reversal in direction compared to the other wavelengths.
The slowest wavelength component will have the longest wavelength and will propagate at the lowest speed among the various wavelength components generated by the disturbance.
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Two narrow slits 50μm apart are illuminated with light of wavelength 620 nm. The light shines on a screen 1.2 m distant. What is the angle of the m=2 bright fringe? Express your answer in radians. \ Incorrect; Try Again; 19 attempts remaining Part B How far is this fringe from the center of the pattern? Express your answer with the appropriate units. X Incorrect; Try Again; 19 attempts remaining
The distance of the m=2 bright fringe from the center of the pattern is approximately 0.1488 meters.
To find the angle of the m=2 bright fringe in a double-slit interference pattern, we can use the formula:
sin(theta) = m * (lambda) / d
where theta is the angle of the fringe, m is the order of the fringe, lambda is the wavelength of the light, and d is the distance between the slits.
Given that the distance between the slits is 50 μm (50 x 10^(-6) m) and the wavelength of the light is 620 nm (620 x 10^(-9) m), and we want to find the angle of the m=2 bright fringe.
Plugging the values into the formula:
sin(theta) = 2 * (620 x 10^(-9) m) / (50 x 10^(-6) m)
Now we can solve for theta by taking the inverse sine (sin^(-1)) of both sides:
theta = sin^(-1)[2 * (620 x 10^(-9) m) / (50 x 10^(-6) m)]
Calculating the value using a calculator:
theta ≈ 0.248 radians
So the angle of the m=2 bright fringe is approximately 0.248 radians.
For Part B, to find the distance of this fringe from the center of the pattern, we can use the formula:
y = m * (lambda) * L / d
where y is the distance from the center, m is the order of the fringe, lambda is the wavelength of the light, L is the distance from the slits to the screen, and d is the distance between the slits.
Given that the distance from the slits to the screen is 1.2 m, and we want to find the distance of the m=2 bright fringe.
Plugging the values into the formula:
y = 2 * (620 x 10^(-9) m) * (1.2 m) / (50 x 10^(-6) m)
Calculating the value:
y ≈ 0.1488 m
So the distance of the m=2 bright fringe from the center of the pattern is approximately 0.1488 meters.
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An object is launched with an initial velocity of 3.5 m/s at an initial angle of 52
∘
above the ground from a height of 0.0 m. After the object has traveled for 0.5 seconds, how far in the x-direction has the object traveled? 1.8 m 3.5 m 1.1 m 7.0 m 2.2 m
The object has traveled 0.9 meters in the x-direction after 0.5 seconds because the horizontal component of the initial velocity is 1.8 meters per second.
The horizontal component of the initial velocity is the part of the velocity that is parallel to the x-axis. The vertical component of the initial velocity is the part of the velocity that is parallel to the y-axis.
The distance traveled in the x-direction after 0.5 seconds is simply the horizontal component of the initial velocity multiplied by the time.
The horizontal component of the initial velocity is:
v_x = v_0 * cos(theta) = 3.5 m/s * cos(52°)
v_x = 1.8 m/s
The distance traveled in the x-direction after 0.5 seconds is:
x = v_x * t = 1.8 m/s * 0.5 s
x = 0.9 m
Therefore, the object has traveled 0.9 m in the x-direction after 0.5 seconds.
The other answer choices are incorrect. 3.5 m is the total initial velocity, not the horizontal component. 1.1 m and 7.0 m are the horizontal distances traveled after 1.0 and 2.0 seconds, respectively. 2.2 m is the horizontal distance traveled after 1.0 second.
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A highway curve of radius 591.21 m is designed for traffic moving at a speed of 71.09 km/h. What is the correct banking angle of the road?
The correct banking angle of the road is approximately 3.85 degrees.
The correct banking angle of a road can be determined using the equation:
θ = arctan((v^2 / (r * g)))
where:
θ is the banking angle of the road,
v is the velocity of the vehicle,
r is the radius of the curve, and
g is the acceleration due to gravity.
Let's plug in the given values:
v = 71.09 km/h = 19.74 m/s (converted to m/s)
r = 591.21 m
g = 9.8 m/s²
θ = arctan((19.74^2 / (591.21 * 9.8)))
Calculating the expression:
θ = arctan(0.067092)
Now, evaluating the arctan:
θ ≈ 0.0672 radians
To convert this angle to degrees, we can multiply it by the conversion factor:
θ ≈ 0.0672 * (180/π)
θ ≈ 3.85 degrees
Therefore, the correct banking angle of the road is approximately 3.85 degrees.
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Two pieces of metal are being joined by braze welding. However, the filler metal, when applied to the preheated base metal, gathers in a ball and will not wet, or tin, the base metal. This condition is caused by A. applying too much flux. B. applying too little flux. C. overheating the base metal. D. underheating the base metal.
Two pieces of metal are being joined by braze welding . However, the filler metal, when applied to the preheated base metal, gathers in a ball and will not wet, or tin, the base metal. This condition is caused by underheating the base metal.
What is Braze welding?
Braze welding, often known as bronze welding, is a process that is used to join pieces of metal together. This is accomplished by heating the metal pieces to be joined to the point where the braze metal added to the joint melts and joins the pieces. The key advantage of braze welding is that it allows two different metals to be joined together.
Braze welding is used in a variety of applications, including the creation of decorative objects and in the construction of machinery, among other things. Braze welding is especially useful in joining metals that have vastly different melting points and cannot be joined using other welding techniques.
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An aircraft has a wing area of 20 m² and whose wings resemble the NACA 23012 with no flaps and is flying horizontally (0° angle of attack) at a constant speed of 250 km/h. To gain height the pilot adjusts the controls so that the angle of attack becomes 10°. Take the density of the air as 1.23 kg/m³. Determine the total power required to execute this action at the same constant speed.
The total power required to execute this action at the same constant speed is 296143.1776 Watt.
In order to gain height, the pilot will adjust the controls so that the angle of attack becomes 10°. The aircraft has a wing area of 20 m² and its wings resemble the NACA 23012 with no flaps, and is flying horizontally (0° angle of attack) at a constant speed of 250 km/h.
We will use the following formula to calculate the total power required:
P = 0.5ρV³SCD
Where:
P is power
ρ is density of air
V is velocity
S is the wing area
CD is the coefficient of drag for NACA 23012
At an angle of attack of 0°, the coefficient of drag for the NACA 23012 wing is 0.023. At an angle of attack of 10°, the coefficient of lift is 0.041.
Let's look at the math behind the calculations:
P = 0.5 * 1.23 * (250 * 1000 / 3600)³ * 20 * (0.023 + 0.041 * 10² / (1 + 10²))
= 296143.1776 Watt
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a) In outer space, far from other objects, block 1 of mass 37 kg is at position (4,11,0)m, and block 2 of mass 1140 kg is at position ⟨22,11,0⟩m. What is the (vector) gravitational force acting on block 2 due to block 1 ? It helps to make a sketch of the situation. Tries 0/10 At 4.225 scconds after noon both blocks were at rest at the positions given above. At 4.55 seconds after noon, what is the (vector) momentum of block 2 ?
P
2
=⟨,⟩ Tries 0/10 At 4.55 seconds after noon, What is the (vectot) momentum of block 17
P
1
=⟨,… Tries 0/10 At 4.55 seconds after noon, which one of the following statements is true? Block 2 is moving faster than block 1 Block 1 and block 2 have the same speed. Block 1 is moving faster than block 2 .
the vector gravitational force acting on block 2 due to block 1 is approximately 1.2674 x 10^(-7) N in magnitude.
To determine the vector gravitational force acting on block 2 due to block 1, we can use Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (approximately 6.67430 x 10^(-11) N m^2/kg^2), m1 and m2 are the masses of the two blocks, and r is the distance between their centers of mass.
Given that the mass of block 1 is 37 kg and its position is (4, 11, 0) m, and the mass of block 2 is 1140 kg and its position is (22, 11, 0) m, we can calculate the distance between their centers of mass:
r = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
r = √((22 - 4)^2 + (11 - 11)^2 + (0 - 0)^2)
r = √(18^2 + 0^2 + 0^2)
r = 18 m
Now we can calculate the gravitational force:
F = (6.67430 x 10^(-11) N m^2/kg^2) * (37 kg * 1140 kg) / (18 m)^2
F ≈ 1.2674 x 10^(-7) N
Therefore, the vector gravitational force acting on block 2 due to block 1 is approximately 1.2674 x 10^(-7) N in magnitude.
For the second part of the question, to determine the momentum of block 2 at 4.55 seconds after noon, we need to know its velocity. Without that information, we cannot calculate the momentum. Please provide the velocity of block 2 at that time so that we can assist you further.
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The gravitational force acting on block 2 from block 1 depends on the distances between them but can't be calculated with the given information. The momentum of block 2 at any given time would be zero due to no external forces. The comparison of speed of block 1 and block 2 can't be made without information about their velocities.
Explanation:The gravitational force acting on block 2 due to block 1 can be calculated using Newton's law of gravitation. This law states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
To calculate the vector force, we would need to know the exact position of these two blocks in three dimensional space. This would involve calculating the distance vector from block 1 to block 2, then use this to calculate the force vector by multiplying the magnitude of the gravitational force (calculated using Newton's law) by the unit vector in the direction of the distance vector. However, given the current details, a precise calculation can't be performed.
As for the momentum of block 2 at 4.55 seconds after noon, it can be calculated using the equation p = mv , where p is momentum, m is mass and v is velocity. Since no external forces are mentioned, block 2 would remain at rest, meaning its velocity is 0 and thus its momentum is also 0.
The question about whether block 1 is moving faster than block 2, or vice versa or both have the same speed, cannot be answered without knowing details about their velocities.
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The shells fired from an artillery piece have a muzzle speed of 180 m/s, and the target is at a horizontal distance of 1.40 km. Two possible angles can be used to hit the target. Find the smallest angle relative to the horizontal at which the gun should be aimed?(Hint: sin2θ=2sinθcosθ)
To hit a target at a horizontal distance of 1.40 km, the gun should be aimed at an angle of approximately 18.7 degrees relative to the horizontal. This angle can be calculated using the horizontal range equation and the given muzzle speed of 180 m/s.
To find the smallest angle relative to the horizontal at which the gun should be aimed, we can use the equation for the horizontal range of a projectile:
R = (v^2 * sin(2θ)) / g,
where R is the horizontal distance, v is the muzzle speed, θ is the angle of projection, and g is the acceleration due to gravity.
Given that the horizontal distance is 1.40 km (or 1400 m) and the muzzle speed is 180 m/s, we can rearrange the equation to solve for the angle θ:
θ = 0.5 * arcsin((R * g) / (v^2)).
Substituting the values, we have:
θ = 0.5 * arcsin((1400 * 9.8) / (180^2)) ≈ 0.5 * arcsin(0.609) ≈ 0.5 * 0.654 ≈ 0.327 radians.
Converting the angle to degrees, we find that the smallest angle relative to the horizontal at which the gun should be aimed is approximately 18.7 degrees.
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(a) Calculate the height (in m ) of a cliff if it takes 2.19 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.05 m/s. (Enter a number.) m (b) How long (in s) would it take to reach the ground if it is thrown straight down with the same speed? (Enter a number.)
The Time can't be zero so it will take 4.37 seconds to reach the ground.
Initial velocity = 8.05 m/s
Time taken = 2.19 s
We have to find the height of the cliff.
From the kinematic equation:v = u + at
v = final velocity
u = initial velocity
a = acceleration
t = time
Taken
The initial velocity as u is positive as it is in the upward direction.
Using first equation of motion we can find acceleration.
a = (v - u) / ta = (0 - 8.05) / 2.19a = -3.68 m/s²
Now, we can find the distance covered (height of the cliff) using third equation of motion.
s = ut + 1/2 at²s = 8.05 x 2.19 + 1/2 (-3.68) x (2.19)²s = 17.59 m
So, the height of the cliff is 17.59 m.
Now, for part (b) the initial velocity is also 8.05 m/s but the direction is downwards so it is negative.
Using third equation of motion we can find time taken.
0 = -8.05 x t + 1/2 (-3.68) x t²0 = t(-8.05 - 1.84t)t² - 4.37t = 0t(t - 4.37) = 0So, t = 0 or 4.37 s.
Time can't be zero so it will take 4.37 seconds to reach the ground.
So, the answer is: (a) 17.59 m (b) 4.37 seconds
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A wave train, which has a wave height of H=4.5 m in deep water and a wave period of T=8 s, approaches the shore with an 8^∘
angle of incidence in deep water. Determine the wave height, group velocity, and angle of incidence at depths d=5,10,15, and 20 m.
θd = sin^(-1) (sin(θ) / √(c / (c - d)))
where θd is the angle of incidence at the given depth, θ is the angle of incidence in deep water (8°), and d is the depth.
By applying these equations, we can calculate the wave height, group velocity, and angle of incidence at depths of 5, 10, 15, and 20 m
When a wave train approaches the shore, the wave height, group velocity, and angle of incidence change at different depths. To determine these values at depths of 5, 10, 15, and 20 m, we need to use the dispersion relation for deep water waves. In deep water, the wave speed is given by the formula:
c = √(gT/2π),
where c is the wave speed, g is the acceleration due to gravity, and T is the wave period.
1. Wave height: The wave height at a specific depth can be determined using the equation:
Hd = H / √(c / (c - d))
where Hd is the wave height at the given depth, H is the wave height in deep water (4.5 m), and d is the depth.
2. Group velocity: The group velocity can be calculated using the equation:
Vg = c / (2π)
where Vg is the group velocity and c is the wave speed.
3. Angle of incidence: The angle of incidence at a specific depth can be determined using the equation:
θd = sin^(-1) (sin(θ) / √(c / (c - d)))
where θd is the angle of incidence at the given depth, θ is the angle of incidence in deep water (8°), and d is the depth.
By applying these equations, we can calculate the wave height, group velocity, and angle of incidence at depths of 5, 10, 15, and 20 m.
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At the same moment from the top of a building 3.0×10^2
m tall, one rock is dropped and one is thrown downward with an initial velocity of 10 m/s. Both of them experience negligible air resistance. How much EARLIER does the thrown rock strike the ground? Your answer: They land at exacly the same time. 0.865 0.525 0.955 0.675
The thrown rock takes approximately 5.94 s to reach the ground. The correct answer is: 1.88 s.
The time it takes for an object to fall freely from a height h can be calculated using the equation: t = √(2h/g)
For the dropped rock: h = 3.0 × 10^2 m
t_dropped = √(2(3.0 × 10^2)/9.8)
t_dropped ≈ √(600/9.8)
t_dropped ≈ 7.82 s
For the thrown rock: h = 3.0 × 10^2 m
v_initial = 10 m/s
g = 9.8 m/s^2
The time it takes to reach the ground can be calculated using the equation: h = v_initial * t_thrown + (1/2) * g * t_thrown^2
3.0 × 10^2 = 10 * t_thrown + (1/2) * 9.8 * t_thrown^2
Simplifying the equation:4.9 * t_thrown^2 + 10 * t_thrown - 3.0 × 10^2 = 0
Solving this quadratic equation, we find two solutions for t_thrown: t_thrown ≈ -13.18 s and t_thrown ≈ 5.94 s.
Comparing the times, we find that the thrown rock strikes the ground approximately 7.82 s - 5.94 s = 1.88 s earlier than the dropped rock.
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(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 54 m? (b) How long will it be in the air? (a) Number Units
(a) The ball must be thrown vertically from ground level to rise to a maximum height of 54 m at a speed of 32.55 m/s . (b) The ball will be in the air for 5.52 sec.
(a) We can use the equation,
v^2 - u^2 = 2gh
Here, v = final velocity = 0 (at maximum height)
u = initial velocity
g = acceleration due to gravity = 9.8 m/s^2
h = maximum height = 54 m
Plugging in the values and solving for u,
u^2 = 2gh - v^2
= 2 × 9.8 × 54 - 0
= 1058.8u = ±√1058.8
= 32.55 m/s (rounded to two decimal places)
Therefore, ball must be thrown vertically from ground level to rise to a maximum height of 54 m at a speed of 32.55 m/s .
(b)We can use the formula,
s = ut + 1/2gt^2
Here, s = maximum height = 54 m
u = initial velocity = 32.55 m/s
g = acceleration due to gravity = 9.8 m/s^2
t = time taken
Plugging in the values,
54 = 32.55t + 1/2 × 9.8 × t^2
Simplifying,4.9t^2 + 32.55t - 54 = 0
Solving this quadratic equation, we get, t = 5.52 s
Therefore the ball will be in the air for 5.52 sec.
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suitcase is 20.0 N. (a) Draw a free-body diagram of the suitcase. No file chosen This answer has not been graded yet. (b) What angle does the strap make with the horizontal (in degrees)? (c) What is the magnitude of the normal force that the ground exerts on the suitcase (in N)? N rolling friction is independent of the angle of the strap. (e) What is the maximum acceleration of the suitcase if the woman can exert a maximum force of 38.7 N ? (Enter the magnitude in m/s
2
.) m/s
2
F_ applied is the force applied by the woman (38.7 N), F_ friction is the force of rolling friction, and m is the mass of the suitcase.
In this case, the vertical component is the weight of the suitcase (20.0 N) and the horizontal component is the tension in the strap. The magnitude of the normal force (N) exerted by the ground on the suitcase is equal to the weight of the suitcase, which is given as 20.0 N. Rolling friction is independent of the angle of the strap. The maximum acceleration of the suitcase can be calculated using Newton's second law of motion. The angle that the strap makes with the horizontal can be determined using trigonometry.
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A car is moving at 9 m/s when it accelerates at 1.9 m/s2 for 27 seconds. What was
the final speed?
The final speed of the car after acceleration at 1.9 m/s^2 for 27 seconds is approximately 60.3 m/s.
The final speed of the car can be calculated using the formula for uniformly accelerated motion:
The final speed of the car can be determined using the formula:
final speed = initial speed + (acceleration * time)
Given:
Initial speed (u) = 9 m/s
Acceleration (a) = 1.9 m/s^2
Time (t) = 27 seconds
Plugging in these values into the formula, we can calculate the final speed:
final speed = 9 m/s + (1.9 m/s^2 * 27 s)
final speed = 9 m/s + 51.3 m/s
final speed ≈ 60.3 m/s
Therefore, the final speed of the car after accelerating at 1.9 m/s^2 for 27 seconds is approximately 60.3 m/s.
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(3 peinti) Compute the diop in blood pressure along a 25−cm length of artery of radius 4 thmm. Arsume that the antery camss biood at a rate of 7 fiterimin. (viscosty of blood =0.04 poise)
the diop in blood pressure along a 25−cm length of artery of radius 4 thmm is approximately 37.8 kPa.
Diop in blood pressure along a 25−cm length of artery of radius 4 thmm can be computed as follows.Diop in blood pressure along a length of an artery is the change in the pressure of the blood as it moves from one end to another end of the artery. It is denoted by ΔP.
The formula to calculate ΔP is given by:ΔP = 4 × Q × L × η / π × r⁴WhereQ = flow rate of the bloodL = length of the arteryr = radius of the arteryη = viscosity of the bloodWe know that the length of the artery, L = 25 cm = 0.25 m
Radius of the artery, r = 4 µm = 4 × 10⁻⁶ m
Flow rate of the blood, Q = 7 L/min = 7 × 10⁻³ m³/sViscosity of the blood, η = 0.04 poise = 0.04 × 10⁻² Pa.s
Therefore,ΔP = 4 × Q × L × η / π × r⁴= 4 × 7 × 10⁻³ × 0.25 × 0.04 × 10⁻² / π × (4 × 10⁻⁶)⁴= 3.78 × 10⁴ Pa≈ 37.8 kPa,
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Find the following for path D in Figure 3.56 : (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, expllieitly show how you follow the steps of the analytical method of vector addition. Figure 3.56 The various knes represent paths taken by different people wallong in a city All blocks are 120 m on a side. 14. Find the following for path D in Figure 3.56 : (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitiy showhow: you follow the steps of the analytical method of vector addition.
a) The total distance traveled along path D is 1,080 m. b) The magnitude of displacement is calculated using the Pythagorean theorem is 648.07 m and the direction is approximately 68.2° above the positive x-axis.
To find the total distance traveled along path D in Figure 3.56, we need to determine the length of each segment and sum them up. According to the given information, all blocks are 120 m on a side. By carefully following the path, we can determine the lengths of each segment:
Segment AB: The path moves right, covering 3 blocks, so the distance traveled is 3 * 120 m = 360 m.
Segment BC: The path moves up, covering 2 blocks, so the distance traveled is 2 * 120 m = 240 m.
Segment CD: The path moves left, covering 1 block, so the distance traveled is 1 * 120 m = 120 m.
Segment DE: The path moves up, covering 3 blocks, so the distance traveled is 3 * 120 m = 360 m.
Therefore, the total distance traveled along path D is 360 m + 240 m + 120 m + 360 m = 1,080 m.
To find the displacement from start to finish, we need to calculate the magnitude and direction. We can follow the steps of the analytical method of vector addition:
Break down each segment into its x (horizontal) and y (vertical) components.
AB: x-component = 360 m, y-component = 0 m
BC: x-component = 0 m, y-component = 240 m
CD: x-component = -120 m, y-component = 0 m
DE: x-component = 0 m, y-component = 360 m
Sum up the x-components: 360 m - 120 m = 240 m
Sum up the y-components: 240 m + 360 m = 600 m
The magnitude of displacement is calculated using the Pythagorean theorem: √(240 m^2 + 600 m^2) ≈ 648.07 m.
To find the direction, we can use trigonometry. The angle θ can be found by taking the inverse tangent of the ratio of the y-component to the x-component: θ = tan^(-1)(600 m / 240 m) ≈ 68.2°.
Therefore, the magnitude of displacement is approximately 648.07 m, and the direction is approximately 68.2° above the positive x-axis.
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A group of students observes that a wooden block (m=0.40 kg) on the end of a string with a radius of 0.7 meters makes 13 rotations in 20.7 seconds when twirled. What is the block's tangential (linear) speed? Part B What is the block's angular speed?
The tangential speed of a block on a string can be calculated using the formula (2π * radius * number of rotations) divided by time, while the angular speed is (2π * number of rotations) divided by time.
Part A:
To find the block's tangential (linear) speed, we can use the formula:
Tangential speed = (2πr * number of rotations) / time
Given that the radius (r) is 0.7 meters, the number of rotations is 13, and the time is 20.7 seconds, we can plug in these values to calculate the tangential speed:
Tangential speed = (2π * 0.7 * 13) / 20.7
Calculating the above expression gives us the tangential speed of the block.
Part B:
The angular speed of the block can be found using the formula:
Angular speed = (2π * number of rotations) / time
Using the given values of 13 rotations and 20.7 seconds, we can substitute them into the formula to calculate the angular speed:
Angular speed = (2π * 13) / 20.7
This calculation will give us the angular speed of the block.
It's important to note that the tangential speed is the linear speed at the edge of the circular path, while the angular speed is the rate at which the object rotates, expressed in radians per second.
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A geostationary satellite carries a C-band transponder which transmits 15 watts into an antenna with an on-axis gain of 32 dB. An earth station is in the center of the antenna beam from the satellite, at a distance of 38,500 km. For a frequency of 4.2GHz a. Calculate the incident flux density at the earth station in watts per square meter and in dBW/m
2
. b. The earth station has an antenna with a circular aperture 3 m in diameter and an aperture efficiency of 62%. Calculate the received power level in watts and in dBW at the antenna output port. c. Calculate the on-axis gain of the antenna in decibels?
a. Flux Density (dBW/m^2) ≈ -84.6 dBW/m^2.
b. Received Power (dBW) ≈ -89.6 dBW.
c. On-axis Gain (G_on-axis) ≈ 58.28 dB.
a. To calculate the incident flux density at the Earth station, we can use the formula:
Flux Density (W/m^2) = (Transmitted Power * Antenna Gain) / (4 * pi * Distance^2)
Given:
Transmitted Power (Pt) = 15 watts
Antenna Gain (G) = 32 dB (decibels)
Distance (D) = 38,500 km = 38,500,000 meters
Converting the antenna gain from decibels to linear scale:
Antenna Gain (G_linear) = 10^(G/10)
G_linear = 10^(32/10)
G_linear = 10^3.2
G_linear = 1584.89
Substituting the values into the formula:
Flux Density (W/m^2) = (15 * 1584.89) / (4 * pi * (38,500,000)^2)
Calculating the flux density:
Flux Density (W/m^2) = 0.00002733 watts per square meter
To convert the flux density from watts per square meter to dBW/m^2:
Flux Density (dBW/m^2) = 10 * log10(Flux Density)
Calculating the flux density in dBW/m^2:
Flux Density (dBW/m^2) = 10 * log10(0.00002733)
Flux Density (dBW/m^2) ≈ -84.6 dBW/m^2
b. To calculate the received power level at the antenna output port, we can use the formula:
Received Power (Pr) = (Flux Density * Aperture Area * Aperture Efficiency)
Given:
Aperture Diameter (D) = 3 meters
Aperture Efficiency (η) = 62% = 0.62
Calculating the Aperture Area (A):
Aperture Area (A) = pi * (D/2)^2
A = 3.1416 * (3/2)^2
A ≈ 7.0686 square meters
Substituting the values into the formula:
Received Power (Pr) = (0.00002733 * 7.0686 * 0.62)
Calculating the received power level:
Received Power (Pr) ≈ 0.0001215 watts
To convert the received power from watts to dBW:
Received Power (dBW) = 10 * log10(Received Power)
Calculating the received power level in dBW:
Received Power (dBW) = 10 * log10(0.0001215)
Received Power (dBW) ≈ -89.6 dBW
c. The on-axis gain of the antenna can be calculated using the formula:
On-axis Gain (G_on-axis) = (4 * pi * Aperture Efficiency * Antenna Gain) / (Wavelength^2)
Given:
Frequency (f) = 4.2 GHz = 4.2 * 10^9 Hz
Calculating the Wavelength (λ):
Wavelength (λ) = Speed of Light / Frequency
λ = (3 * 10^8 meters/second) / (4.2 * 10^9 Hz)
Substituting the values into the formula:
On-axis Gain (G_on-axis) = (4 * pi * 0.62 * 1584.89) / ((3 * 10^8 / (4.2 * 10^9))^2)
Calculating the on-axis gain:
On-axis Gain (G_on-axis) ≈ 58.28 dB
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A cable exerts a constant upward tension of magnitude $1.15 \times 10^4 \mathrm{~N}$ on a $1.13 \times 10^3 \mathrm{~kg}$ elevator as it rises through a vertical distance of $62.0 \mathrm{~m}$.
(a) Find the work done by the tension force on the elevator
(b) Find the work done by gravity on the elevator
Work done by tension force is 4.80×10⁶ J and the work done by gravity on the elevator is 6.96×10⁵ J.
A. The work done by the tension force on the elevator can be determined by calculating the change in kinetic energy of the elevator.
This is because the tension force is in the same direction as the displacement of the elevator, so it does positive work on the elevator.
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
Thus, we can write:
Work done by tension force = Change in kinetic energy of elevator
The initial kinetic energy of the elevator is zero, since it starts from rest.
The final kinetic energy of the elevator can be calculated using the equation:
K.E. = 1/2 mv^2
where m is the mass of the elevator and v is its final velocity.
We can use the equation of motion:
v^2 = u^2 + 2as
where u is the initial velocity (zero),
a is the acceleration, and
s is the displacement.
We can solve for a to get:
a = v^2/2s
Substituting the given values, we get:
a = (2(62)/(1.13×10³)) m/s²
a = 0.979 m/s²
Substituting this value of acceleration into the equation of motion,
we get:
v^2 = 2as
v^2 = 2(0.979)(62)
v = 27.8 m/s
Substituting the values of mass and final velocity into the equation for kinetic energy,
we get:
K.E. = 1/2(1.13×10³)(27.8)
K.E. = 4.80×10⁶ J
Thus, the work done by the tension force is:
Work done by tension force = K.E. - 0
Work done by tension force = 4.80×10⁶ J
B. The work done by gravity on the elevator can be determined using the equation:
Work done by gravity = mgh
where m is the mass of the elevator,
g is the acceleration due to gravity, and
h is the vertical distance through which the elevator rises.
Substituting the given values, we get:
Work done by gravity = (1.13×10³)(9.81)(62.0)
Work done by gravity = 6.96×10⁵ J
Thus, the work done by gravity on the elevator is 6.96×10⁵ J.
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