The net force applied on the walls of the container, due to pressure and immersion under water, is approximately 21709.15 N.
Given, Area of the container = 160 cm², Pressure of the gas, P = 2 bars, Depth of immersion of the container under water, h = 54 m, Acceleration due to gravity, g = 9.8 m/s². We are to calculate the net force applied on the walls of the container. Let us consider a horizontal section of the container at a depth of h under water. The area of this section is the same as the area of the container. The force acting on this section due to atmospheric pressure is given by, F₁ = P × A, where A is the area of the section. The force acting on this section due to the water above it is given by, F₂ = ρghA, where ρ is the density of water, g is the acceleration due to gravity and h is the depth of the section below the water surface. The force acting on this section due to the gas inside is given by, F₃ = PA, where P is the pressure of the gas inside and A is the area of the section.Since the section is in equilibrium, the net force acting on it must be zero. Hence, we can write, F₁ + F₂ + F₃ = 0.P × A + ρghA + PA = 0.P + ρgh + P = 0.P = -ρgh/2. Since the pressure cannot be negative, we take the magnitude of P, i.e., P = ρgh/2. Substituting the given values, we get, P = (1000 kg/m³) × (9.8 m/s²) × (54 m) / 2P = 264600 PaThe force acting on the section due to the gas inside is given by, F₃ = PA = (264600 Pa) × (160 × 10⁻⁴ m²)F₃ = 42.336 N. The net force acting on the walls of the container is the force acting on each section of the container. Since the area of each section is the same, the net force acting on the walls of the container is given by,F = nF₃, where n is the number of sections in the container. Since the area of the container is given to be 160 cm², we can assume that it is a cube with sides of length 4.48 cm (approximately). Let us divide each side into 8 equal parts. This gives us a total of 8³ = 512 sections in the container. Hence, the net force acting on the walls of the container is, F = 512 × 42.336 N ≈ 21709.15 N (rounded off to three significant digits). Therefore, the net force applied on the walls of the container is approximately 21709.15 N.For more questions on net force
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A 1.89 m tall woman stands 1.05 m from a lens with focal length 78.8 cm. How tall (in m ) is her image formed by the lens? Be sure to include the sign to indicate orientation!
The height of the image formed by the lens is 4.0487 m.
A 1.89 m tall woman stands 1.05 m from a lens with a focal length of 78.8 cm. We can determine the height of her image formed by the lens using the following steps:
Given Data:
Height of Woman, h = 1.89 m
Distance of Woman from Lens, u = 1.05 m
Focal Length of Lens, f = 78.8 cm = 0.788 m
Formula Used:
Magnification produced by the lens, m = h' / h, where h' is the height of the image produced by the lens
Magnification produced by the lens, m = - v / u, where v is the distance of the image from the lens
Solution:
We first need to convert the focal length of the lens from centimeters to meters:
Focal length, f = 78.8 cm = 0.788 m
The height of the image formed by the lens is given by:
Magnification produced by the lens, m = h' / h - v / u
Here, h = 1.89 m, u = 1.05 m, and f = 0.788 m:
m = f / (f - u)
On substituting the given values:
m = 0.788 / (0.788 - 1.05)
m = -2.1382
The magnification is negative, indicating that the image is inverted. Hence, the height of the image formed by the lens is given by:
Magnification produced by lens, m = - v / u = -h' / h
Therefore,
-2.1382 = -h' / 1.89
⇒ h' = 2.1382 × 1.89
h' = 4.0487 m
The height of the image formed by the lens is 4.0487 m, and its orientation is inverted or upside down.
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A girl stands on a scale in an elevator, In which if the following conditions would the scale read the LARGEST welght (i.e., largest normal force)? When the elevator is accelerating upward. When the elevator is accelerating downward. When the elevator is stationary. When the elevator is moving upward at a constant velocity. When the elevator is moving downward at a constant velocity. Vowng Smed Work Revert Saloastaiesoanse Which of the following is true? The kinetic friction force is greater than the maximum static friction force. The static friction force on an object resting on an inclined plane is the same regardless of the angle of incline, as long as the block remains at re The kinetic friction force on an object depends on its speed. The force of friction is in the direction opposite to an object's motion or potential motion.
The normal force experienced by the girl on the scale is equal to the force of gravity acting on her.
It can be calculated by using the equation: Fnorm = mg where m is the mass of the girl, g is the acceleration due to gravity, and F norm is the normal force on the girl. The normal force is what is read by the scale. Now, let's examine the different conditions: When the elevator is accelerating upward :In this situation, the scale would read a larger weight than the actual weight of the girl. When the elevator is moving upward at a constant velocity :In this situation, the scale would read the actual weight of the girl.
This is because the force of gravity acting on the girl is equal to the force of the elevator moving upward, which results in a net force acting on the girl that is equal to the force of gravity. When the elevator is moving downward at a constant velocity:In this situation, the scale would read the actual weight of the girl. This is because the force of gravity acting on the girl is equal to the force of the elevator moving downward, which results in a net force acting on the girl that is equal to the force of gravity.
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During a baseball game, a baseball is struck, at ground level, by a batter. The ball leaves the baseball bat with an initial speed v0=25.7m/sv0=25.7m/s at an angle θ=32.5∘θ=32.5∘ above the horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Ignore air resistance throughout this problem.
question: Calculate the horizontal distance, xmax, in meters, that the ball has traveled when it returns to ground level.
Answer:
The horizontal distance (xmax) that the ball has traveled when it returns to ground level is approximately 62.45 meters.
Explanation:
To calculate the horizontal distance (xmax) that the ball has traveled when it returns to ground level, we can use the equations of projectile motion.
First, we need to determine the time it takes for the ball to reach its maximum height.
The time of flight is given by the equation: t = (2 * v0 * sin(θ)) / g
Where: v0 = initial speed of the ball (25.7 m/s)
θ = angle above the horizontal (32.5°)
g = acceleration due to gravity (9.8 m/s^2)
Plugging in the values, we get: t = (2 * 25.7 * sin(32.5°)) / 9.8
t ≈ 2.92 seconds
Next, we can use the time of flight to find the horizontal distance traveled. The horizontal distance is given by the equation:
xmax = v0 * cos(θ) * t
Plugging in the values, we get: xmax = 25.7 * cos(32.5°) * 2.92
xmax ≈ 62.45 meters
Therefore, the horizontal distance (xmax) that the ball has traveled when it returns to ground level is approximately 62.45 meters.
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A.ball.is.thrown directly downward with an initial speed of 8.60 m/s, from a height of 29.8 m. After what time interval does it strike the ground? Your Fesporse differs from the correct answer by more than 10%. Double check your calculations. 5
The ball will strike the ground after approximately 1.93 seconds.
To determine the time interval it takes for the ball to strike the ground, we can use the equations of motion.
Since the ball is thrown directly downward, its initial velocity (u) is -8.60 m/s (negative because it is downward), acceleration (a) due to gravity is -9.8 m/s² (negative due to downward motion), and the initial height (h) is 29.8 m. We can use the equation h = ut + (1/2)at² to find the time (t).
Rearranging the equation gives t = (-u ± √(u² - 2ah))/a. Substituting the given values, we find t ≈ 1.93 s.
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At some time during the motion of a particle, the acceleration vector (its "tail" is
attached to the "velocity vector’s tip") makes an angle of 45° relative to the direction of the
velocity vector.
(a) In the "next instant of time" (Δis very small), the particle will move:
A. In a curved path (not a straight line) with constant speed.
B. In a straight line with decreasing speed.
C. In a curved path (not a straight line) with increasing speed.
(b) Sometime later during the motion of the particle, the acceleration vector makes an angle of
90° relative to the direction of the velocity vector. In the "next instant of time" (Δis very
small), the particle will move:
A. In a circular path with constant speed.
B. In a curved path (not a straight line) with decreasing speed.
C. In a straight line with increasing speed.
(c) After one minute (60 seconds) of motion from its start, the toy car battery "goes dead" (it
lost all of its electrical power), and the toy truck comes to a stop 10 seconds later because of
friction in the wheels.
How fast (speed) was the toy truck moving at the moment the battery lost its power?
A. 210 m/s
B. 0.11 m/s
C. 3.8 × 105 m/s
(d) What is the magnitude of the acceleration (deceleration) of the toy truck to bring it to a stop
after the battery "goes dead?" Assume that this acceleration (deceleration) is constant.
A. 0.34 m ⁄s2
B. 9.8 m ⁄s2
C. 21 m ⁄s2
(a) In the next instant of time, the particle will move in a curved path with increasing speed when the acceleration vector makes an angle of 45° relative to the velocity vector.
(b) In the next instant of time, the particle will move in a straight line with increasing speed when the acceleration vector makes an angle of 90° relative to the velocity vector.
(c) The speed of the truck, when the battery lost its power, was 0 m/s.
(d) The magnitude of the acceleration to bring the toy truck to a stop after the battery goes dead is 0 m/s².
(a) In the "next instant of time" (Δ is very small), the particle will move in a curved path (not a straight line) with increasing speed when the acceleration vector makes an angle of 45° relative to the direction of the velocity vector.
(b) Sometime later during the motion of the particle, the acceleration vector makes an angle of 90° relative to the direction of the velocity vector.
In the "next instant of time" (Δ is very small), the particle will move in a straight line with increasing speed when the acceleration vector makes an angle of 90° relative to the direction of the velocity vector.
(c) The truck's initial velocity, u = 0
Acceleration, a = -v / t (where v is the final velocity, t is the time taken)
The truck comes to rest after 10 seconds since the battery goes dead, i.e., t = 10 s
The final velocity of the truck, v = 0
The acceleration of the truck, a = -v / t = 0 / 10 = 0 m/s²
Thus, the speed of the truck when the battery lost its power was 0 m/s.
(d) The initial velocity, u = 0
The final velocity, v = 0
The time taken, t = 10 s
The acceleration of the truck,
a = (v - u) / t = (0 - 0) / 10 = 0 m/s²
Thus, the magnitude of the acceleration (deceleration) of the toy truck to bring it to a stop after the battery "goes dead" is 0 m/s².
Answer: B. 9.8 m ⁄s².
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A baseball thrown from the outfield is released from shoulder height at an initial velocity of 29.4 m/s and an initial angle of 30.0
∘
with respect to the horizontal. The ball follows its trajectory for a total of 3.00 s before being caught by another player at an equal shoulder-height level. For this problem, air resistance is negligible. Find the range of the ball. In other words, what is Δx for the ball? Hint: find the components of the initial velocity vector, and use v
0x
in an appropriate kinematic equation. Also, you already have the time of flight!
The range of the baseball is 76.5 meters.
find the range of the baseball, we need to calculate the horizontal distance traveled by the ball, which is represented by Δx.
Initial velocity of the baseball, v₀ = 29.4 m/s,
Initial angle, θ = 30.0 degrees,
Time of flight, t = 3.00 s.
We can start by calculating the horizontal component of the initial velocity, v₀x, using the formula:
v₀x = v₀ * cos(θ).
Substituting the given values:
v₀x = 29.4 m/s * cos(30.0 degrees).
Calculating v₀x:
v₀x ≈ 25.5 m/s.
we can use the formula for displacement in the horizontal direction to find the range (Δx):
Δx = v₀x * t.
Substituting the values:
Δx = 25.5 m/s * 3.00 s.
Calculating Δx:
Δx = 76.5 m.
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A police radar unit transmits microwaves of frequency 3.00×10
10
Hz and their speed in air is 3.00×10
8
m/s. Suppose a car is receding from the stationary police car at a speed of 170 km/h. 1. [15\%] What is the frequency difference between the original transmitted signal and the signal received after it bounces on the receding car (the reflected signal)? [Hint: the reflected signal is Doppler-shifted twice!] 2. [5\%] Suppose the police car is now moving at a speed of 80 km/h in the same direction as the other vehicle. What is the difference in frequency between the transmitted and reflected signals?
1. The frequency difference between the original transmitted signal and the signal received after it bounces on the receding car (the reflected signal) is 2.99 x 10¹⁰ Hz.
2. The difference in frequency between the transmitted and reflected signals is -3.00 x 10¹⁰ Hz.
The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the source of the waves. Doppler effect is used by police radar units to determine the speed of moving vehicles.
The frequency of the radar signal is slightly altered due to the motion of the target vehicle.2. The formula for frequency shift due to Doppler effect is given as:
f' = f(v + vr) / (v - vs)
where, f' is the frequency of the reflected signal,
f is the frequency of the transmitted signal,
v is the speed of the waves in air,
vr is the speed of the target vehicle (negative for receding), and
vs is the speed of the stationary police car.
1.
Given that, the frequency of the transmitted microwaves, f = 3.00 x 10¹⁰ Hz
The speed of the waves in air, v = 3.00 x 10⁸ m/s
The speed of the target vehicle, vr = -47.22 m/s (-170 km/h to m/s)
The Doppler shift occurs twice, once when the signal hits the car and once when it reflects back.
Therefore, the frequency shift will be twice.
Using the Doppler effect formula, we can find the frequency of the reflected signal as:
f' = 2f(v + vr) / (v + vs) = 2f(v + vr) / v
since vs = 0f' = 2 × 3.00 x 10¹⁰ × (3.00 x 10⁸ - 47.22) / 3.00 x 10⁸ = 5.07 x 10⁸ Hz
The frequency difference between the transmitted and reflected signals is:
Δf = f' - f = 5.07 x 10⁸ - 3.00 x 10¹⁰ = -2.99 x 10¹⁰ Hz
Therefore, the frequency difference between the original transmitted signal and the signal received after it bounces on the receding car (the reflected signal) is 2.99 x 10¹⁰ Hz.
2.
Given that, the speed of the police car, vs = 22.22 m/s (80 km/h to m/s)
In this case, the police car is moving in the same direction as the target vehicle, so the relative speed between the target vehicle and the radar gun is:
vr = vr1 - vr2 = -47.22 - 22.22 = -69.44 m/s
Using the Doppler effect formula:
f' = f(v + vr) / (v - vs)f' = 3.00 x 10¹⁰ × (3.00 x 10⁸ - 69.44) / (3.00 x 10⁸ - 22.22) = 2.98 x 10⁸ Hz
The difference in frequency between the transmitted and reflected signals is:
Δf = f' - f = 2.98 x 10⁸ - 3.00 x 10¹⁰ = -3.00 x 10¹⁰ Hz
Therefore, the difference in frequency between the transmitted and reflected signals is -3.00 x 10¹⁰ Hz.
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An asteroid with mass X kilograms (kg) is moving eastward at Y meters per second (m/s) collides with another asteroid, also with mass Xkg, which is moving northward at Zm/s. They stick together and keep moving through space. Draw a diagram of this interaction, labeling all the relevant objects, quantities, and velocity vectors before and after the collision. [c] What is the velocity (magnitude and direction) of the two asteroids after they collide? Your answer should be significant to two digits.
The velocity (magnitude and direction) of the two asteroids after the collision depends on their initial velocities and masses and can be calculated using the principles of conservation of momentum.
Before the collision, you have two asteroids: one moving eastward with a velocity vector labeled as [tex]\(\vec{V_1}\)[/tex] and another moving northward with a velocity vector labeled as [tex]\(\vec{V_2}\)[/tex]. Both asteroids have the same mass, labeled as X.
After the collision, the two asteroids stick together and move as a single object. The velocity vector of the combined asteroids after the collision is labeled as [tex]\(\vec{V_{\text{final}}}\)[/tex].
To determine the magnitude and direction of the velocity of the combined asteroids after the collision, you would need to calculate the resultant velocity vector by considering the conservation of momentum. The magnitude of [tex]\(\vec{V_{\text{final}}}\)[/tex] can be found by calculating the vector sum of [tex]\(\vec{V_1}\) and \(\vec{V_2}\)[/tex], and the direction can be determined by the angle between the resultant vector and a reference axis.
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A proton accelerates from rest in a uniform electric field of 670 N/C. At one later moment, its speed is 1.10Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. m/s
2
(b) Over what time interval does the proton reach this speed? s (c) How far does it move in this time interval? m (d) What is its kinetic energy at the end of this interval?
The acceleration of the proton is 6.43 × 10¹⁴ m/s².
The time interval required to achieve the final velocity of the proton is 1.7 × 10⁻⁹ s.
The distance traveled by the proton in this time interval is 3.57 × 10⁻⁷ m.
The kinetic energy of the proton at the end of this interval is 1.01 × 10⁻¹¹ J.
Given that a proton accelerates from rest in a uniform electric field of 670 N/C. At one later moment, its speed is 1.10 Mm/s.
(a)
The electric force acting on the proton is, F = qE
Where, q = charge on the proton = 1.6 × 10⁻¹⁹ C.E = electric field = 670 N/CF = ma
Where, a = acceleration of the proton
m = mass of the proton = 1.67 × 10⁻²⁷ kg.
Putting these values in the above equation, we get
a = F/m = (qE)/m = (1.6 × 10⁻¹⁹ × 670)/1.67 × 10⁻²⁷= 6.43 × 10¹⁴ m/s².
(b)
The acceleration of the proton is a = 6.43 × 10¹⁴ m/s² and its initial velocity is u = 0.
The final velocity of the proton is v = 1.10 Mm/s = 1.1 × 10⁶ m/s.
Let t be the time interval required to achieve the final velocity of the proton, then
v = u + at1.1 × 10⁶ = 0 + (6.43 × 10¹⁴)tt = v/a = (1.1 × 10⁶)/(6.43 × 10¹⁴)= 1.7 × 10⁻⁹ s.
(c)
The distance traveled by the proton in time t is given by,
s = ut + (1/2)at²= 0 + (1/2)at²= (1/2) × (6.43 × 10¹⁴) × (1.7 × 10⁻⁹)²= 3.57 × 10⁻⁷ m.
(d)
The kinetic energy of the proton at the end of this interval is given by,
K.E = (1/2)mv²
Where,v = 1.1 × 10⁶ m/sm = 1.67 × 10⁻²⁷ kg.
Putting these values in the above equation, we get
K.E = (1/2) × 1.67 × 10⁻²⁷ × (1.1 × 10⁶)²= 1.01 × 10⁻¹¹ J
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A football is kicked at 25.0 m/s at an angle of 33.0
∘
above the horizontal. Assume it starts 1.00 m above ground level. (Neglect any effects due to air resistance.) Describe the velocity vector of the ball in terms of its γ - and y-components, v
x
and v
y
, respectively, when it is 4.00 m above ground level and still moving upward. You attach a meter stick to an oak tree, such that the top of the meter stick is 1.27 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.181 seconds to pass the length of the meter stick, how high h
0
above the ground was the acorn before it fell, assuming that the acorn did not run into any branches or leaves on the way down? While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.95 m/s. The stone subsequently falls to the ground, which is 15.9 m below the point where the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use g=9.81 m/s
2
for the acceleration due to gravity. impact speed: How much time is the stone in the air? elapsed time:
1.The impact speed of the stone is approximately 19.6 m/s.
At 4.00 m above ground level and still moving upward, the velocity vector of the ball in terms of its γ- and y-components, v x and v y , respectively is:
v x = 25.0 m/s cos (33.0°) ≈ 20.98 m/sv y = 25.0 m/s sin (33.0°) ≈ 13.51 m/s2.
For the acorn to pass the length of the meter stick (1.27 meters) in 0.181 seconds, the acorn was this high h0 above the ground before it fell: h0 = 0.5 gt2
= 0.5 (9.81 m/s2)(0.181 s)2
≈ 0.157 m3.
To find the time the stone is in the air, we can use the fact that the time taken for the stone to reach its maximum height is the same as the time it takes for the stone to fall from its maximum height to the ground.
Using the vertical motion of the stone, we have:
v = v0 + gt, where v0 = 6.95 m/s and g = 9.81 m/s2.
When the stone reaches its maximum height, its final velocity is zero,
so we have:
0 = v0 - gt, or t = v0 / g = 6.95 m/s / 9.81 m/s2 = 0.707 s.
The total time the stone is in the air is twice this time: t total = 2t = 1.41 s.4.
To find the impact speed of the stone, we can use the equation:
v2 = v02 + 2gh,
where h = 15.9 m is the height of the building and v0 = 6.95 m/s is the initial velocity of the stone.
Rearranging this equation to solve for v,
we have:
v = sqrt(v02 + 2gh) = sqrt((6.95 m/s)2 + 2(9.81 m/s2)(15.9 m)) ≈ 19.6 m/s.,
the impact speed of the stone is approximately 19.6 m/s.
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You are free-floating in space and feel no acceleration. You notice a clock sitting on a planet surface. What do you notice about the clock compared to your watch? Would you expect the same thing for a clock in an accelerating spaceship and why?
The time dilation effect will be less significant for an accelerating spaceship than on the planet's surface due to the relatively lower magnitude of acceleration compared to the gravitational force of a planet.
If an individual is free-floating in space and feels no acceleration and notices a clock sitting on a planet surface, they will notice that the clock is ticking slower compared to their watch. The ticking of the clock will be slower because of the gravitational force which causes time dilation, and the closer to the planet surface, the slower time will move. If the individual is in an accelerating spaceship, the clock will appear to tick faster than their watch. This occurs due to the acceleration of the spaceship that causes time dilation.
According to the theory of general relativity, acceleration and gravity are equivalent. As a result, any acceleration in an accelerating spaceship can cause time dilation. Suppose an individual on an accelerating spaceship looks at a clock sitting on the surface. In that case, they will notice the same time dilation effect as in the case of a clock sitting on a planet surface. Thus, the clock will tick slower or faster based on the direction of the acceleration. If the acceleration is towards the planet, the clock will tick slower, but if it is in the opposite direction, it will tick faster.
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You throw a rock from the upper edge of a 95.0−m vertical dam with a speed of 29.0 m/s at 65.0
∘
above the horizon. 1) How long after throwing the rock wilt you see it hit the water? (Neglect any effects due to air resistance.) (Express your answer to three significant figures.) 2) How far from the bottom of the dam does the rock tand? (Neglect any effects due to air resistance.) (Express your answer to three significant figures
The horizontal distance from the bottom of the dam where the rock hit the water is 149.06 m (approx).Answer: 149.06 m. Initial velocity u = 29.0 m/s, Initial angle θ = 65°, Initial position, h = 95.0 m, Acceleration due to gravity, g = 9.81 m/s²
Let's first find the time taken by the rock to hit the water
.From the above motion, we know that h = ut sin θ - ½ g t²Where,h = initial height u = initial velocity θ = initial angle above the horizon (in radian)g = acceleration due to gravity t = time taken to hit the ground.
Here, θ = 65°= 65° × π/180= 1.1345 radh = 95.0 mu = 29.0 m/st = ?.
By substituting all the given values in above equation, we get95.0 = 29.0 × sin 1.1345 × t - ½ × 9.81 × t²95.0 = 29.0 × 0.940 × t - 4.905 t²95.0 = 27.26 t - 4.905 t²4.905 t² - 27.26 t + 95.0 = 0.
This is a quadratic equation of the form:ax² + bx + c = 0Where,a = 4.905b = -27.26c = 95.0.
Using the quadratic formula, we get,t = [27.26 ± √(27.26² - 4 × 4.905 × 95.0)] / 2 × 4.905= [27.26 ± 34.9] / 9.81= 6.435 s or 3.088 s [Neglecting the negative value].
Therefore, the time taken by the rock to hit the water is 6.44 s (approx).Answer: 6.44 s2.
Now, let's find the horizontal distance covered by the rock when it hit the water.
The horizontal distance covered by the rock is given by,x = ut cos θ × t.
Here,x = horizontal distance u = initial velocityθ = initial angle above the horizon (in radian)t = time taken to hit the ground.
By substituting the values of u, θ, and t in the above equation, we get,x = 29.0 × cos 1.1345 × 6.44= 149.06 m.
Therefore, the horizontal distance from the bottom of the dam where the rock hit the water is 149.06 m (approx).Answer: 149.06 m
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A spherical, non-conducting shell of inner radius r
1
=7 cm and outer radius r
2
=18 cm carries a total charge Q= uniformly throughout the volume of the shell. What is the magnitude of the electric field at a distance r=11 cm the shell? (k=1/4πϵ
0
=8.99×10
9
N⋅m
2
/C
2
)
The magnitude of the electric field at a distance r = 11 cm from the shell
To find the magnitude of the electric field at a distance r = 11 cm from the shell, we can use Gauss's law and the concept of electric flux.
Inner radius of the shell, r1 = 7 cm = 0.07 m
Outer radius of the shell, r2 = 18 cm = 0.18 m
Total charge of the shell, Q = uniformly distributed
Distance from the shell, r = 11 cm = 0.11 m
Electric constant, k = 1 / (4πε0) = 8.99 × 10^9 N·m^2/C^2
The electric field at a point outside the shell is equivalent to the electric field created by a point charge located at the center of the shell, considering the charge inside the shell.
Since the charge is uniformly distributed throughout the volume of the shell, the electric field inside the shell is zero.
To calculate the electric field at a distance r = 11 cm from the shell, we consider a Gaussian surface in the form of a sphere centered at the center of the shell with radius r = 11 cm.
The electric flux through this Gaussian surface is given by:
Φ = Q_in / ε0
Since the electric field inside the shell is zero, all the charge Q is enclosed within the Gaussian surface. Therefore, Q_in = Q.
Using Gauss's law:
Φ = E * A = Q / ε0
The electric field E at a distance r from the shell is uniform and has the same magnitude at all points on the Gaussian surface. The area A of the Gaussian surface is 4πr^2.
Substituting the values into the equation:
E * 4πr^2 = Q / ε0
Rearranging the equation to solve for E:
E = (Q / (4πr^2 * ε0))
Substituting the given values:
E = (Q / (4π * (0.11 m)^2 * 8.99 × 10^9 N·m^2/C^2))
Therefore, the magnitude of the electric field at a distance r = 11 cm from the shell is given by the equation above.
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. How far apart are two conducting plates that have an electric field strength of 4.50×10
3
V/m between them, if their potential difference is 15.0kV ?
The distance between the two conducting plates is approximately 3.33 meters.
To find the distance between the two conducting plates, we can use the formula:
Electric field strength (E) = Voltage (V) / Distance (d)
Given:
Electric field strength (E) = 4.50 × 10³ V/mPotential difference (V) = 15.0 kV = 15.0 × 10³ VWe can rearrange the formula to solve for the distance (d):
d = V / E
Substituting the given values:
d = (15.0 × 10³ V) / (4.50 × 10³ V/m)
Simplifying the expression:
d = 3.33 m
Therefore, the distance between the two conducting plates is approximately 3.33 meters.
The correct format of the question should be:
How far apart are two conducting plates that have an electric field strength of 4.50×10³V/m between them, if their potential difference is 15.0kV ?
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You are on a train traveling east at speed of 12 m/s with respect to the ground.
1)
If you walk east toward the front of the train, with a speed of 2.4 m/s with respect to the train, what is your velocity with respect to the ground?
m/s east
2)
If you walk west toward the back of the train, with a speed of 3.6 m/s with respect to the train, what is your velocity with respect to the ground?
m/s east
3)
Your friend is sitting on another train traveling west at 15 m/s. As you walk toward the back of your train at 3.6 m/s, what is your velocity with respect to your friend?
m/s east
Velocity of the train with respect to the ground = 12 m/s (east)
Velocity of walking with respect to the train = 2.4 m/s (east)
Adding these velocities together, we get:
Velocity with respect to the ground = 12 m/s + 2.4 m/s = 14.4 m/s (east)
Therefore, your velocity with respect to the ground when walking east toward the front of the train is 14.4 m/s east.
To determine your velocity with respect to the ground when walking west toward the back of the train, we can subtract your velocity with respect to the train from the velocity of the train with respect to the ground.Velocity of the train with respect to the ground = 12 m/s (east)
Velocity of walking with respect to the train = 3.6 m/s (west)
Subtracting these velocities, we get:
Velocity with respect to the ground = 12 m/s - 3.6 m/s = 8.4 m/s (east)
Therefore, your velocity with respect to the ground when walking west toward the back of the train is 8.4 m/s east.
To determine your velocity with respect to your friend, we need to consider their relative motion. Since your friend is on another train traveling west, their velocity with respect to the ground is in the opposite direction to your train's velocity.Velocity of the train with respect to the ground = 12 m/s (east)
Velocity of your friend's train with respect to the ground = 15 m/s (west)
Velocity of walking with respect to the train = 3.6 m/s (west)
To find your velocity with respect to your friend, we subtract the velocity of their train from the sum of your velocity with respect to the train and the velocity of your train with respect to the ground:
Velocity with respect to your friend = Velocity of walking with respect to the train + Velocity of the train with respect to the ground - Velocity of your friend's train with respect to the ground
Velocity with respect to your friend = 3.6 m/s + 12 m/s - 15 m/s = 0.6 m/s (east)
Therefore, your velocity with respect to your friend is 0.6 m/s east as you walk toward the back of your train.
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What is the weight of a 67.4 kg astronaut on the Moon where the value of ' g ' is only 78\% the strength of ' g ' on the Earth? Express your answer in N (i.e. Newtons). Use two digits of precision for your answer. Your Answer: Answer Hide hint for Question 10 Remember that at the surface of a planet , the weight is calculated from Wlocal =m×glocal
The weight of a 67.4 kg astronaut on the Moon where the value of ' g ' is only 78% the strength of ' g ' on the Earth is 67.4 kg.
The weight of an astronaut on the Moon can be calculated using the formula:
Weight ([tex]W_{local[/tex]) = mass (m) * local gravity ([tex]g_{local[/tex])
Given:
Mass of the astronaut (m) = 67.4 kg
Local gravity on the Moon ([tex]g_{local[/tex]) = 78% of the strength of the Earth's gravity (g)
First, let's calculate the local gravity on the Moon:
[tex]g_{local[/tex]= 78% * g
To find the weight of the astronaut on the Moon, we can substitute the values into the formula:
[tex]W_{local[/tex]= m * [tex]g_{local[/tex]
[tex]W_{local[/tex]= 67.4 kg * (78% * g)
[tex]W_{local[/tex]= 67.4 kg * (0.78 * g)
To calculate the weight in Newtons, we need to know the value of the Earth's gravity (g). The average value of the Earth's gravity is approximately 9.8 m/s².
Now, let's calculate the weight on the Moon:
W_local = 67.4 kg * (0.78 * 9.8 m/s²)
≈ 503.28 N
Therefore, the weight of a 67.4 kg astronaut on the Moon, where the local gravity is 78% the strength of Earth's gravity, is approximately 503.28 N (Newtons).
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A 343 kg mechanized loader places 4.58 kg of scientific equipment into the bay of a research vessel. a) How much work does the loader do to raise the mass of equipment 2.30 m to the bay landing? b) The transfer of the equipment to the bay landing does not go well and the equipment falls from rest on the edge of the landing to the floor below. What is the kinetic energy of the equipment when it hits the floor?
a) The work done is 99.32 J. b) The kinetic energy is 229.5 J when it hits the floor.
a) The work done by the loader in raising the mass of equipment to the bay landing can be calculated using the formula; Work = mgh where; m = 4.58 kg, g = 9.8 m/s², h = 2.3 m
Work = (4.58)(9.8)(2.3) = 99.32 J
Therefore, the loader does 99.32 J of work to raise the mass of equipment to the bay landing.
b) The potential energy of the equipment at the landing can be calculated using the formula; Potential energy = mgh where; m = 4.58 kg, g = 9.8 m/s², h = 2.3 m
Potential energy = (4.58)(9.8)(2.3) = 99.32 J
Since no energy is lost in the transfer of the equipment, the potential energy gained will be converted to kinetic energy as the equipment falls. Therefore, the kinetic energy of the equipment when it hits the floor can be calculated using the formula;
Kinetic energy = ½mv² where; m = 4.58 kg, v = velocity of equipment before it hits the floor. The velocity of the equipment before it hits the floor can be calculated by equating the kinetic energy of the equipment to the potential energy it had at the landing as given by the formula;
Potential energy = Kinetic energy
4.58 * 9.8 * 2.3 = ½(4.58)v²
v² = 100.198
v = √(100.198)
≈ 10.01 m/s
The kinetic energy of the equipment when it hits the floor is given by;
Kinetic energy = ½mv²
= ½(4.58)(10.01)²
≈ 229.5 J
Therefore, the kinetic energy of the equipment when it hits the floor is 229.5 J.
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Two point charges are located on the x-axis: a charge of +4.80nC at x=0 and an unknown charge q at x=0.500 m. No other charges are nearby. If the electric field is zero at the point x=1.00 m, what is q ? nC
To calculate the unknown charge q on x = 0.5 m on the x-axis given that there are two point charges present on the x-axis, we use the concept of Electric field formula.
Let's assume the unknown charge q to be at point P (x = 0.5 m) on the x-axis. The distance between the charges on the x-axis is 0.5 m from one charge and 0.5 m from another charge. The electric field due to the charge
where,q₁ = +4.80 nCr₁ = distance between charge
4.80 nC and point P = 0.5 m.
Substituting the values in equation (1),
we getE₁ = 8.99 x 10⁹ x (4.8 x 10⁻⁹) / (0.5)²E₁ = 1720.16 N/C
Similarly, the electric field due to the charge q is given by
E₂ = kq₂ / r₂² --- (2) where,q₂ = charge on point P = q,r₂ = distance
between point P and charge +4.80 nC = 0.5 m.
Substituting the values in equation (2),
we getE₂ = 8.99 x 10⁹ x q / (0.5)²E₂ = 17980.80q N/C
The total electric field at point P is given by the algebraic sum of E₁ and E₂. We know that the net electric field at
x = 1.00 m is zero, i.e., E = 0,E = E₁ + E₂ = 0∴ E₁ =
E₂andq = - E₁ x (0.5)² / (8.99 x 10⁹)q = - (1720.16) x
(0.5)² / (8.99 x 10⁹)q = - 4.80 nC , the unknown charge q is -4.80 nC.
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A planet is found orbiting an identical star to our sun. It orbits at a semi-major axis of 1.2 AU. How long does it take to complete an orbit?
The planet takes 1.438 years to complete one orbit around the star.
The time taken by a planet to complete one orbit around its star is known as the orbital period. The orbital period of a planet is dependent on its distance from the star, and Kepler's third law can be used to calculate it.
The semi-major axis of a planet's orbit is represented by a. The square of the orbital period (P) of a planet is proportional to the cube of its semi-major axis (a) and can be expressed as follows:
P² = a³ (Kepler's third law)
The semi-major axis of the planet's orbit is 1.2 AU.
The semi-major axis of the Earth's orbit is 1 AU. As a result, the planet's semi-major axis is slightly larger than the Earth's semi-major axis.
Using Kepler's third law,
P² = a³
P² = (1.2)³
P² = 1.44 x 1.44 x 1.44
P² = 2.0736
P = √2.0736
P = 1.438 years
Therefore, the time taken by the planet to complete an orbit around its star is 1.438 years.
The planet takes 1.438 years to complete one orbit around the star.
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Before working this problem, review Conceptual Example 14. A pellet gun is fired straight downward from the edge of a cliff that is 19.2 m above the ground, The pellet strikes the ground with a speed of 30.5 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?
If the pellet gun is fired straight upward from the edge of the cliff that is 19.2 m above the ground, the pellet would reach a height of approximately 49.2 m above the cliff edge before falling back down.
When the pellet gun is fired straight downward, the initial velocity is 30.5 m/s in the downward direction. We can use the kinematic equation for vertical motion to find the time it takes for the pellet to hit the ground. Since the initial velocity is in the negative direction, we can use the equation:
h = v_i * t + (1/2) * g * t²
where h is the height above the ground, v_i is the initial velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s²). Plugging in the known values:
0 = 30.5 * t - (1/2) * 9.8 * t²
Solving this quadratic equation, we find two solutions: t = 0 s (which represents the time at the top of the trajectory) and t ≈ 2.20 s (which represents the time it takes for the pellet to hit the ground).
Now, if the gun is fired straight upward, the initial velocity would be positive 30.5 m/s. Using the same kinematic equation, we can calculate the maximum height reached by the pellet. The final velocity at the maximum height is 0 m/s since the pellet momentarily stops before falling back down. Thus, the equation becomes:
0 = 30.5 * t - (1/2) * 9.8 * t²
Solving for t, we find t ≈ 3.13 s. Plugging this value back into the equation to find the maximum height (h) reached by the pellet, we have:
h = 30.5 * 3.13 - (1/2) * 9.8 * (3.13)²
h ≈ 49.2 m
Therefore, if the pellet gun is fired straight upward, the pellet would reach a height of approximately 49.2 m above the cliff edge before falling back down.
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What potential difference (take the absolute value) is needed to accelerate the following particles from rest to 0.09 c, where c is the speed of light? a) a particle with a charge 4 e and a mass 6u ? V b) a particle with a charge 94 e and a mass 220 u? V
The potential difference required, use the formula V = K.E. / Q. Substitute the values of charge and mass to calculate the potential differences for the given particles.
To determine the potential difference required to accelerate the particles, we can use the formula for the kinetic energy of a particle:
K.E. = (1/2)mv^2
where m is the mass of the particle and v is its velocity.
a) For a particle with a charge of 4e and a mass of 6u, we need to find the kinetic energy when its velocity is 0.09c. Since the speed of light, c, is approximately 3 x 10^8 m/s, the velocity of the particle is 0.09c = 0.09(3 x 10^8) = 2.7 x 10^7 m/s. Substituting the values into the formula, we get:
K.E. = (1/2)(6u)(2.7 x 10^7)^2
Now, we know that the potential difference, V, is equal to the change in electric potential energy (ΔPE) divided by the charge (Q). Since the particle is initially at rest, the initial electric potential energy is zero. Therefore, the potential difference needed is equal to the kinetic energy divided by the charge:
V = K.E. / (4e)
b) For a particle with a charge of 94e and a mass of 220u, we follow the same steps as above, but substitute the values accordingly:
V = K.E. / (94e)
By calculating the respective values for V using the given formulas and the provided charge and mass values, we can determine the potential differences required for the given particles to accelerate to 0.09c.
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For the open loop transfer function: σ(s)H(s)=
(x+1)(0.1x+1)
1
. Draw the magnitude and phase Bode plots of the the frequency response.
To draw the magnitude and phase Bode plots of the frequency response, we need to analyze the given open loop transfer function. The transfer function is represented as σ(s)H(s) = (x+1)(0.1x+1)/1.
1. First, let's simplify the transfer function:
σ(s)H(s) = (x+1)(0.1x+1)/1
= 0.1x^2 + 1.1x + 1
2. To create the Bode plot, we need to rewrite the transfer function in terms of ω, where ω is the angular frequency:
σ(jω)H(jω) = 0.1(jω)^2 + 1.1(jω) + 1
3. To plot the magnitude Bode plot, we take the absolute value of the transfer function and convert it to decibels:
|σ(jω)H(jω)| = √(0.1^2ω^4 + 1.1^2ω^2 + 1^2)
To plot this, we need to take the logarithm of the magnitude and multiply by 20 to get the result in decibels.
4. To plot the phase Bode plot, we calculate the phase angle of the transfer function:
∠(σ(jω)H(jω)) = atan2(Imaginary part, Real part) where Imaginary part and Real part are obtained from the transfer function.
By following these steps, we can plot the magnitude and phase Bode plots of the frequency response. Remember that the magnitude plot shows how the amplitude of the output changes with frequency, while the phase plot indicates the phase shift between the input and output signals at different frequencies.
Please note that the specific values of x are not given in the question, so we cannot provide an exact plot. However, the steps outlined above can be applied to any open loop transfer function to create the Bode plots.
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The surface of a 2 m long flat plate is maintained at 50°C. Water at a temperature of 10°C and a velocity of 0.6 m/s flows over the surface. Calculate the heat transfer per unit width of the plate. What would be the heat transfer per unit width of the plate if a wire was placed near the leading edge of the plate to induce turbulance over its entire length?
According to the question the heat transfer per unit width of the plate is 2874.4 W/m.
To calculate the heat transfer per unit width of the plate, we can use the convective heat transfer equation:
[tex]\[ Q = h \cdot A \cdot \Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat transfer per unit width of the plate (in watts per meter, W/m),
- [tex]\( h \)[/tex] is the convective heat transfer coefficient (in watts per square meter per Kelvin, W/(m²·K)),
- [tex]\( A \)[/tex] is the surface area of the plate (in square meters, m²), and
- [tex]\( \Delta T \)[/tex] is the temperature difference between the surface of the plate and the fluid (in Kelvin, K).
Given:
- Length of the plate, [tex]\( L = 2 \, \text{m} \)[/tex]
- Temperature of the plate surface, [tex]\( T_{\text{plate}} = 50 \, \text{°C} = 323.15 \, \text{K} \)[/tex]
- Temperature of the fluid, [tex]\( T_{\text{fluid}} = 10 \, \text{°C} = 283.15 \, \text{K} \)[/tex]
- Fluid velocity, [tex]\( V = 0.6 \, \text{m/s} \)[/tex]
First, let's calculate the convective heat transfer coefficient, [tex]\( h \)[/tex], using the Dittus-Boelter equation for forced convection over a flat plate:
[tex]\[ h = 0.023 \cdot \left( \frac{{\rho \cdot V \cdot c_p}}{{\mu}} \right)^{0.8} \cdot \left( \frac{{k}}{{D_h}} \right)^{0.4} \][/tex]
where:
- [tex]\( \rho \)[/tex] is the fluid density (in kg/m³)
- [tex]\( c_p \)[/tex] is the fluid specific heat capacity (in J/(kg·K))
- [tex]\( \mu \)[/tex] is the fluid dynamic viscosity (in kg/(m·s))
- [tex]\( k \)[/tex] is the fluid thermal conductivity (in W/(m·K))
- [tex]\( D_h \)[/tex] is the hydraulic diameter (in meters, m)
Since the fluid is water, we can use the following properties at 10°C (283.15 K):
- [tex]\( \rho = 998 \, \text{kg/m³} \)[/tex]
- [tex]\( c_p = 4186 \, \text{J/(kgK)} \)[/tex]
- [tex]\( \mu = 0.001 \, \text{kg/(ms)} \)[/tex]
- [tex]\( k = 0.606 \, \text{W/(mK)} \)[/tex]
The hydraulic diameter [tex]\( D_h \)[/tex] for a flat plate is equal to its thickness, which is not provided. We will assume a thickness of 0.01 m (10 mm).
Substituting the values into the Dittus-Boelter equation:
[tex]\[ h = 0.023 \cdot \left( \frac{{998 \cdot 0.6 \cdot 4186}}{{0.001}} \right)^{0.8} \cdot \left( \frac{{0.606}}{{0.01}} \right)^{0.4} \][/tex]
Simplifying:
[tex]\[ h = 35.86 \, \text{W/(m²·K)} \][/tex]
Next, we calculate the surface area of the plate. Since we have a flat plate with length [tex]\( L = 2 \)[/tex] m and width [tex]\( W = 1 \)[/tex] m (assuming a unit width), the surface area is [tex]\( A = L \times W = 2 \times 1 = 2 \) m².[/tex]
Now, we can calculate the temperature difference [tex]\( \Delta T = T_{\text{plate}} - T_{\text{fluid}} \):[/tex]
[tex]\[ \Delta T = 323.15 - 283.15 = 40 \, \text{K} \][/tex]
Finally, substituting the values into the convective heat transfer equation:
[tex]\[ Q = h \cdot A \cdot \Delta T = 35.86 \times 2 \times 40 = 2874.4 \, \text{W/m} \][/tex]
Therefore, the heat transfer per unit width of the plate is 2874.4 W/m.
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A proton is located at ⟨1,−8,4⟩×10
−10
m. If you want to calculate the electric field at the origin due to the proton, what is the
r
, relative position vector, you will need? (a) ⟨1,−8,4⟩×10
−10
m (b) ⟨−1,8,−4⟩×10
−10
m (c) 9×10
−10
m (d) −9×10
−10
m (e) ⟨−2,16,−8⟩×10
−10
m
The correct Option is B. the relative position vector, r, needed to calculate the electric field at the origin due to the proton is ⟨−1,8,−4⟩×10−10m.
The relative position vector, r, is the vector pointing from the point charge to the point where we want to find the electric field.
Since we want to find the electric field at the origin (0, 0, 0), r is the vector pointing from the charge to the origin.
In this case, the proton is located at ⟨1,−8,4⟩×10−10m.
Therefore, the vector pointing from the proton to the origin is: r = - ⟨1,−8,4⟩×10−10m
To explain why it is negative, note that the vector pointing from the origin to the proton is ⟨1,−8,4⟩×10−10m.
However, we want the vector pointing from the proton to the origin, so we need to flip the direction of this vector, which is achieved by multiplying it by -1.
Therefore, the relative position vector, r, needed to calculate the electric field at the origin due to the proton is: (b) ⟨−1,8,−4⟩×10−10m.
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The current in a single-loop circuit with one resistance R is 4.6 A. When an additional resistance of 2.4Ω is inserted in series with R, the current drops to 3.9 A. What is R ?
We are given that the current in a single-loop circuit with one resistance R is 4.6 A. When an additional resistance of 2.4Ω is inserted in series with R, the current drops to 3.9 A.
We have to determine the value of R. To solve this problem, we'll use Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points.
Ohm's Law equation is expressed as:
V = IR
Where V is voltage,
I is current,
and
R is resistance.
Now, when there is only one resistance R in the circuit, the current flowing through the circuit is given by:
V = IR ... (1)
We are given that the current through the circuit when there is only one resistance R is 4.6A, so we can write equation (1) as:
V = 4.6R ... (2)
Similarly, when a resistance of 2.4Ω is inserted in series with R, the current flowing through the circuit is given by:
V = I(R + 2.4) ... (3)
We are given that the current through the circuit when a resistance of 2.4Ω is inserted in series with R is 3.9A, so we can write equation (3) as:
V = 3.9(R + 2.4) ... (4)
Now, equating equations (2) and (4), we get:
4.6R = 3.9(R + 2.4)4.6R = 3.9R + 9.3640R - 741R = 936R = 936/0.7R = 1337.14 Ω
Therefore, the value of resistance R is 1337.14 Ω.
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The height of a powerful hydroelectric power plant is 245m. Falling water powers 10 hydrogenerators. One 640MW generator uses 358m^3 of water each second. The density of the water is 1000kg/m^3. Acceleration due to gravity is 10m/s^2.
Find coefficient of efficiency of one hydrogenerator. The distance between dam top and generator is 194m. (ats 92%)
The height of a powerful hydroelectric power plant is 245m. Falling water powers 10 hydrogenerators. One 640MW generator uses 358m^3 of water each second. The density of the water is 1000kg/m^3. Acceleration due to gravity is 10m/s^2.The coefficient of efficiency of one hydrogenerator is approximately 92.31% (rounded to 92% as stated).
To find the coefficient of efficiency of one hydrogenerator, we can use the following formula:
Efficiency = (Electrical Power Output / Hydraulic Power Input) × 100
First, let's calculate the hydraulic power input to the hydrogenerator. The hydraulic power input is given by:
Hydraulic Power Input = (Force exerted by water × Distance) / Time
The force exerted by the water can be calculated using the formula:
Force = Density ×Volume × Acceleration due to gravity
Here, the volume of water used per second by the generator is given as 358 m^3/s, and the density of water is 1000 kg/m^3.
Force = (1000 kg/m^3) × (358 m^3/s) × (10 m/s^2)
Force = 3,580,000 N
Next, let's calculate the hydraulic power input by multiplying the force with the distance the water falls:
Hydraulic Power Input = Force × Distance
Hydraulic Power Input = 3,580,000 N × 194 m
Hydraulic Power Input = 693,320,000 N·m/s (or Joules/second, which is equivalent to Watts)
Now, let's calculate the electrical power output of the hydrogenerator, which is given as 640 MW (megawatts). We need to convert it to watts:
Electrical Power Output = 640 MW × 10^6 W/MW
Electrical Power Output = 640,000,000 W
Finally, we can calculate the efficiency:
Efficiency = (Electrical Power Output / Hydraulic Power Input) * 100
Efficiency = (640,000,000 W / 693,320,000 W) * 100
Efficiency ≈ 92.31%
Therefore, the coefficient of efficiency of one hydrogenerator is approximately 92.31% (rounded to 92% as stated).
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Two electric charges, q1 = + 6.94 nC and q2 = + 7.66 nC, are located on the x-axis at x = 0.00 m and x = 1.00 m, respectively. What is the magnitude of the electric field at the point x = 0.416 m, y = 0.416 m?
Substitute the values of |q1|, |q2|, r1, and r2 into the above equations and calculate E1, E2, and |E_total| to find the magnitude of the electric field at the point (0.416 m, 0.416 m).
To find the magnitude of the electric field at the point (0.416 m, 0.416 m), we need to calculate the electric fields produced by each charge and then find the vector sum.
Using Coulomb's law, the electric field produced by q1 at the given point is:
E1 = k * |q1| / r1^2,
where k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), |q1| is the magnitude of charge q1 (6.94 nC), and r1 is the distance between q1 and the point (0.416 m, 0.416 m).
Similarly, the electric field produced by q2 at the given point is:
E2 = k * |q2| / r2^2,
where |q2| is the magnitude of charge q2 (7.66 nC), and r2 is the distance between q2 and the point (0.416 m, 0.416 m).
Finally, the magnitude of the total electric field at the given point is:
|E_total| = √(E1^2 + E2^2).
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A car moving in a straight line starts at x=0 at t= 0 . It passes the point x=25.0 m with a speed of 11.5 m/s at t=3.00 s. It passes the point x=400 Find the average velocity between t=3.00 s and t=20.0 s. m with a speed of 42.0 m/s at t=20.0 s. Part B Find the average acceleration between t=3.00 s and t=20.0 s
In order to determine the average velocity of a car moving in a straight line, the distance traveled and the elapsed time need to be considered.
The distance traveled by the car between the two points can be found by subtracting the initial position from the final position. Let the initial position be x1 and the final position be x2, then the distance traveled between the two positions is Δx = x2 – x1.
From the question, the car passed the point
x=25.0 m with a speed of 11.5 m/s at t=3.00 s,
and it passed the point
x=400 m with a speed of 42.0 m/s at t=20.0 s.
So, the distance traveled between the two positions is given by:
Δx = x2 – x1 = 400 m – 25.0 m = 375 m
The elapsed time is given as t2 – t1 = 20.0 s – 3.00 s = 17.0 s
The average velocity between
t=3.00 s and t=20.0 s is: Vav g = Δx/Δt = 375 m / 17.0 s = 22.1 m/s
Part B The average acceleration of the car between t=3.00 s and t=20.
0 s can be calculated using the formula:
aav g = Δv/Δtwhere Δv is the change in velocity and Δt is the time interval between the two points.
The change in velocity is given as:
v2 – v1 = 42.0 m/s – 11.5 m/s = 30.5 m/
t=3.00 s and t=20.0 s is: aavg = Δv/Δt = 30.5
The average acceleration of the car between
t=3.00 s and t=20.0 s is 1.79 m/s².
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A+2−μC charge is at rest in a magnetic field of 2 T pointing along the +x-axis. What is the force acting on this charge in the magnetic field? −4.00 micronewtons 4.00 micronewtons −2.00 micronewtons 2.00 microNewtons 0 microNewtons QUESTION 6 In a vacuum, red light has a wavelength of 700 nm and violet light has a wavelength of 400 nm. This means that in a vacuum, red light has higher frequency has lower frequency frequency will be energy dependent has higher frequency and moves faster than violet light. has lower frequency and moves faster than violet light.
charge is at rest in a magnetic field of 2 T pointing along the +x-axis.The formula to find the magnetic force on a charge is: Therefore, the correct answer is has lower frequency and moves faster than violet light.
Fmagnetic = q * v * B
Where, q = charge of the particle,
v = velocity of the particle and
B = magnetic field strength.
Given,A+2−μC charge is at rest in a magnetic field of 2 T pointing along the +x-axis.Since the charge is at rest, the velocity of the particle is zero. Hence, the force acting on this charge in the magnetic field will be zero. Thus, the correct answer is 0 microNewtons.In a vacuum, red light has a wavelength of 700 nm and violet light has a wavelength of 400 nm.
This means that in a vacuum, red light has higher frequency has lower frequency frequency will be energy dependent has higher frequency and moves faster than violet light. has lower frequency and moves faster than violet light.Red light has a lower frequency and moves faster than violet light in a vacuum. The speed of light is always the same in a vacuum regardless of the frequency of the light. The difference in wavelength between red and violet light is due to the difference in their frequency. Red light has a longer wavelength and a lower frequency than violet light. The frequency of red light is around 4.3×10¹⁴ Hz while the frequency of violet light is around 7.5×10¹⁴ Hz.
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Air at atmospheric pressure and 5°C flows at a speed of 10 m/s over a surface with a constant temperature of 45°C. At a particular point on the surface, the coefficient of friction is c is found to have a value of 0.01. Find the value of film temperature appropriate to this point and, using Reynolds' analogy (for Pr= 1), estimate the local value of heat flux. By how much would your estimate change if you took account of the fact that Pr#12 (Ans=25°C, 2400W / m², increase by approximately 25%)
we can see the percentage change in the heat flux estimation. In this case, the estimate would increase by approximately 25%.
To find the film temperature at the particular point on the surface, we can use the concept of the thermal boundary layer. The film temperature is defined as the average temperature of the fluid in the thermal boundary layer adjacent to the surface.
Given the surface temperature of 45°C and the air temperature of 5°C, we can use the formula:
Film temperature = (Surface temperature + Air temperature) / 2
Film temperature = (45°C + 5°C) / 2 = 25°C
Therefore, the film temperature at the particular point on the surface is 25°C.
Using Reynolds' analogy, which states that the local heat transfer coefficient (h) is proportional to the local mass transfer coefficient (k), we can estimate the local value of heat flux.
The local heat flux (q) can be estimated using the formula:
q = h * (Surface temperature - Air temperature)
However, in this case, we need to account for the fact that the Prandtl number (Pr) is given as 1, which means the fluid is air-like. For air-like fluids, the local heat flux is related to the local mass transfer coefficient (k) by the equation:
q = 0.023 * k * (Re^0.8) * (Pr^0.4) * (Surface temperature - Air temperature)
Given that the air is flowing over the surface at 10 m/s and the coefficient of friction (c) is 0.01, we can estimate the local heat flux by calculating the local mass transfer coefficient (k) using the equation:
k = c * ρ * V
where ρ is the air density and V is the velocity of the air.
Assuming standard atmospheric conditions and using the air density at 5°C and atmospheric pressure, we can calculate ρ.
Once we have the value of ρ, we can calculate the local mass transfer coefficient (k) and then use the above equation to estimate the local heat flux (q).
Now, if we consider that the Prandtl number (Pr) is actually 12 instead of 1, we need to modify the equation for the local heat flux:
q = 0.023 * k * (Re^0.8) * (Pr^0.4) * (Surface temperature - Air temperature)
with the new value of Pr (Pr = 12).
By substituting the new Pr value, we can estimate the new value of heat flux.
Comparing the two estimates, we can see the percentage change in the heat flux estimation. In this case, the estimate would increase by approximately 25%.
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