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laiz [17]
2 years ago
6

1. A 1000 kg equipment lift travels up a 200 m shaft in 45 seconds. Assuming the speed is constant, what is the

Physics
1 answer:
ANTONII [103]2 years ago
8 0

Answer:

43,555 W

Explanation:

Since the speed is constant, this means that the force applied to lift the equipment is equal to the weight of the equipment. So we can write:

F=mg

where

m = 1000 kg is the mass

g=9.8 m/s^2 is the acceleration due to gravity

So,

F=(1000)(9.8)=9800 N

Then the work done in lifting the equipment is:

W=Fd

where

d = 200 m is the displacement of the equpment

Substituting,

W=(9800)(200)=1.96\cdot 10^6  J

Finally, the power used to lift the equipment is the ratio between the work done and time taken:

P=\frac{W}{t}

where

W=1.96\cdot 10^6 J

t = 45 s is the time taken

Solving,

P=\frac{1.96\cdot 10^6}{45}=43,555 W

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A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of
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(1) The wavelength of the wave is 1.164 m.

(2) The  velocity of the wave is 23.7 m/s.

(3) The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

<h3> Wavelength of the wave</h3>

A general wave equation is given as;

y(x, t) = A sin(Kx - ωt)

<h3>Velocity of the wave</h3>

v = ω/K

From the given wave equation, we have,

y(x, t) = 0.048 sin(5.4x - 128t)

v = ω/K

where;

  • ω corresponds to 128
  • k corresponds to 5.4

v = 128/5.4

v = 23.7 m/s

<h3>Wavelength of the wave</h3>

λ = 2π/K

λ = (2π)/(5.4)

λ = 1.164 m

<h3>Maximum speed of the wave</h3>

v(max) = Aω

where;

  • A is amplitude of the wave
  • ω is angular speed of the wave

v(max) = (0.048)(128)

v(max) = 6.14 m/s

Thus, the wavelength of the wave is 1.164 m.

The  velocity of the wave is 23.7 m/s.

The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

Learn more about wavelength here: brainly.com/question/10728818

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1 year ago
The histogram below shows the number of downloads of a song over time.
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Given data:

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  • A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.
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<em>From Figure:</em>  

        Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).

<em>So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.</em>

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2 years ago
A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and
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<u>Option b. </u>A smaller magnitude of momentum and more kinetic energy.

<h3>What is a momentum?</h3>
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Answer:

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since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

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