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balandron [24]
3 years ago
9

A child on a merry-go-round takes 3.9 s to go around once. What is his angular displacement during a 1.0 s time interval?

Physics
1 answer:
nignag [31]3 years ago
4 0

Answer:

Angular displacement=2π/3.9 rad

Explanation:

Given data

Time t=3.9s

Required

The angular displacement during a 1.0 s time interval

Solution

In 3.9 second the child covers a full circle=2π rad

Angular displacement after 1.0 second is given as:

=\frac{2\pi }{3.9} rad

Angular displacement=2π/3.9 rad

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If a car moving down this road rounds this curve and changes its direction to the right, is this an example of speed, velocity,
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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

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A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m + t/(0.20 s))], where x is in m and t in s.
Len [333]

Corrected and Formatted Question:

A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))], where x is in m and t in s.

(a) In what direction is this wave traveling?

(b) What are the wave speed, frequency, and wavelength?

(c) At t = 0.50 , what is the displacement of the string at x = 0.20 m?

Answer:

The wave is travelling in the negative x direction

The wave speed = 12.0m/s

The frequency = 5Hz

The wavelength = 2.4m

The displacement at t = 0.50s and x = 0.20m is -0.029m

Explanation:

The general wave equation is given by;

y(x, t) = y cos (2\pi(x/λ) - 2\pift)    --------------------------------(i)

Where;

y(x, t) is the displacement of the wave at position x and a given time t

y = amplitude of the wave

f = frequency of the wave

λ = wavelength of the wave

Given;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]   ------------------(ii)

Which can be re-written as;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m)) + 2π(t/(0.20 s))]  -------------(iii)

Comparing equations (i) and (iii) we have that;

=> 2π(x/(2.4 m) = 2π(x/λ)

=> λ = 2.4m

Therefore the wavelength of the wave is 2.4m

Also, still comparing the two equations;

=> 2π(t/(0.20 s) = 2πft

=> f = 1 / 0.20

=> f = 5Hz

Therefore the frequency of the wave is 5Hz

To get the wave speed (v), it is given by;

v = f x λ

Where f = 5Hz and λ = 2.4m

=> v = 5 x 2.4

=> v = 12.0m/s

Therefore, the speed of the wave is 12.0m/s

At t = 0.50s and x = 0.20m;

The displacement, y(x,t) of the string wave is given by

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]

<em>Convert the amplitude of 3.0cm to m</em>

=> 3.0cm = 0.03m

<em>Substitute this back into the equation</em>

=> y(x, t) = (0.03m) × cos[2π(x/(2.4 m) + t/(0.20 s))]

<em>Substitute the values of t and x into the equation above;</em>

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m) + 0.50/(0.20 s))]

<em>Carefully solve the equation</em>

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m)) + 2π(0.50/(0.20 s))]

=> y(x, t) = (0.03m) × cos[0.08π + 5π]

=> y(x, t) = (0.03m) × cos[5.08π]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × -0.9684

=> y(x, t) = 0.029m

Therefore the displacement at those points is -0.029m

Also, the sign of the displacement shows that the direction of the wave is in the negative x direction.

8 0
3 years ago
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