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FinnZ [79.3K]
2 years ago
8

A spring with a mass attached is initially at rest. It is then stretched 2 cm by a physics student. Another physics student with

a similar mass and spring system then stretches their spring to 4 cm. Compare and contrast the potential energies of the springs after they are stretched by the students. *
Physics
1 answer:
Andrew [12]2 years ago
3 0
The student who displaced the spring by 2 cm has less potential energy than the student who displaced the spring by 4 cm, this is because potential energy (elastic) is directly proportionate to extension (displaced amount), so as the amount of displacement of the spring is higher, then the potential energy of the springs is higher and vice versa.
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Alexus [3.1K]
Was there any choice of answers so i can help or something u have to figure out
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3 years ago
The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate
ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}      ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

\theta=\frac{3.7m}{0.162m}=22.83rad

22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}

to calculate the angular speed w we can use\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}

Thus, wf=12.68rad/s

C) We can use our result in B)

\alpha=3.52\frac{rad}{s^2}

I hope this is useful for you

regards

3 0
2 years ago
Read 2 more answers
An example for curvilinear motion.
Shkiper50 [21]

Ball thrown into the air at an angle.

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2 years ago
Suppose the Pathfinder moves at a rate of 0.2m/s for 20 seconds and then turns around and travels at the same speed for 3 second
kotykmax [81]

Answer:

PLEASE BE MORE SPECIFIC

Explanation:

7 0
2 years ago
The sound from a single source can reach point O by two different paths. One path is 20.0 m long and the second path is 21.0 m l
aleksandrvk [35]

Answer:

minimum frequency = 170 Hz

Explanation:

given data

One path long = 20 m

second path long = 21 m

speed of sound = 340 m/s

solution

we get here destructive phase that is path difference of minimum \frac{\lambda}{2}

here  λ is the wavelength of the wave

so path difference will be

21 - 20 = \frac{\lambda}{2}  

λ = 2 m

and

velocity that is express as

velocity = frequency × wavelength    .............1

frequency  = \frac{340}{2}  

minimum frequency = 170 Hz

7 0
3 years ago
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