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FinnZ [79.3K]
3 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]3 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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An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t
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Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

P(z)=P_{0}+\rho _wgh

Now the gauge pressure is given by

P(z)-P_{0}=\rho _wgh

Applying values we get

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8 0
3 years ago
What force is needed to give a 0.25-kg arrow an acceleration of 196 m/s/s
kakasveta [241]
Force (f) = ?

Acceleration (a) = 196 m/s^2

Mass (m) = 0.25 kg

F = (m) • (a)

F = (0.25) • (196)

F = 49 N

Answer : 49 N

I hope that helps you!! Any more questions??

8 0
3 years ago
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it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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