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gavmur [86]
3 years ago
13

In Case 1, a mass M hangs from a vertical spring having spring constant k and is at rest in its equilibrium position. In Case 2

the mass has been lifted a distance D vertically upward. If we define the potential energy in Case 1 to be zero, what is the potential energy of Case 2
Physics
1 answer:
evablogger [386]3 years ago
4 0

Answer: hello your question is incomplete attached below is the complete question

answer :  1/2 KD^2  ( option A )

Explanation:

P.E ( potential energy ) = mgd

In case 1 P.E = 0   i.e. mgd = 0  

Given that in case 2 the Mass M had moved through the Distance D by the compression of the spring

<u>The potential energy of the M in case 2 </u>

= P.E of M at rest + P.E of the spring

= 0 + 1/2 KD^2

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Answer:

28 degree C

Explanation:

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T_1=4.00^{\circ}

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We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

\alpha(T_2-T_1)=\frac{R_2}{R_2}-1

Using the formula

-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1

-5\times 10^{-4}(T_2-4)=0.988-1=-0.012

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3 years ago
A particle accelerator fires a proton into a region with a magnetic field that points in the x-direction. (a) If the proton is m
iVinArrow [24]

Answer:

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<em></em>

Explanation:

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valentinak56 [21]

Answer:

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Answer:

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