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
1 answer:
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:
Option A is correct.
(The faster object encounters more resistance)
Explanation:
Option A is correct. (The faster object encounters more resistance)
Air resistance depends on various factors:
- Speed of the object
- Cross-sectional area of the object
- Shape of the object
Formula:
As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.