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Nana76 [90]
3 years ago
6

A block of mass M is suspended from two identical springs of negligible mass, spring constant k, and unstretched length L. First

one spring is attached to the end of the other spring. The block is then attached to the second spring and slowly lowered to its equilibrium position. The two springs stretched a total distance of X1. Next, the springs are hung side by side. The block is then attached to the end of springs and again slowly lowered to its equilibrium position. The springs each stretch a distance of X2. Which of the following correctly shows the relationship between X1 and X2.
(A)X1=X2
(B)X1=√2X2
(C)X1=2X2
(D)X1=4X2
(E)X1=8X2
Physics
1 answer:
34kurt3 years ago
8 0

Answer:

The answer is (C)X1=2X2

Explanation:

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A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

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\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

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