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inysia [295]
4 years ago
9

A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance . For the next loading, the spring is co

mpressed a distance . How much faster does the second dart leave the gun compared with the first
Physics
1 answer:
Rina8888 [55]4 years ago
7 0

Complete question is;

A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?

Answer:

The second dart leaves the gun two times faster than the first one.

Explanation:

If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;

Potential energy = kinetic energy

Thus;

½kx² = ½mv²

Making velocity "v" the subject, we have;

v = √(kx²/m)

Since the initial distance is "x", thus initial launching velocity is;

v1 = √(kx²/m)

Since next distance is 2x, thus, second launch velocity is;

v2 = √(k(2x)²/m)

Expanding, we have;

v2 = √(4kx²/m)

v2 = 2√(kx²/m)

Comparing this to the one gotten for v1 earlier, we can see that it is double v1.

So, v2 = 2v1

Hence, The second dart leaves the gun two times faster than the first one.

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Two points are located on a rigid wheel that is rotating with decreasing angular velocity about a fixed axis. Point A is located
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d.Both points have the same instantaneous angular velocity.

Explanation:

Let's analyze each option one by one:

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a_t = \alpha r

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e.Each second, point A turns through a greater angle than point B. --> FALSE. The two points cover the same angular displacement in the same time, since they have same angular velocity.

Learn more about circular motion:

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#LearnwithBrainly

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We can solve the problem by using the law of conservation of energy:

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Re-arranging, we find the maximum height:

h=\frac{v^2}{2g} = \frac{(75.0 m/s)^2}{2(9.8 m/s^2)}=287.0 m


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