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Delicious77 [7]
3 years ago
14

While jumping on a trampoline you calculate that at the highest peak of your jump you have 900 joules of gravitational potential

energy. What will be your kinetic energy just before landing back on the trampoline?
Physics
1 answer:
BabaBlast [244]3 years ago
8 0

Jumping on a trampoline is a classic example of conservation of energy, from potential into kinetic. It also shows Hooke's laws and the spring constant. Furthermore, it verifies and illustrates each of Newton's three laws of motion.

<u>Explanation</u>

When we jump on a trampoline, our body has kinetic energy that changes over time. Our kinetic energy is greatest, just before we hit the trampoline on the way down and when you leave the trampoline surface on the way up. Our kinetic energy is 0 when you reach the height of your jump and begin to descend and when are on the trampoline, about to propel upwards.

Potential energy changes along with kinetic energy. At any time, your total energy is equal to your potential energy plus your kinetic energy. As we go up, the kinetic energy converts into potential energy.

Hooke's law is another form of potential energy. Just as the trampoline is about to propel us up, your kinetic energy is 0 but your potential energy is maximized, even though we are at a minimum height. This is because our potential energy is related to the spring constant and Hooke's Law.

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What potential difference is needed to give a helium nucleus (q=2e) 85.0 kev of kinetic energy?
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E=q \Delta V
where q=2e is the charge of the helium nucleus, and \Delta V is the potential difference applied to it.
Since we know the kinetic energy, we have
E=K=85~keV=q \Delta V
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How does the ratio of tin to copper affect the properties of the alloy bronze?
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2 years ago
You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)
Ilia_Sergeevich [38]

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

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a = g = 9.8 m/s^2

Part b)

Now from kinematics equation we will have

y = v_i t + \frac{1}{2} at^2

now we have

y = 25 m

so from above equation

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t = 2.26 s

Part c)

If we throw the rock horizontally by speed 20 m/s

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Part D)

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angle of projection = 65 degree

now we have

v_x = vcos\theta

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v_y = vsin\theta

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PART E)

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so here we can use kinematics in Y direction

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