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ivann1987 [24]
3 years ago
15

A rock falls from a tower that is 320 feet high. As it is​ falling, its height is given by the formula h equals 320 minus 16 t s

quared . How many seconds​ (in tenths) will it take for the rock to hit the ground ​(hequals​0)?
Physics
1 answer:
RSB [31]3 years ago
4 0

Answer:

a. 4.5secs

Explanation:

From the question, the equation describing the height is given by

h(t)=320-16t^{2}\\

at the point when the rock hit the ground, the height,h will be zero.

Hence we can have

h(t)=320-16t^{2}\\\\at  h(t)=0\\0=320-16t^{2}\\hence \\320=16t^{2}\\t^{2}=\frac{320}{16}\\ t^{2}=20\\t=\sqrt{20}\\ t=4.47secs\\t=4.5sec\\

hence the rock will hit the ground in 4.5secs

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Answer:

The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.

Solution:

As per the question:

Radius of an alveolus, R = 200\mu m = 200\times 10^{- 6}\ m

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Now,

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3 years ago
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BabaBlast [244]

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.

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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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Answer:

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Explanation:

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Answer:

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