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bagirrra123 [75]
3 years ago
8

Our body contains ______________ potential energy from food we have eaten.

Physics
1 answer:
Alisiya [41]3 years ago
5 0

Answer: Our body contains chemical potential energy from food we have eaten.

This chemical potential energy is transformed into the kinetic energy of our hands and arms as we rub our hands together.

As our hands move past each other and rub against each other, friction allows the kinetic energy to be transformed into thermal energy on the surface of our hands.

Explanation:

It is true that our food contains chemical bonds and these bonds have potential energy stored. So, when we eat food then our body acquires chemical potential energy.

When we rub our hands and arms then they form kinetic energy as atoms present within the skin of our hands come into motion. This rubbing of hands leads to the formation of heat which means thermal energy is being generated.

Thus, we can conclude that our body contains chemical potential energy from food we have eaten.

This chemical potential energy is transformed into the kinetic energy of our hands and arms as we rub our hands together.

As our hands move past each other and rub against each other, friction allows the kinetic energy to be transformed into thermal energy on the surface of our hands.

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A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th
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Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

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3 years ago
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let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz

The frequency of the simple pendulum is given as;

F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m

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