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3 years ago

An example of potential energy is a ball sitting _____ of the stairs.

Physics

1 answer:

expeople1 [14]3 years ago

3 0

Answer:

at the top

Explanation:

Potential energy is the stored energy, mechanical energy,

or energy possessed by by virtue of the position of an object.an example of potential energy is the energy that a ball possesses by virtue of its sitting at the top of the stairs it being about to roll down the stairs.

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Compared to all other forms of energy, wind power is most similar to tidal wave power because _____.

SOVA2 [1]

It is because they are both renewable sources of energy

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3 years ago

Read 2 more answers

Why is space black if the sun is shining

faltersainse [42]

Answer:One star can't light up a whole universe

Explanation:It is like saying one light can feel up the whole town which it can't do.

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3 years ago

A 65-kg woman in an elevator is accelerating upward at a rate of 0.6 m/s2. The gravitational force is ___ N?

Nastasia [14]

Answer:

The gravitational force is 130.

Explanation:

During this problem you have to multiply the 65 and the 0.6.

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3 years ago

Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

solmaris [256]

Answer:

$\lambda = 3.68 *10^{-36} \ m$

$\lambda_p = 1.28*10^{-14} \ m$

Explanation:

From the question we are told that

The mass of the person is $m = 180 \ kg$

The speed of the person is $v = 1 \ m/s$

The energy of the proton is $E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{h}{m * v }$

Here h is the Planck constant with the value

$h = 6.62607015 * 10^{-34} J \cdot s$

$\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }$

=> $\lambda = 3.68 *10^{-36} \ m$

Generally the energy of the proton is mathematically represented as

$E_p = \frac{1}{2} * m_p * v^2_p$

Here $m_p$ is the mass of proton with value $m_p = 1.67 *10^{-27} \ kg$

=> $8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2$

=> $v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }$

=> $v = 3.09529 *10^{7} \ m/s$

$\lambda_p = \frac{h}{m_p * v_p }$

so $\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }$

=> $\lambda_p = 1.28*10^{-14} \ m$

5 0

3 years ago

A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the

Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

$a_c =w^2r = \frac{v^2}{r}$

in this case we know the speed of the runner

$v =8.00\ m/s$

The radius "r" will be the distance from the runner to the center of the track

$r = 28.2\ m$

$a_c = \frac{8^2}{28.2}\ m/s^2$

$a_c = 2.27\ m/s^2$

The answer is the last option

3 0

3 years ago