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maria [59]
3 years ago
15

Exactly how fast would the average human have to run in order to actually ran across water without breaking surface tension? Thi

s is just a little thing I'm curious about and would like to see if anyone knows the answer.
Physics
2 answers:
Mila [183]3 years ago
7 0

Answer:

It wouldn't work no matter how fat a person runs

Explanation:

The reason for this is because of the weight of a human and that a human can't possibly get to that kind of speed.

lina2011 [118]3 years ago
7 0

Answer:

It's something weird like  67 mph

Not sure if that's exactly right, but i hope it helps

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Calculate the TOTAL mechanical energy of pendulum is it swings from his highest point to its lowest point. Pendulum mass is 4
Lerok [7]

Answer:

its should be 2.0 and 4.5 on it

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3 years ago
What did physicist James Joule's famous paddle wheel experiment demonstrate?
PilotLPTM [1.2K]

<u>Answer </u>

A. that the initial gravitational potential energy of the masses transformed into kinetic energy of the paddles and then to thermal energy in the water


<u>Explanation</u>

James Joule allowed some water to fall from a height of 1 foot. the water would turn a paddle wheel at the bottom causing a temperature of water to raise.

The height form which the water fell, mass and the temperature of water was measured and used to calculate mechanical equivalent of heat.

From the choices given the best answer is A. that the initial gravitational potential energy of the masses transformed into kinetic energy of the paddles and then to thermal energy in the water.

7 0
4 years ago
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A box with a mass of 7 kilograms is pushed up a ramp to a height of 5 meters. What is the work done against the force of gravity
zloy xaker [14]
A. 3.4 x 10 2 its right i guessed it
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4 years ago
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A spherical drop of water carrying a charge of 42 pC has a potential of 620 V at its surface (with V = 0 at infinity). (a) What
iren [92.7K]

Answer:

0.0006091222 m

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\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

Electric potential is given by (at r = R)

V=\dfrac{q}{4\pi\epsilon R}\\\Rightarrow R=\dfrac{q}{4\pi\epsilon V}\\\Rightarrow R=\dfrac{42\times 10^{-12}}{4\pi\times 8.85\times 10^{-12}\times 620}\\\Rightarrow R=0.0006091222\ m

The radius of the drop is 0.0006091222 m

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The density is 4.76 gcm^-3
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3 years ago
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