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stiv31 [10]
3 years ago
10

Helpppppppppppp meeeeeeeeeeeeeeeee

Physics
2 answers:
Trava [24]3 years ago
6 0

kinetic energy to gravitational potential energy

Irina18 [472]3 years ago
6 0

Answer:

I don't know if I'm correct but it might be

The second option with kinetic energy.

And again idk if im correct.

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A
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Small evidence is also called what?
siniylev [52]
Small evidence is also called trace evidence.
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3 years ago
The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the
natta225 [31]

Answer:

50 N

Explanation:

Efficiency of a machine can't be more than 1, so I assume you mean 40%.  (Remember, efficiency and mechanical advantage are not the same).

Efficiency is the ratio of work out of a system to the work in to the system.

e = Wout / Win

Work is force times distance, so:

e = (Fout × Dout) / (Fin × Din)

Rearranging:

Fin = (Fout × Dout) / (e × Din)

Fin = (Fout / e) × (Dout / Din)

Fin = (Fout / e) / (Din / Dout)

We know that e = 0.40, and Fout = 120 N.  Since there are 6 pulleys, we also know that Din/Dout = 6.

F = (120 N / 0.4) / 6

F = 50 N

5 0
3 years ago
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
Illusion [34]

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

6 0
3 years ago
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