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2 years ago

Describe Newton's law of gravitation and its application?

Physics

1 answer:

Jlenok [28]2 years ago

4 0

Answer:

Newton's law of gravitation states that every object in this universe attracts every other object with a force called gravity,

when the distance between the two object is increased the gravitational force decreases by its square,

and when the mass an object increases the gravitational force also increases,

therefore, gravitational force is directly proportional to the mass of the two objects and is square inversely proportional to distance between them.

Hope it helped u,

pls mark as the brainliest and put thanks

^_^

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6. An elevator takes 33k) of energy to lift a group of people up 6 m. What is the total mass of

Xelga [282]

Answer:

350 Kg

Explanation:

We use the Work Formula:

W = Force . Distance

33.000 Joules = Force . 6

Force = 5500 Newtons

Force = Mass . Acceleration

5500 = Mass . 10

Mass = 550

550 - 200 = 350 Kg

5 0

3 years ago

I need help with my physics homework agh! Please help it's due tomorrow. <br>

storchak [24]

Rubbing both pieces cause each piece to have a negative charge.

When two parts have the same they repel each other, so holding one piece up tot he end of the other piece would push it away.

Because one piece is held in the middle by a string, it would rotate the piece in a circle.

If they held the piece to the other end of the one held by a string it would start to rotate in the opposite direction.

4 0

3 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$