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Vladimir79 [104]

3 years ago

14

2. Who was the first person to"measure gravity?" (Hint: By "measuring gravity, we mean

1 answer:

babunello [35]3 years ago

4 0

Answer:

The first person to "measure gravity" is sir Issac newton

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Electromagnetic induction

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Answer:

Electromagnetic Induction or Induction is a process in which a conductor is put in a particular position and magnetic field keeps varying or magnetic field is stationary and a conductor is moving. This produces a Voltage or EMF (Electromotive Force) across the electrical conductor.

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State one of Kepler's laws of gravitation

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All planets move around the sun

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3 years ago

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6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t

spin [16.1K]

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

$p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s$

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

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2 years ago

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Two iron weights, one twice the mass of the other, are dropped from the top of a building. Compared with the lighter weight, the

Whitepunk [10]

The twice as heavy weight will hit the ground with more force, or impact.

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2 years ago

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If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you

Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, $Q=121\ nC=121\times 10^{-9}\ C$

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

$Q=ne$

e is the charge on electron

$n=\dfrac{Q}{e}$

$n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}$

$n=7.5625\times 10^{11}$

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

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3 years ago