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lapo4ka [179]
3 years ago
6

A 5kg particle moving at a speed of 10m/s to the right makes an elastic collision with a wall and rebounds backward calculate th

e magnitude of the impulse of the body
Physics
1 answer:
worty [1.4K]3 years ago
7 0

Answer:

The magnitude of the impulse experienced by the particle is 100 kg.m/s.

Explanation:

Given;

mass of the particle, m = 5 kg

initial velocity of the particle, v₁ = 10 m/s

assuming the particle rebounds with same velocity backwards, v₂ = - 10 m/s

The impulse experienced by the particle is the change in linear momentum;

J = ΔP = mv₁ - mv₂

J = m(v₁ - v₂)

J = 5 (10 - (-10))

J = 5 (10 + 10)

J = 5(20)

J = 100 kg.m/s

Therefore, the magnitude of the impulse experienced by the particle is 100 kg.m/s.

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Answer:

V = IR

Explanation:

Required

Which equation represents ohm's law?

Literally, ohm's law implies that current (I) is directly proportional to voltage (V) and inversely proportional to resistance (R).

Mathematically, this can be represented as:

I\ \alpha\ \frac{V}{R}

Convert the expression to an equation

I\ =\ \frac{V}{R}

Multiply both sides by R to make V the subject

I\ * R\ =\ \frac{V}{R} * R

I\ * R\ =V

Reorder

V = I\ * R

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Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?
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The correct option is A.
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A 60-kg block initially at rest on a frictionless horizontal surface is acted upon by a force of 3.0 N for a distance of 7.0 m.
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On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m
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Answer:

F = 1.489*10^{-7}  N

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Force between these two objects is given by:

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