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4 years ago

How fast does a 3 kg rocket accelerate if a force of 140 N is applied?

Physics

1 answer:

elixir [45]4 years ago

3 0

<h3>Answer:</h3>

46.67 m/s²

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of a rocket = 3 kg
Force applied = 140 N

we are supposed to calculate the acceleration of the rocket;

We are going to use the second Newton's law of motion;
According to Newton's second Law of motion force is equivalent to the product of mass and acceleration.
This is from the fact that the resultant force and the rate of change in linear momentum are directly proportional.

Therefore;

F = Ma

In this case;

Rearranging the formula

a = F ÷ m

= 140 N ÷ 3 kg

= 46.67 m/s²

Thus, the acceleration of the rocket is 46.67 m/s²

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Answer: (a) $F=7(10)^{-7}N$

(b) $F=1.344(10)^{-6}N$

(c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

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According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

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Where:

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$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

Knowing this, let's begin with the answers:

<h2 /><h2>(a) Gravitational force Father exertes on baby</h2>

Using equation (1) and taking into account the mass of the father $m_{1}=100kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between them $r=0.2m$ , the force $F_{F}$ exerted by the father is:

$F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}$ (2)

$F_{F}=0.0000007N=7(10)^{-7}N$ (3)

<h2>(b) Gravitational force Jupiter exertes on baby</h2>

Using again equation (1) but this time taking into account the mass of Jupiter $m_{J}=1.898(10)^{27}kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between Jupiter and Earth (where the baby is) $r_{E}=6.29(10)^{11}m$ , the force $F_{J}$ exerted by the Jupiter is:

$F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}$ (4)

$F_{J}=0.000001344N=1.344(10)^{-6}N$ (5)

<h2>(c) Comparison</h2>

Now, comparing both forces:

$F_{J}=0.000001344N=1.344(10)^{-6}N$ and $F_{F}=0.0000007N=7(10)^{-7}N$ we can see $F_{J}$ is greater than $F_{F}$ . However, the difference is quite small as well as the force exerted on the baby.

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