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coldgirl [10]
3 years ago
14

At the surface of Mars the acceleration due to gravity is 3.8m/s . A book weighs 34 N at the surface of the earth. What is it’s

mass on the earth’s surface? What are it’s mass and weight on mar’s surface?
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
8 0

Answer:

On the surface of Mars:

Mass, m = 3.47 kg

Weight, W' = 13.186 N

Solution:

As per the question:

Acceleration due to gravity at the surface of Mars, g' = 3.8\ m/s^{2}

Weight of book on the Earth's surface, W = 34 N

Now,

We know that:

Weight, W = mg

where

m = mass of the body

a = acceleration due to gravity on the earth's surface = 9.8\ m/s^{2}

Thus the mass of the body on the surface of the earth, m:

m = \frac{W}{g} = \frac{34}{9.8} = 3.47\ kg

<em>Also, we know that mass is constant and it is the weight of the body that varies with the acceleration due to gravity.</em>

Now,

Mass of the body on the surface of Mars will be same and is equal to m = 3.47 kg

Weight of the book on the surface of Mars, W' = mg'

W' = 3.47\times 3.8 = 13.186\ N

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Answer:

100 m/s

Explanation:

Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁

Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is  completely stationary. Let v₁ is speed of bond's boat.

It is the concept of the conservation of momentum. It remains conserved. So,

m_1u_1+m_2u_2=m_1v_1+m_2v_2

Putting all the values, we get :

0+(2m_1)50=m_1v_1+(2m_2)(0)\\\\100m_1=m_1v_1\\\\v_1=100\ m/s

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3 years ago
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Answer:

A. 91 meters north

Explanation:

Take +y to be north.

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a = 0 m/s²

t = 7 s

Find: Δy

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Answer:

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=> 33=22+(a)(10)

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Given data

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*The current flows through the resistor is I = 5 A

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\begin{gathered} R=\frac{10}{5} \\ =2\text{ ohm} \end{gathered}

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1 year ago
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