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3 years ago

According to Newton’s first law of motion, what is required to make an object at rest move

Physics

2 answers:

pogonyaev3 years ago

6 0

To make an object at rest move you would need to apply force to the object.

zimovet [89]3 years ago

3 0

Newton's first law states that an object will remain at rest or move at a constant speed in a straight line unless it is acted on by an unbalanced force.

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Low pressure systems

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4 years ago

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4 years ago

A golfer, driving a golf ball off the tee, gives the ball a velocity of 38 m/sec. The mass of the ball is 0.045 kg, and the dura

atroni [7]

Answer:

$\Delta p=1.71\frac{kg\cdot m}{s}$

Explanation:

The momentum of a body is defined as the product of its mass and its velocity at a given time. Therefore the change in the momentum of the ball is given by the difference between the final momentum and the initial momentum:

$\Delta p=p_f-p_i\\\Delta p=mv_f-mv_i\\\Delta p=m(v_f-v_i)\\\Delta p=0.045kg(38\frac{m}{s}-0\frac{m}{s})\\\Delta p=1.71\frac{kg\cdot m}{s}$

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4 years ago

In what ways are earth and a bar magnet alike?

xenn [34]

One way is the both have magnetic fields.

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3 years ago

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population sy

Genrish500 [490]

The question is incomplete. The complete question is :

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is 5.97×1024 kg and that of the Moon is 7.34×1022 kg. The radius of the Moon’s orbit is about 3.84×105 m.

Solution :

Given :

Mass of earth, $M_e = 5.97 \times 10^{24} \ kg$

Mass of moon, $M_m = 7.34 \times 10^{22} \ kg$

Mass of each human, $m_p =65 \ kg$

Therefore mass of total population, $M_p = 65 \times 7 \times 10^{9} \ kg$

$M_p = 4.55 \times 10^{11} \ kg$

Let the earth is at the origin of the coordinate system. Then,

Since $M_e>> M_p$

$M_m>> M_p$

Hence if we shift all the population on the moon there will be negligible change in the mass of the moon and earth. Hence there will not be any significant shift on the centre of mass. i.e.

$X_{cm} = \frac{5.97 \times 10^{24}+ 7.34 \times 10^{22} \times 3.84 \times 10^5}{5.97 \times 10^{24}+ 7.34 \times 10^{22}}$

$= 4.68 \times 10^6 \ m$

$4.68 \times 10^3 \ km$ from the earth.

3 0

3 years ago