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spin [16.1K]
3 years ago
14

Compare the gravitational force on a 4 kg

Physics
1 answer:
Eddi Din [679]3 years ago
6 0

Answer:

  • The gravitational force by the Earth on the object, and by the object on the Earth is

  • F = GMm/R²

  • = 6.674×10−11 m3⋅kg−1⋅s−2 × 6 × 10^24 kg × 44.5 kg/(6.4 × 10^6 m)²

  • = 435 N

<u>Please note that the ration between the gravitation force 435 and the mass 44.5</u>

  • should be gravitational acceleration

  • 435/44.5 = 9.78

  • I attribute the discrepancy between 9.78 and the usual 9.81 to rounding off in the

  • Earth's weight and radius.

The mass of the Moon is M / 81.3.

The radius of the Moon is R × 0.27.

The gravitational force on the moon would be

G(M/81.3)m/(R×0.27)² = 0.17×GMm/R²

The gravitational force on the moon is smaller by the factor of about 0.17.

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vitfil [10]

Answer:

Option C

Explanation:

v= u + at

20 = 5 + a(5)

15= a(5)

a= 3 m/s²

Force = mass × acceleration

= 10 × 3

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What is science and physics?​
Ilya [14]

Answer:

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5 0
3 years ago
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A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
3 years ago
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