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3 years ago

What is the gravitational force between Mars and the sun? 7.43 × 1030 N 1.79 × 1026 N 1.65 × 1021 N 3.76 × 1032 N

Physics

1 answer:

VMariaS [17]3 years ago

4 0

The gravitational force between Mars and the Sun is $1.65\cdot 10^{21} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we have:

$m_1 = 1.99\cdot 10^{30} kg$ is the mass of the Sun

$m_2 = 6.39\cdot 10^{23} kg$ is the mass of Mars

$r=229\cdot 10^6 km = 229\cdot 10^9 m$ is the average distance Mars-Sun

Substituting into the equation, we find the gravitational force:

$F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(6.39\cdot 10^{23})}{(229\cdot 10^9)^2}=1.62\cdot 10^{21} N$

So, the closest answer is

$1.65\cdot 10^{21} N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

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trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

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a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

$V=\frac{MU+mu}{M+m}$

Substituting values:

$V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s$

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

$V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134$

The coefficient of kinetic friction between the sled and the snow is 0.0134

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3 years ago

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2 years ago

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Answer:

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Explanations

600N(downwards)

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