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Nonamiya [84]
3 years ago
5

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

#LearnwithBrainly

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Explanation:

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3 0
3 years ago
A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.4 m/s, a typical raindrop speed. How fast
kenny6666 [7]

The final velocity after the collision is 8.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, if we consider the system to be isolated (=no external unbalanced forces), the total momentum of the raindrop+mosquito must be conserved before and after the collision.

If the collision is perfectly inelastic, moreover, the raindrop and the mosquito stick together and travel at the same velocity v after the collision.

Mathematically:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 is the mass of the first mosquito

u_1 = 0 is the initial velocity of the mosquito

m_2 = 50 m_1 is the mass of the raindrop

u_2 = 8.4 m/s is the initial velocity of the raindrop

v is the final combined velocity of the raindrop+mosquito

Re-arranging the equation and substituting, we find:  

m_1 u_1 + 50 m_1 u_2 = (m_1 + 50 m_1) v\\50 m_1 u_2 = 51 m_1 v\\50 u_2 = 51 v\\v=\frac{50}{51}u_2 = \frac{50}{51}(8.4)=8.2 m/s

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4 0
3 years ago
For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this
ikadub [295]

Answer:

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Explanation:

7 0
3 years ago
A 16 kg object cuts 4 km in 25 minutes, find the applied force on the object?
GenaCL600 [577]

Answer:

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Explanation:

7 0
2 years ago
A 12-kg piece of metal displaces 1.6 L of water when submerged. Part A Find its density. Express your answer to two significant
Tatiana [17]

Answer:

ρ = 7500 kg/m³

Explanation:

Given that

mass ,m = 12 kg

Displace volume ,V= 1.6 L

We know that

1000 m ³ = 1 L

Therefore V= 0.0016 m ³

When metal piece is fully submerged

We know that

mass = Density x volume

m=\rho \times V

Now by putting the values in the above equation

\rho=\dfrac{12}{0.0016}\ kg/m^3

ρ = 7500 kg/m³

Therefore the density of the metal piece will be  7500 kg/m³.

6 0
3 years ago
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