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Nonamiya [84]
4 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]4 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:

(a) Electric field at the surface of a sphere is given as:

E = kQ/r²

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To find charge Q, we make Q subject of the formula:

Q = (E * r²)/k

Hence, charge, Q, at the surface of the earth, having radius, r = 6.371 * 10⁵ and electric field, E = 150 N/C is:

Q = [150 * (6.371 * 10⁶)²] / (9 * 10⁹)

Q = 6.765 * 10⁵ C

Since we're told that the charge at the earth's surface is negative,

Q = -6.765 * 10⁵ C

Surface charge density, σ, given as:

σ = |Q|/A

Where

|Q| = magnitude of charge

A = surface area.

Surface area, A, of the earth is given as:

A = 4πr²

A = 4π * (6.371 * 10⁶)²

A = 510064471909788 m²

σ = 6.765 * 10⁵/510064471909788

σ = 1.33 * 10⁻⁹ C/m²

(b) At a height 5km from the earth's surface, the electric field will be:

E = kQ/(r + 5km)²

r + 5km = 6376km = 6.376 * 10⁶m

=> E = (9 * 10⁹ * 6.765 * 10⁵)/(6.376 * 10⁶)²

E = 149.77 N/C

The difference between the electric field at the surface of the earth and at a height of 5km is:

159 - 149.77 N/C = 0.23 N/C

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Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise fro
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Answer:

a)Fnx = -26.92 N

b)Fny= 8.4 N

c)Fex = 26.92 N

d)Fey = - 8.4 N

e)β = 18.34° clockwise from the positive x axis

Explanation:

Look at the attached graphic

a) What is the x component of the net force acting on the object?

Fnx:x component of the net force acting on the object

Fnx= F₁x+ F₂x

F₁x=  33.4*sin 23.8° = 13.48 N

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Fnx =  13.48 N- 40.4 N= -26.92 N

b) What is the y component of the net force acting on the object?

Fny= F₁y+ F₂y

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Fny=30.6 N-22.2 N = 8.4 N

c) What is the x component of the equilibrant?

Fex: component of the equilibrant

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Fex = 26.92 N

d) What is the y component of the equilibrant?

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Fey = - 8.4 N

e) What is the magnitude of the equilibrant?

F_{e} :  equilibrant force

F_{e} = \sqrt{(F_{e}x)^{2}+{(F_{e}y)^{2} }

F_{e} = \sqrt{(26.92)^{2}+{(-8.4)^{2} }

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f) What is the angle the equilibrant makes with the x axis?

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β = -18.34° or β = 18.34° clockwise from the positive x axis

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