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trapecia [35]
3 years ago
7

The Moon's center is 3.9x10 m from Earth's center. The Moon is 1.5x10^8 km from the Sun's center. If the mass of the Moon is 7.3

x10^22 kg, find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon
Physics
1 answer:
nika2105 [10]3 years ago
3 0

Explanation:

It is given that The Moon's center is 3.9x10⁸ m from Earth's center. The moon 1.5x10⁸ km from the Sun's center. We need to find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon.

The gravitational force is given by :

F=\dfrac{Gm_em_m}{r^2}

It means F\propto \dfrac{1}{r^2}

So,

\dfrac{F_1}{F_2}=\dfrac{r_2}{r_1}

r₁ = 3.9x10⁸ km

r₂= 1.5x10⁸ km

So,

\dfrac{F_1}{F_2}=\dfrac{1.5\times 10^8}{3.9\times 10^8}\\\\\dfrac{F_1}{F_2}=\dfrac{5}{13}

Hence, the ratio of the gravitational forces exerted by Earth and the Sun on the Moon is 5:13.

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Answer:

70 Joules

Explanation:

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Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?
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The equilibrium conditions allow to find the results for the balance forces are:

  • F₁ = 225.4 N
  • F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

            ∑ F = 0

            ∑ τ = 0

           

Where F are the forces and τ the torques.

The torque  is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

           F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

         F₂ 2 - W₁ 1 - W₂ 1.5 = 0\frac{W_1  \ 1 + W_2 \ 1.5}{2}

Let's calculate F₂

         F₂ = \frac{W_1 \ 1 + W_2 \ 1.5 }{2}  

         F₂ = (m g + M g 1.5)/ 2

         F₂ = \frac{(12 + 68 \ 1.5 ) \  9.8}{2}  

         F₂ = 558.6 N

We substitute in the translational equilibrium equation.

         F₁ = W₁ + W₂ - F₂

         F₁ = (m + M) g - F₂

         F₁ = (12 +68) 9.8 - 558.6

         F₁ = 225.4 N

In conclusion using the equilibrium conditions we can find the forces of the balance are:

  • F₁ = 225.4 N
  • F2 = 558.6 N

Learn more here:  brainly.com/question/12830892

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Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final
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Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B =  -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

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-0.0760 M/s of B -----------> x

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⇒ Determine the final concentration of A  after 4s

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A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

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