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Drupady [299]
3 years ago
12

The earth and moon are seperated by a

Physics
1 answer:
cestrela7 [59]3 years ago
7 0

Answer:

4674568 m or 4675 km

Explanation:

Let x be the distance from the center of the mass system to the center of the earth. This means the distance from the center of the mass system to the center of the moon is 3.85*10^8 - x. For the masses to balance out, we have the following equation:

m_e*x = m_m*(3.85*10^8 - x)

where m_e is the mass of Earth, and m_m is mass of the moon

5.98 * 10^{24}x = 7.35*10^{22}(3.85*10^8 - x)

We can divide both sides by 10^{22}

598x = 7.35*3.85*10^8 - 7.35x

605.35x = 2829750000

x = 2829750000/605.35 \approx 4674568 m or 4675 km

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Electrolytes serve several purposes in the body, including _________
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Nerve and muscle function and fluid balance

Explanation:

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4 0
2 years ago
A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The
almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

             Initial momentum = Final momentum

                                       Pi = Pf

                  M*vs_e = m*vc_e + 4m*vm_e

Where,

                  M = m + 4m = 5m

                  vc_e = Velocity of command relative to earth

                  vm_e = Velocity of motor relative to earth  

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

                  vm_e =  v_c/m + vc_e        

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

                  5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

                  5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

                  5m*vs_e -  4m*(v_c/m) = 5m*vc_e

                   vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

                  vc_e = 3760 - 0.8*(90)

                  vc_e = 3,688 km/h

5 0
3 years ago
Read 2 more answers
A person travelled in the -x direction at a constant speed of 2.5 m/s. The person's final position is -10.0 m, and the initial p
Nostrana [21]

Answer:

(c) time required to travel = 8 sec

Explanation:

We have given the final position = -10 m on x axis

And the initial position =10 m

So total distance = 10-(-10)=20 m

The speed is given as 2.5 m/sec

We have tof ind the time required by the person to travel

Time is given by t=\frac{distance}{speed}=\frac{20}{2.5}=8 sec

So the option (c) is correct option

4 0
3 years ago
What net force is required to accelerate a car at a rate of 5 m/s ^ 2 if the car has a mass of 3, 000 kg ?
Olenka [21]

Answer:

The answer is 15000kgms^-2

Explanation:

Actually here we are using the formula F=ma

3 0
2 years ago
Explain why a diverging lens is used to correct nearsightedness (difficulty seeing objects far away).
Darina [25.2K]

Answer:

so that it can diverge the light to make sure that it focused on the ratina and the image is formed.

Explanation:

nearsightedness is when the light is focused in front of the ratina and for an image to be formed in the eye, the light must be focused on the ratina so to correct that we use the diverging lenses so that it will diverge the light and allow the cornea and the lens to converge it so it is focused on the ratina.

8 0
3 years ago
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