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Elena-2011 [213]
2 years ago
15

Determine the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Eart

h is 1/60 of its value at the Earth's surface.
Physics
1 answer:
Temka [501]2 years ago
5 0

The distance from the Earth's center to the point outside the Earth is 55800 Km

<h3>How to determine the distance from the surface of the Earth</h3>
  • Acceleration due to gravity of Earth = 9.8 m/s²
  • Acceleration due to gravity of the poin (g) = 1/60 × 9.8 = 0.163 m/s²
  • Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
  • Mass of the Earth (M) = 5.97×10²⁴ Kg
  • Distance from the surface of the Earth (r) =?

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by g

r² = GM / g

Take the square root of both sides

r = √(GM / g)

r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]

r = 4.94×10⁷ m

Divide by 1000 to express in Km

r = 4.94×10⁷ / 1000

r = 4.94×10⁴ Km

<h3>How to determine the distance from the center of the Earth</h3>
  • Distance from the surface of the Earth (r) = 4.94×10⁴ Km
  • Radius of the Earth (R) = 6400 Km
  • Distance from the centre of the Earth =?

Distance from the centre of the Earth = R + r

Distance from the centre of the Earth = 6400 + 4.94×10⁴

Distance from the centre of the Earth = 55800 Km

Learn more about gravitational force:

brainly.com/question/21500344

#SPJ1

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ASHA 777 [7]

Explanation:

Speed of the marathon runner, v = 9.51 mi/hr

Distance covered by the runner, d = 26.220 mile

Let t is the time taken by the marathon runner. We know that the speed of the runner is given by total distance divided by total time taken. Mathematically, it is given by :

v=\dfrac{d}{t}

t=\dfrac{d}{v}

t=\dfrac{26.220\ mi}{9.51\ mi/hr}

t = 2.75 hours

Since, 1 hour = 60 minutes

t = 165 minutes

Since, 1 minute = 60 seconds

t = 9900 seconds

Hence, this is the required solution.

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3 years ago
Based on our study of electromagnets, what would be the best thing to do to a generator in order to produce more electricity?
Keith_Richards [23]

Answer:

A

Explanation:

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4 0
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A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches
Luden [163]
Yes, the volume of the cylinder will remain constant. As the radius decreases, the height will increase to make sure that the volume is kept the same.
We have been given a value of dr/dt and are required to find dh/dt
Because the volume is constant, we can plug it into the formula for the volume of the cylinder and rearrange it to make h the subject:
128 = πr²h
h = 128/πr²
Now we differentiate both sides:
dh/dr = -256/πr³
Applying the chain rule:
dh/dt = dh/dr x dr/dt
dh/dt = (-256/πr³) x -0.05
dh/dt = 64/5πr³; substituting the value of r
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4 years ago
A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

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Answer:

a. stay the same for very long

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It is rare for any motion to stay the same for a very long time. The force applied on a body causes changes in the magnitude of motion.

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