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lys-0071 [83]
3 years ago
8

When we kept the Earth's mass the same, but shrank its size, we saw that had an effect on its escape speed. Albert Einstein used

to perform what he called "thought experiments". Let's do a thought experiment in which the Earth's mass stays the same, but we keep packing that mass into a smaller and smaller size. Would we ever get to a point where there is no possible way for a spacecraft to escape the Earth? If so, why? (Hint: There's an important fact in Chapter 1 which is relevant to answering this question correctly.)
Physics
1 answer:
nadezda [96]3 years ago
7 0

Answer:

Check the explanation

Explanation:

The escape velocity is the velocity needed by any object to overcome the gravitational force of the planet on which it’s present. Now we know that the gravitational force is directly proportional to the mass of the planet and inversely proportional to the distance of the object from the center of planet.

If we keep the mass of earth constant and decrease the size of the earth than this will decrease the distance between the object and the center of the earth and thus the gravitational force that will act on the body will increase substantially which will in turn increase the value of the escape velocity.

The value of escape velocity will keep on increasing as the size of the earth will shrink till it reaches to a point when the value of escape velocity becomes more than the speed of light and since it’s impossible to travel with a speed greater than the speed of light and therefore at this point it will become impossible for a spacecraft to escape the earth.

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Where is the diagram? What is the question?
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Three cars collide, a 1500 kg sports car, a 1750 kg family car, and a 1200 kg compact car. Which experiences the greatest change
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Answer:

d

Explanation:

This is because momentum is defined as p = mv

delta p = Force *time

neither velocity nor time is given so a conclusion cannot be made on which has the greatest momentum change.

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You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
3 years ago
When work is done and a force is transferred which choice describes the movements of the object?
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When work is done and a force is transferred an object must move
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3 years ago
A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potentialenergy func
OlgaM077 [116]

Answer:

A= 148.92  m/s²

Explanation:

Given that

U(x,y) = (6.00  )x²  - (3.75  )y ³

m= 0.04 kg

Now force in the x-direction

Fx= - dU/dx

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dx= 12 x

When x=0.4 m

dU/dx= 12 x 0.4 = 4.8

So we can say that

Fx= - 4.8 N

From Newtons law

F= m a

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a = -120 m/s²

Acceleration in x direction ,a = -120 m/s²

In y -direction

F= - dU/dy

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dy = 0 - 3.75 x 3 y²

When y = 0.56 m

dU/dy = - 3.75 x 3 x 0.56 x 0.56

dU/dy = - 3.52

So we can say that force in y -direction

F= 3.52 N

F= m a'

3.52 = 0.04 x a'

a'=88.2 m/s²

acceleration in y direction is 88.2 m/s²

The resultant acceleration

A=\sqrt{a^2+a'^2}

A=\sqrt{120^2+88.2^2}

A= 148.92  m/s²

7 0
3 years ago
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