The image illustrates that as the distance between two objects increases, the force of gravity ____________. A) decreases. B) in

Answer:

The equivalent distance in kilometers is 4012 × km.

Explanation:

It's known that 1 millimeter converts to meters. Then, 1 meter converts to kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

1 mm = m

1 m = km

Thus, 1 mm = × km = km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

4012 mm = 4012 × km.

The distance therefore is 4012 × km.

The result is -15.625 m/s².

Acceleration signifies the alteration of velocity over a specified duration. It can be calculated with this formula:

Where:

vf = final velocity

vi = initial velocity

t = time

Let’s examine the information provided in your query:

Initially, the vehicle was traveling at 25 m/s before coming to a halt. Thus, it was in motion and subsequently ceased moving, indicating that the final velocity is 0 m/s.

However, we notice that the problem does not provide a time value. We need to determine the time taken from when it was in motion to when it reached the traffic light located 20 m away.

The time can be calculated using the kinematics equation:

We derive the equation by substituting the known values first.

The duration from when it was in motion until it stopped is 1.6s. Now we can utilize this in our acceleration calculation.

It is important to note that the acceleration is negative, indicating the vehicle slowed down.

Response:

83%

Clarification:

At the surface, the weight can be expressed as:

W = GMm / R²

where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.

When in orbit, the weight is given by:

w = GMm / (R+h)²

where h indicates the shuttle's altitude above Earth's surface.

The weight ratio is as follows:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

For R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

Thus, the shuttle maintains 83% of its weight as it orbits.

The entire and thorough solution is included.