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UNO [17]
2 years ago
13

Give an example of mass making a difference in the amount of gravitational energy. Tell how you know the gravitational energy is

different and your example
Please help due today!!
Physics
1 answer:
Andreas93 [3]2 years ago
5 0

Answer:

The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.

Examples of Gravitational Energy

A raised weight.

Water that is behind a dam.

A car that is parked at the top of a hill.

A yoyo before it is released.

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In a Little League baseball game, the 145 g ball enters the strike zone with a speed of 17.0 m/s . The batter hits the ball, and
VMariaS [17]

Hi there!

Impulse = Change in momentum

I = Δp = mΔv = m(vf - vi)

Where:

m = mass of object (kg)

vf = final velocity (m/s)

vi = initial velocity (m/s)

Begin by converting grams to kilograms:

1 kg = 1000g ⇒ 145g = .145kg

Now, plug in the given values. Remember to assign directions since velocity is a vector. Let the initial direction be positive and the opposite be negative.

I = (.145)(-20 - 17) = -5.365 Ns

The magnitude is the absolute value, so:

|-5.365| = 5.365 Ns

4 0
2 years ago
PLEASE HELP ASAP!! CORRECT ANSWER ONLY PLEASE!!
Liula [17]
The answer is Trend Line.
5 0
3 years ago
Read 2 more answers
If we want to describe work, we must have
tresset_1 [31]

The answer is C

Hope that helps!

Good luck :)

8 0
3 years ago
Read 2 more answers
A train travelled 500 meters in 25 seconds
KIM [24]

\huge\mathfrak\red {Answer:}

<h2>Speed = Distance/Time</h2>

If a train travelled 500 meters in 25 seconds then,

Speed = 500m/25sec

<h2>→ 20 m/sec</h2>

\mathfrak\purple {Hope\: this\: helps\: you...}

7 0
2 years ago
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If a spring is stretched 4m from its starting length when 20n of force is applied, then how much work (in joules) is done by the
lys-0071 [83]

ANSWER:

250 J

STEP-BY-STEP EXPLANATION:

F = 20N is required to stretch the spring by 4 meters

We know that the force is equal to:

F=k\cdot x

We solve for k (spring constant):

k=\frac{F}{x}=\frac{20}{4}=5\text{ N/m}

The work done in stretching the spring is given by the following equation (in this case the stretch is 10 meters:

\begin{gathered} W=\frac{1}{2}k\cdot x^2 \\ \text{ Replacing} \\ W=\frac{1}{2}\cdot5\cdot10^2 \\ W=250\text{ J} \end{gathered}

The work required is 250 joules.

5 0
1 year ago
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