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UNO [17]
3 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]3 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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What is velocity and it’s SI Unit?
svp [43]

Answer:

Velocity is the displacement (change In position of an an object it is similar to distance ) divided by the time taken. SI unit m/s (metre per second)

3 0
3 years ago
Two forces P and Q act on an object of mass 15.0 kg with Q being the larger of the two forces. When both forces are directed to
Rom4ik [11]

Answer:

P=6.25N  and Q=16.25N

Explanation:

In order to solve this problem we must first draw a free body diagram for both situation, (see attached picture).

Now, we need to analyze the two free body diagrams. So let's analyze the first diagram. Since the body is accelerated, then the sum of forces is equal to mass times acceleration, so we get:

\Sigma F=ma

We can assume there will be only the two mentioned forces P and Q, so

the sum of forces will be:

P+Q=ma

P+Q=(15kg)(1.50m/s^{2})

P+Q=22.5N

We can do the same analysis for the second free body diagram:

\Sigma F=ma

Q-P=(15kg)(0.7m/s^{2})

Q-P=10.5N

so now we have a system of equations we can solve by elimination:

Q+P=22.5N

Q-P=10.5N

Now, we can add the two equations together so the P force is eliminated, so we get:

2Q=32.5N

now we can solve for Q:

Q=\frac{32.5N}{2}

so

Q=16.25N

Now we can use any of the equations to find P.

Q+P=22.5N

P=22.5N-Q

when substituting for Q we get:

P=22.5N-16.25N

so

P=6.25N

5 0
4 years ago
Assuming 100% efficient energy conversion, how much water stored behind a 50 centimetre high hydroelectric dam would be required
11Alexandr11 [23.1K]

Complete question is;

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes

Answer:

Amount of water required to charge the battery = 7.35 m³

Explanation:

The formula for Potential energy of the water at that height = mgh

Where;

m = mass of the water

g = acceleration due to gravity = 9.8 m/s²

H = height of water = 50 cm = 0.5 m

We know that in density, m = ρV

Where;

ρ = density of water = 1000 kg/m³

V = volume of water

So, potential energy is now given as;

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Now, formula for energy of the battery is given as;

E = qV

We are given;

q = 50 A.min = 50 × 60 = 3,000 C

V = 12 V

Thus;

qV = 3,000 × 12 = 36,000 J

E = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery.

Thus;

(4900V) = (36,000)

4900V = 36,000

V = 36,000/4900

V = 7.35 m³

5 0
3 years ago
An object exhibits SHM with an angular frequency w = 4.0 s-1 and is released from its maximum displacement of A = 0.50 m at t =
vivado [14]

Explanation:

It is given that,

Angular frequency, \omega=4\ s^{-1}

Maximum displacement, A = 0.5 m at t = 0 s

We need to find the time at which it reaches its maximum speed. Firstly, we will find the maximum velocity of the object that is exhibiting SHM.

v_{max}=A\times \omega

v_{max}=0.5\times 4

v_{max}=2\ m/s............(1)

Acceleration of the object, a=\omega^2A

a=4^2\times 0.5

a=8\ m/s^2...............(2)

Using first equation of motion we can calculate the time taken to reach maximum speed.

v=u+at

t=\dfrac{v-u}{a}

t=\dfrac{2-0}{8}

t = 0.25 s

So, the object will take 0.25 seconds to reach its maximum speed. Hence, this is the required solution.

4 0
3 years ago
) Calculate current passing in an electrical circuit if you know that the voltage is 8 volts and the resistance is 10 ohms
Svetach [21]

Explanation:

<em>Hey</em><em>,</em><em> </em><em>there</em><em>!</em>

<em>Here</em><em>,</em><em> </em><em>In</em><em> </em><em>question</em><em> </em><em>given</em><em> </em><em>that</em><em>, </em>

<em>potential</em><em> </em><em>difference</em><em> </em><em>(</em><em>V</em><em>)</em><em>=</em><em> </em><em>8</em><em>V</em>

<em>resistance</em><em> </em><em>(</em><em>R</em><em>)</em><em>=</em><em> </em><em>1</em><em>0</em><em> </em><em>ohm</em>

<em>Now</em><em>,</em>

<em>According</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>Ohm's</em><em> </em><em>law</em><em>,</em>

<em>V</em><em>=</em><em> </em><em>R</em><em>×</em><em>I</em><em> </em><em> </em><em> </em><em> </em><em>{</em><em> </em><em>where</em><em> </em><em>I</em><em> </em><em>=</em><em> </em><em>current</em><em>}</em>

<em>or</em><em>,</em><em> </em><em>I</em><em> </em><em>=</em><em> </em><em>V</em><em>/</em><em>R</em>

<em>or</em><em>,</em><em> </em><em>I</em><em> </em><em>=</em><em> </em><em>8</em><em>/</em><em>1</em><em>0</em>

<em>Therefore</em><em>, </em><em> </em><em>current</em><em> </em><em>is</em><em> </em><em>4</em><em>/</em><em>5</em><em> </em><em>A</em><em> </em><em>or</em><em> </em><em>0</em><em>.</em><em>8</em><em> </em><em>A</em><em>.</em>

<em>(</em><em>A</em><em>=</em><em> </em><em>ampere</em><em> </em><em>=</em><em> </em><em>unit</em><em> </em><em>of</em><em> </em><em>current</em><em>)</em><em>.</em>

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em> </em>

6 0
4 years ago
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