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Virty [35]
3 years ago
12

What is potential difference​

Physics
2 answers:
lisabon 2012 [21]3 years ago
5 0

Answer:

VOLTAGE

Explanation:

Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. ... A potential difference of one Volt is equal to one Joule of energy being used by one Coulomb of charge when it flows between two points in a circuit

Ahat [919]3 years ago
3 0
Difference in the amount of energy
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Inez thinks it is important to be honest with her friends.

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Explanation:

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borishaifa [10]

Answer:

7.5s

Explanation:

Given parameters:

Velocity  = 30m/s

Deceleration  = 4m/s²

Unknown:

Time it takes for the car to come to complete rest  = ?

Solution:

 To solve this problem, we use the kinematics expression below:

        v  = u + at

 Since this is a deceleration

         v  = u - at

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

         v - u  = -at

         0  - 30  = -4 x t

              -30  = -4t

               t  = 7.5s

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2 years ago
A 50 kg bear climbed on the tree branch 10 meters above the ground. If the bear descends to 5 meters above the ground, its poten
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A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is
OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

U = \frac{GMm}{r}

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}

U_2= - 5.1180*10^9J

Point B )

We now use the average radius distance from the earth.

U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}

U_2= -5.8925*10^9J

Then,

\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)

\Delta U = 774.5*10^6

By the law of conservation of energy we know that,

\Delta U = \frac{1}{2}mv^2

clearing v,

v= \sqrt{2 \Delta U/m}

v= \sqrt{2*774.5*10^6 /94.2}

v= 4055.08m/s

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0
3 years ago
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