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jarptica [38.1K]
3 years ago
5

If a 2 x 10^-4C test charge is given 6.5J of energy, determine the electric potential difference.

Physics
1 answer:
Gwar [14]3 years ago
8 0

Answer:

The electric potential difference is 32500 volt.

Explanation:

Given that,

Chargeq=2\times10^{-4}C

Energy = 6.5 J

We need to calculate the electric potential difference

Potential difference :

Potential difference is equal to the energy divide by charge.

Using formula of potential difference

V=\dfrac{E}{Q}

V=\dfrac{6.5}{2\times10^{-4}}

V=32500\ volt

Hence, The electric potential difference is 32500 volt.

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3. An electric motor is used to lift a 6.0 kg of mass through a height of 1 metre. The energy it uses is measure on an energy me
olga55 [171]

Answer:

Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.



Figure 1. (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock.

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PEg gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

Converting Between Potential Energy and Kinetic Energy

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PEg to KE without explicitly considering the intermediate step of work. (See Example 2.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.

More precisely, we define the change in gravitational potential energy ΔPEg to be ΔPEg = mgh, where, for simplicity, we denote the change in height by h rather than the usual Δh. Note that h is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is

mgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 Jmgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 J

Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the

5 0
3 years ago
Find initial velocity and time in air​
kkurt [141]

Answer:

The answer to your question is: v = 121.46 m/s

Explanation:

Data

h = 30 m

vo = ?

dx = 300 m

Formula

t = \sqrt{\frac{2h}{g}}

speed = distance / time

Process

t = \sqrt{\frac{2(30)}{9.81}} \\

t= 2.47 s

speed = 300 / 2.47

speed = v = 121.46 m/s

6 0
3 years ago
How is momentum conserved in a system in which two satellites connect?
STALIN [3.7K]
I think the correct answer from the choices listed above is the first option. The one satellite has all the momentum before they connect, and then afterwards they share it. <span>For a collision occurring between object 1 and object 2 in an isolated system, the total </span>momentum<span> of the two objects before the collision is equal to the total </span>momentum<span> of the two objects after the collision.</span>
4 0
3 years ago
Read 2 more answers
If you used 650 W of power lifting a 300 N weight in 2 seconds how high did you lift?
dimulka [17.4K]

Answer:

4.33m

Explanation:

Power = work done/ time

work done = power × time =650 × 2 = 1300J

work done = force × distance

distance = work done/force

distance = 1300/300 = 4.33m

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3 years ago
A horse pulls a wagon that weighs 10 times more than the horse. The horse pulls with a force of 500N to move the wagon forward.
____ [38]

Answer:

f. 500N

Explanation:

The force exerted by the wagon on the horse is 500N directed in the opposite path .

This is premised on the Newton's third law of motion which states that "action and reaction forces are equal and opposite in direction".

So, as the action force of 500N is used by the horse to drag the wagon, the wagon also exerts an oppositely directed force of 500N on the horse.

This way, the acceleration can continue.

5 0
3 years ago
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