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ser-zykov [4K]
3 years ago
13

To complete a project, 200,000 Joules of work is needed. The time taken to complete the project is 20 seconds. How much power is

needed?
Physics
2 answers:
Leno4ka [110]3 years ago
5 0

Answer:

10,000

Explanation:

200,000/20 = power needed

200,000/20 = 10,000

Hope this Helps!

MArishka [77]3 years ago
3 0

Answer:

10,000 Watts

Explanation:

Power is the ratio of work to time, or work over time. It can be found using the following formula.

p=w/t

where p is the power, w is the work, and t is the time.

The project needed 200,000 Joules, and it took 20 seconds to complete the project. Therefore, w=200,000 and t=20. Substitute these values into the formula.

p=200,000/20

Divide

p=10,000

Add appropriate units. The units for power is Watts.

p=10,000 Watts

10,000 Watts of power is needed.

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kirza4 [7]

Answer:

A. Distance over which the force is applied

Explanation:

As we know that in pulley system the mass of the car is balanced by the tension in the string

so here we will have

T = r \times F

so here in order to decrease the force needed to lift the car we have to increase Distance over which the force is applied

So here if we increase the distance over which force is applied then it will reduce the effort applied by us in this pulley system as the torque will be more if the distance is more.

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3 years ago
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FrozenT [24]
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8 0
2 years ago
A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. T
larisa [96]

Answer:

a. Decreases

b. Increases

c. Remains the same

d. Increases

Explanation:

a. Capacitance is given by c= Ak€/d

where A is conductivity plate with Area

K is a constant

€ is dielectric with permittivity.

d is the distance

b. Potential difference is given by

V = Ed, since, the electric field remains the

same, the potential diterence also increases with increase in distance.

Since the capacitance depends upon the distance, and all the other factors are kept constant, the capacitance decreases.

c. Electric field remains the same because charge on the

plate remains the same.

d. since electric field remains the same and capacitance decreases, the energy increases.

E= 1/2c * Q^2

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