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jek_recluse [69]
3 years ago
10

You push a refrigerator with a force of 100 N. If you move the refrigerator a distance of 5 m while you are pushing, how much wo

rk do you do?
Physics
1 answer:
ddd [48]3 years ago
7 0

The work done to push the refrigerator is 500 Nm.

Explanation:

Work done is the measure of force required to move any object from one point to another. So it is calculated as the product of force and displacement.

If the force increases the work done will increase and similarly, the increase in displacement increases the work done. So to push the refrigerator work should be done on the object and not by the object.

As the force is 100 N and the displacement is 5 m then, work done can be measured as

Work = Force × Displacement

Work = 100 × 5 = 500 Nm

So the work done to push the refrigerator is 500 Nm.

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Answer:

emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.

Direction is counter clockwise.

Explanation:

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3 years ago
What is the main difference between velocity and speed?
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Speed is scalar, meaning it's only going to be like 74 mph, doesn't matter which direction. Velocity is a vector, meaning it has direction. You can go -74mph when talking about velocity, not speed.
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2 years ago
Crystal can ride her bike 10 miles in 2.1 hours. What is her velocity
masha68 [24]

Answer:

v = 4.76 m/s

Explanation:

Given,

The distance traveled by her bike, d = 10 miles

The time of her travel, t = 2.1 m/s

The velocity of an object is defined as the distance traveled by the object to the time of travel. Therefore,

                          V = d/t   m/s

                              = 10 / 2.1

                              = 4.76 m/s

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6 0
2 years ago
A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb,
beks73 [17]

Answer:

a) 0.477 W/m²

b) 13.407 N/C

c) 18.96 N/C

Explanation:

P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2

∴ Average intensity is 0.477 W/m²

b) Rms value

\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C

∴ Rms value of the electric field is 13.407 N/C

c) Peak value

E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C

∴ Peak value of the electric field is 18.96 N/C

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