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Kaylis [27]
4 years ago
9

Power can be defined as the distance over which work was done. how much work can be done in a given time. all the work in a give

n area. the energy required to do work.
Physics
1 answer:
Artist 52 [7]4 years ago
4 0

Power is defined as rate of work done or rate of energy

so here we can say that

Power = \frac{Work}{time}

so here correct answer must be

how much work can be done in a given time.

So we need to find the work done in a given period of time and then we will have to find the ratio of total work done and total time taken.

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Forces are needed to lift, turn, move, open, close, push, pull, and so on. When you throw a ball, you are using force to make the ball move through the air. More than one force can act on an object at the same time.

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Explanation:

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3 years ago
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Which collects light in a reflecting telescope?
Alecsey [184]
Concave lens. These are used in making the objectives of reflection telescopes
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4 years ago
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On a cold day, a heat pump absorbs heat from the outside air at 14°F (−10°C) and transfers it into a home at a temperature of 86
Leto [7]

Answer:

6.575

Explanation:

T1 = 30C = 30 + 273 = 303 K

T2 = - 10 C = - 10 + 273 = 263 K

The coefficient of performance of heat pump

k = T2 / (T1 - T2)

k = 263 / (303 - 263) = 6.575

7 0
4 years ago
A ball is dropped from rest at a point 12 m above the ground into a smooth, frictionless chute. The ball exits the chute 2 m abo
Nonamiya [84]

Answer:

29,7 m

Explanation:

We need to devide the problem in two parts:

A)  Energy

B) MRUV

<u>Energy:</u>

Since no friction between pint (1) and (2), then the energy conservatets:

Energy = constant ----> Ek(cinética) + Ep(potencial) = constant

Ek1 + Ep1 = Ek2 + Ep2

Ek1 = 0  ; because V1 is zero (the ball is "dropped")

Ep1 = m*g*H1

Ep2= m*g*H2

Then:

Ek2  = m*g*(H1-H2)

By definition of cinetic energy:

m*(V2)²/2 = m*g*(H1-H2) --->  V2 = \sqrt{(2*g*(H1-H2)}

Replaced values:  V2 = 14,0 m/s

<u>MRUV:</u>

The decomposition of the velocity (V2), gives a for the horizontal component:

V2x = V2*cos(α)

Then the traveled distance is:

X = V2*cos(α)*t.... but what time?

The time what takes the ball hit the ground.

Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²

In the vertical  axis:

Y3 = 0 ; Y2 = H2 = 2 m

Reeplacing:

-2 = 14*t + (1/2)*(-9,81)*t²

solving the ecuation, the only positive solution is:

t = 2,99 sec ≈ 3 sec

Then, for the distance:

X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m

6 0
4 years ago
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