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stealth61 [152]
3 years ago
12

Explain the law of conservation of energy then give an example that shows the slot to be true in your example explain where ther

e is potential energy and where there is kinetic energy also explain if any energy is lost to friction
Physics
1 answer:
Orlov [11]3 years ago
3 0
<h2>Answer:</h2>

<em><u>The energy of a system always remains constant as it can be converted from one type two another type. It can never be lost or created.</u></em>

<h3>Explanation:</h3>

Lets take an example of the boy riding a bicycle.

The boy is using the biochemical energy in his body to run the pedals up and down in a cycle form.

The biochemical energy is converted into the potential energy in the pedal.

As the pedal moves, it gives energy to move the wheels of the cycle and consequently the cycle will move forward.

Hence the potential energy is converted into the kinetic energy.

If the road is not clear, a lot of energy is needed to move the wheels or cycle as compared to clear road. Because some of the energy will be converted into heat during friction.

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Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0
3 years ago
A box of weight 280 N is being pulled to the right by a pulling force vec F of magnitude 50 N. The box is moving at a constant s
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Answer:

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5 0
3 years ago
Derived quantities depend on...........physical quantity​
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Derived quantities depend on.( fundamental)..........physical quantity

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7 0
3 years ago
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If the plotted points on a speed-time graph do not form a straight line, what do you know about the object's acceleration?
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3 years ago
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Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a hori
LenaWriter [7]

Answer:

Explanation:

Sam mass=75kg

Height is 50m

20° frictionless slope

Horizontal force on Sam is 200N

According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.

Therefore

Wg - Ww =∆K.E

Note initial the body was at rest at top of the slope.

Then, ∆K.E is K.E(final) - K.E(initial)

K.E Is given as ½mv²

Since initial velocity is zero then, K.E(initial ) is zero

Therefore, ∆K.E=½mVf²

Wg is work done by gravity and it is given by using P.E formulas

Wg=mgh

Wg=75×9.8×50

Wg=36750J

Ww is work done by wind and it's is given by using formulae for work

Work=force × distance

Ww=horizontal force × horizontal distance

Using Trig.

TanX=opposite/adjacent

Tan20=h/x

x=h/tan20

x=50/tan20

x=137.37m

Then,

Ww=F×x

Ww=200×137.37

We=27474J

Now applying the formula

Wg - Ww =∆K.E

36750 - 27474 =½×75×Vf²

9276=37.5Vf²

Vf²=9275/37.5

Vf²= 247.36

Vf=√247.36

Vf=15.73m/s

3 0
4 years ago
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