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SpyIntel [72]
2 years ago
6

I need help with questions b and d, that’s all. Thank you.

Physics
1 answer:
Mkey [24]2 years ago
7 0

b).  The power depends on the RATE at which work is done.  

Power = (Work or Energy) / (time)

So to calculate it, you have to know how much work is done AND how much time that takes.

In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A.  But the question doesn't tell us anywhere how much time that takes.  So there's NO WAY to calculate the power needed to do it.

The more power is used, the faster the car is lifted.  The less power is used, the slower the car creeps up the first hill.  If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill.  That would just take a long time, but it could be done if the power is small enough.

Without knowing the time, we can't calculate the power.

...

d).  Kinetic energy = (1/2) · (mass) · (speed squared)

On the way up, the car stops when it reaches point-A.  

On the way down, the car leaves point-A from "rest".

WHILE it's at point-A, it has <u><em>no speed</em></u>.  So it has no (<em>zero</em>) kinetic energy.

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A boat with a horizontal tow rope pulls a waterskier. She skis off to the side, the rope makes an angle of 15° with the forward
dsp73

Answer:

<h3>13,976.23Joules</h3>

Explanation:

Workdone by the rope is expressed using the formula;

W = Fd sin(theta)

F is the tension in the rope = 180

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theta is the angle of inclination = 15°

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4 0
2 years ago
A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a
IgorLugansk [536]

Answer:1.5

Explanation:

Given

mass of first  cart m_1=6 kg

initial Velocity u_1=3 m/s

mass of second cart m_2=3 kg

u_2=0 m/s

In the absence of External Force we can conserve momentum

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v=\frac{m_1u_1+m_2u_2}{m_1+m_2}

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Final kinetic Energy of two masses

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K.E._2=18 J

Initial Kinetic Energy

K.E._1=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2

K.E._1=\frac{1}{2}6\times 3^2+0

K.E._1=27 J

ratio =\frac{K.E._1}{K.E._2}=\frac{27}{18}=1.5

5 0
3 years ago
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