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Sergeu [11.5K]
4 years ago
14

A 4 kg kitten is sliding at 12 m/s on a horizontal frictionless surface. A constant force is applied that slows it with an accel

eration of 3 m/s/s. How much work must this force do to stop the kitten
Physics
1 answer:
Liono4ka [1.6K]4 years ago
3 0

Answer:

288 Joules

Explanation:

Work= delta kinetic energy. To find the work required to stop the kitten, you have to find the delta kinetic energy between when the kitten is moving and when it is stopped. Since it is not moving when it's stopped, it has no kinetic energy. So to find the work, just find the kinetic energy of the kitten moving at 12m/s.

KE=1/2((m)(v^2)), therefore

delta KE=(1/2)*((4kg)*((12m/s)^2))-0

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Answer:

wool maybe, I was a bit confused in silk and wool but wool

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Properties can be observed only when the substance in a sample of matter are changing into different substance
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It states a key concept in chemistry.


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3 years ago
Define principal focus of concave mirror.Where should we place a candle in front of a concave mirror to get an enlarged,erect im
Mars2501 [29]

Answer

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3 0
4 years ago
A child pushes a toy box across the floor in 5 seconds. If he did 30 J of work on the toy box, what amount of power was required
s2008m [1.1K]
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4 0
3 years ago
Read 2 more answers
Two point charges exert a 7.35 N force on each other. What will the force become if the distance between them is increased by a
I am Lyosha [343]

Answer :

New force becomes, F' = 1.83 N

Explanation:

Let two point charges exert a force of 7.35 N force on each other. The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

q_1\ and\ q_2 are charges

r is the distance between charges if the distance between them is increased by a factor of 2, r' = 2r

New force is given by :

F'=\dfrac{kq^2}{r'^2}

F'=\dfrac{kq^2}{(2r)^2}

F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}

F'=\dfrac{1}{4}\times 7.35

F' = 1.83 N

So, the new force between charges will be 1.83 N. Therefore, this is the required solution.          

3 0
3 years ago
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