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kumpel [21]
3 years ago
10

Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. Th

e child pushing the sled is exerting a horizontal a force of 8.00 N. There is a force of friction of 5.00 N. What is the acceleration of the sled? Round the answer to the nearest hundredth

Physics
2 answers:
irinina [24]3 years ago
5 0
I'm not quite sure what happens to Fay so I didn't finish but hope it helps

bulgar [2K]3 years ago
5 0

Answer:

The weight is: 294 N

The normal force is 286 N

The acceleration is: -0.38 m/s²

Explanation:

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A tiny pig is dropped from the top of a building and lads safely on a trampoline. The tiny pig will have an increase in what typ
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Answer:

The kinetic energy will have an increase as it falls.

Explanation:

The energy of a particle falling freely will be the sum of its kinetic energy and its potential energy.

When the pig is just dropped, the total enrgy will be only its potential energy. As it falls down its stored potetial energy will be converted into kineitic energy. Just before it touches the trampoline its total energy will be solely kinteic energy.

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If you pluck a guitar string hard the sound waves it gives off will have __________ amplitudes.
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Answer:

large

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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
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Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

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