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borishaifa [10]
3 years ago
5

Why does a skateboard coasting on a flat surface slow down and eventually comes to a stop

Physics
2 answers:
Maurinko [17]3 years ago
5 0

It depends on what skateboard. It will stop mainly because some  skateboards need a hill or curvy platform to function.

AveGali [126]3 years ago
5 0
The skateboard will slow down because of kinetic and static friction. Kinetic friction will cause it to slow down while it’s moving. It will eventually slow down to a point that its velocity can’t overpower the static friction, and it will stop.
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A sailcraft is stalled on a windless day. A fan is attached to the craft and blows air into the sail which bounces backward upon
mylen [45]

Answer:

Impulse = change in momentum w bounce

There are 2 impulses acting. Recoil of the fan going the negative direction and the impulse of the air bouncing off the sail. The greater impulse will bounce so the direction will be to the right moving the craft.

7 0
2 years ago
A lever has an input arm 50 centimeters long and an output arm 40 centimeters long. What is the mechanical advantage of the leve
taurus [48]

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4 0
3 years ago
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Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

5 0
3 years ago
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I found this!! maybe this will help :)

8 0
2 years ago
What did scientists predict the big bang should have left ??
ELEN [110]
<span>radiation, hydrogen, and helium </span>
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