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OLga [1]
3 years ago
11

The back shelf of a car has a physics textbook on it. The coefficient of static friction of 0.44 between the book and the surfac

e. If the car is initially moving at a speed of 16 m/s, in how short a distance can the car be stopped at constant acceleration without causing the book to slide forward (and hit the driver in the back of the head)? (This is why you should never keep packages on the back shelf of your car.)
Physics
1 answer:
Law Incorporation [45]3 years ago
8 0

Answer:29.68 m

Explanation :  

Given

coefficient of friction \mu =0.44

Car initial speed=16 m/s

The maximum acceleration that book can bear without sliding is \mu g

a_{max}=0.44\times 9.8=4.312 m/s^2

using v^2-u^2=2as

Final velocity v=0

-(16)^2=2(-4.31)s

s=\frac{256}{2\times 4.312}

s=29.68 m

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Answer:

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The _____ deflects these winds to the right in the northern hemisphere and to the Southern Hemisphere
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Coriolis Effect

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The Coriolis effect is responsible for the deflection of winds to the right in the Northern hemisphere and to the right in the Southern hemisphere. It is an effect that occurs because of the rotation of the earth around its axis.

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Does the moon actually change shape as it revolves/rotates around the Earth? Explain your answer
rusak2 [61]

<u>Answer:</u>

<em>The moon doesn’t change shape on its own.</em>

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It just reflects off the light from the sun. Due to tidal locking phenomenon one face of the moon permanently faces the sun. Because of the changes in position of moon with respect to the sun the moon is lighted up variably giving rise to various phases like new moon, full moon, crescent etc.

3 0
3 years ago
The linear charge density on the inner conductor is and the linear charge density on the outer conductor is
Lubov Fominskaja [6]

Complete Question

The complete question is  shown on the first uploaded image (reference for Photobucket )

Answer:

The  electric field is  E = -1.3 *10^{-4} \ N/C

Explanation:

 From the question we are told that

    The linear charge density on the inner conductor is  \lambda _i  =  -26.8 nC/m  =  -26.8 *10^{-9} C/m

    The linear charge density on the outer conductor is

  \lambda_o  = -60.0 nC/m  =  -60.0 *10^{-9} \ C/m

     The position of interest is r =  37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field  at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

         E (2 \pi r l) =  \frac{ \lambda_i }{\epsilon_o}

=>    E =  \frac{\lambda _i }{2 \pi *  \epsilon_o * r}

substituting values  

       E =  \frac{ -26 *10^{-9} }{2 * 3.142 *  8.85 *10^{-12} * 0.0373}

       E = -1.3 *10^{-4} \ N/C

The  negative  sign tell us that the direction of the electric field is radially inwards

    =>   |E| = 1.3 *10^{-4} \ N/C

5 0
3 years ago
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