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lesya [120]
3 years ago
8

A 35 n object is on a 25 degree incline. the force of friction up the incline is 8.0 N.

Physics
1 answer:
luda_lava [24]3 years ago
4 0
The downwards component of weight of the object would be 35sin25 degrees = 14.8

F=14.8 - 8 = 6.8 N                                         m= 35/9.8= 3.57 kg
F=mA
Therefore,
3.57A= 6.8 N
=> A= 6.8/3.57
=> A= 1.902 ms^-2


F(max)= (U)R                       where (U)= coefficient of friction
                                                   and R= Normal reaction force

R= 35cos25
   = 31.72 N
Since, F(max)= 8
          8= (U)* 31.72
      =>(U)= 8/31.72
      =>(U)= 0.25
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Answer:

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v^2=u^2+2as

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's' is the distance it covers

Thus for maximum height applying the values in the equation we get

0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters

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v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s

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Answer:

B)

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The value the scale shows is the reaction force to the normal force (they are equal by Newton's 3rd Law) that the scale exerts on Eric.

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F=W-N=ma

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Block A of 5 kg with a speed of 3 m/s collides with block B of 10 kg that is stationary. After the collision, block B travels wi
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Answer:

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as well.

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3. Incorrect

Explanation

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4. incorrect

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5.  Incorrect

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8 0
3 years ago
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