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ICE Princess25 [194]
3 years ago
15

Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimu

m necessary. find the magnitude of the frictional force between the person and the wall.
Physics
2 answers:
Naddika [18.5K]3 years ago
7 0
In the magnitud is in good
Kamila [148]3 years ago
7 0

Answer:

friction force must be equal to the magnitude of weight of the object

Explanation:

When the object is stuck against the wall

then we will have

F_n = \frac{mv^2}{r}

also we know that

mg = F_f

so if the block is at rest and not sliding then in vertical direction the force must be balanced

so here we can say

F_g = F_f

so friction force must be equal to the magnitude of weight of the object

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Answer:

F = 6,000 N

Explanation:

F = m × a

F = 2,000 × 3

F = 6,000 kg m/s²

F = 6,000 N

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What does the energy of an electromagnetic wave depend on
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When a car of mass 1167 kg accelerates from 10.0 m/s to some final speed, 4.00  10 5J of work are done. Find this final speed.
Akimi4 [234]
  • Mass=1167kg
  • Initial velocity=u=10m/s
  • Acceleration=a=4m/s^2
  • Work done=105J=W
  • Final velocity=v=?
  • Force=F
  • Distance=d

Apply Newton's second law

\\ \tt\hookrightarrow F=ma

\\ \tt\hookrightarrow F=1167(4)=4668N

Now

\\ \tt\hookrightarrow W=Fd

\\ \tt\hookrightarrow d=\dfrac{W}{F}

\\ \tt\hookrightarrow d=\dfrac{105}{4668}

\\ \tt\hookrightarrow d=0.022m

Now

  • d be s

According to third equation of kinematics

\\ \tt\hookrightarrow v^2=u^2+2as

\\ \tt\hookrightarrow v^2=10^2+2(4)(0.022)

\\ \tt\hookrightarrow v^2=100+8(0.022)

\\ \tt\hookrightarrow v^2=100+0.176

\\ \tt\hookrightarrow v^2=100.176

\\ \tt\hookrightarrow v=10.001m/s

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2 years ago
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Answer:

2.96×10⁸ m/s

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Speed = distance / time

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