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katovenus [111]
3 years ago
10

A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that th

e crate just begins to move when your force exceeds 321 N . After that you must reduce your push to 205 N to keep it moving at a steady 22.3 cm/s . Part A What is the coefficient of static friction between the crate and the floor? μsμ s = nothing Request Answer Part B
Physics
1 answer:
iragen [17]3 years ago
8 0

Answer:

 μ = 0.725

Explanation:

This problem refers to Newton's second law.

       F = ma

Let's write the equations on each axis

Y Axis

      N-W = 0

     N = W

    N = mg

X axis

   F-fr = ma

With the body not started moving its acceleration is zero

  F-fr = 0

  F = fr

The friction force equation is

  fr = μ N

  fr = μ m g

Let's replace and calculate

   F = μ m g

   μ = F / mg

   μ = 321 /45.2 9.8

   μ = 0.725

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2 years ago
The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values
Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

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solution

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we know sound level  β  = 10 log(I/I_{0})        ......................a

put here value (I/I_{0}) = 10^−12 W/m² and  β = 115

115  = 10 log(I/10^−12)

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power = 0.316228 × 4π (2.30)²

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and

at 10 dB intensity is 1 × 10^–11 W/m²

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Answer:

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