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4vir4ik [10]
3 years ago
12

The kinetic friction coefficient between a cabinet and the floor is 0.3. Mass of the cabinet is 300kg. A man pushes the cabinet

along the floor with a 1000N force. Find the acceleration of the cabinet. g
Physics
1 answer:
podryga [215]3 years ago
7 0

Answer:

<h2>0.39m/s^2</h2>

Explanation:

Step one:

given data

mass m= 300kg

applied force F= 1000N

coefficient of friction μ= 0.3

Step two:

The net force Fn= applied force-friction force  

Fn=F-F1

F1= limiting force

F1=μ*m*g

F1=0.3*300*9.81

F1=882.9N

the Net force= 1000-882.9

Fn=117.1N

Step three:

we know that

F=ma

Fnet=ma

a= Fnet/m

a=117.1/300

a=0.39m/s^2

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Using the equation for decay, calculate the amount left of a radioactive sample amount N0 if the decay constant is 0.00125 secon
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The amount left of a radioactive sample amount N0 if the decay constant is 0.00125 seconds and the time is 180 seconds is 0.7999 N.

<h3>What is half-life?</h3>

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A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied
Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that  the minimum force that must be applied on the <em>book is</em>

  • F=44N

From the question we are told

A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the  Force  is mathematically given as

F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\

F=44N

Therefore

the minimum force that must be applied on the <em>book is</em>

F=44N

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