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Elden [556K]
3 years ago
6

A 100 N box sits on a 30 degree incline. If the static coefficient of friction is 0.1, what is the magnitude of the static frict

ional force acting on the box? Select one: a. 5.0 N
b. 8.7 N
c. 10N

Physics
1 answer:
Semmy [17]3 years ago
8 0

Answer:

f = 8.7 N                        

Explanation:

It is given that,

Weight of the box, W = 100 N

It is inclined at an angle of 30 degrees.

The coefficient of friction, \mu_s=0.1

We need to find the magnitude of the static frictional force acting on the box. Let f is the frictional force. It is given by :

f=\mu_s\times N

N is the normal force

f=\mu_s\times mg\ cos\theta  

f=0.1\times 100\times \ cos(30)  

f = 8.66 N

or

f = 8.7 N

So, the the static frictional force acting on the box is 8.7 N. Hence, this is the required solution.

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In a pig caller can produce a sound intensity level of 107 dB. How many pig callers would be needed to generate an intensity lev
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Answer:

20 pig callers

Explanation:

Given that:

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we have:

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Taking the logarithm function;

10 \ log \bigg(\dfrac{I}{I_o} \bigg) = 13 \ dB

where;

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log \bigg(\dfrac{I}{I_o} \bigg) = 1.3

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2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially
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Answer:

 E_particle = 1,129 10⁻²⁰ J / particle

  T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

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Gas b

         E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

          E_total = E_a + E_b

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The average energy among the entire mass is

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               T = E / k

     

               T = 1,129 10⁻²⁰ / 1,381 10⁻²³

               T = 8.175 102 K

               T= 817.5 K

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