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alexira [117]
3 years ago
5

Bob is pulling a 30 kg filing cabinet with a force of 200 N , but the filing cabinet refuses to move. The coefficient of static

friction between the filing cabinet and the floor is 0.80. What is the magnitude of the friction force on the filing cabinet?
Physics
1 answer:
Arada [10]3 years ago
5 0

Answer:

<em>Magnitude of the Frictional force is 200 N</em>

Explanation:

The frictional force is the force that tries to oppose relative motion between two surfaces that are contacting. The coefficient of static friction is the coefficient of friction of a body that is not moving.

Newton's third law of motion states that action and reaction forces are equal and opposite. So the frictional force felt on the filing cabinet will be equal to the applied force pulling the cabinet.

Frictional force = Force applied

Force applied  = 200 N

Therefore, the magnitude of the friction force on the filing cabinet is 200 N

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Write down one fact about each of the following crust, mantle, outer, core, &amp; inner core
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Answer:

Hope this help you!!

Explanation:

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Mantle : The mantle is the mostly-solid bulk of Earth's interior. The mantle lies between Earth's dense, super-heated core and its thin outer layer, the crust. The mantle is about 1,802 miles thick, and makes up a whopping 84% of Earth's total volume

Outer Core : The outer core is the third layer of the Earth. It is the only liquid layer, and is mainly made up of the metals iron and nickel, as well as small amounts of other substances. The outer core is responsible for Earth's magnetic field. As Earth spins on its axis, the iron inside the liquid outer core moves around.

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2 years ago
Fatima is watching her pet cat, Winter, napping in the sun. Fatima is curious about the heart rate of Winter when she is napping
svetoff [14.1K]

Answer:

Explanation:

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A cat's heart rate changes while it is napping.

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3 years ago
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The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

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Explanation:

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