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Lubov Fominskaja [6]
3 years ago
9

20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?

Physics
1 answer:
Tresset [83]3 years ago
7 0
Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance 
so Work done = 0.085 x 20g x 28
 = 466.956 J

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Riley is cleaning his room(a once a year activity) and pulls the Hoover with a force of 8 N a total distance of 20 metres. How m
Radda [10]

Answer:

160J

Explanation:

Given force = 8N and total distance = 20 meters

Workdone = force x distance

= 8 x 20

= 160J

Therefore, workdone by Riley in pulling the hoover is 160J

7 0
3 years ago
All of the following equations are statements of the ideal gas law except
elena55 [62]

Answer:

a. P = nRTV

Explanation:

The question is incomplete. Here is the complete question.

"All of the following equations are statements of the ideal gas law except a. P = nRTV b. PV/T = nR c. P/n = RT/v d. R = PV/nT"

Ideal gas equation is an equation that describes the nature of an ideal gas. The molecule of an ideal gas moves at a particular velocity depending on the temperature. This gases collides with one another elastically. The collision that an ideal gas experience is a perfectly elastic collision.

The ideal gas equation is expressed as shown:

PV = nRT where:

P is the pressure of the gas

V is the volume

n is the number of moles

R is the ideal gas constant

T is the temperature.

Based on the formula given for an ideal gas, it can be inferred that the equation. P = nRTV is not a statement of an ideal gas equation.

The remaining option will results to an ideal gas equation if they are cross multipled.

7 0
4 years ago
Please help with 4 and 5, thank you :)
allochka39001 [22]

Answer: #4

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6 0
3 years ago
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A
ArbitrLikvidat [17]

Answer:

The mass of the solid cylinder is m  =  1612.5  \  kg

Explanation:

From the question we are told that

   The radius of the grinding wheel is R =  0.330 \ m

   The  tangential force is F_t =  250 \ N

    The angular acceleration is  \alpha  =  0.940 \ rad/s^2

The torque experienced by the wheel is mathematically represented as

     \tau  =  I  *  \alpha

Where  I  is the moment of inertia

The torque experienced by the wheel can also be  mathematically represented as

       \tau  =  F_t  * r

substituting values

       \tau  =  250 * 0.330

      \tau  = 82.5  \ N\cdot m

So

   82.5  =  I *  \alpha

    82.5  =  I *  0.940

So

   I  =  87.8 \ kg \cdot m^2

This moment of inertia can be mathematically evaluated as

     I  =  \frac{1}{2} * m* r^2

substituting values

  87.8  =  \frac{1}{2} * m* (0.330)^2

=>   m  =  1612.5  \  kg

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3 years ago
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They prefer leaves i believe
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