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disa [49]
3 years ago
9

If I rub a polythene piece with a wool , is there a transfer of mass from wool to polythene?​

Physics
1 answer:
kirill [66]3 years ago
8 0

Yes, As a result, wool is positively charged while polythene is negatively charged. As a result, 1.87 1012 electrons have been transported from wool to polythene. As a result, only a sliver of mass is transferred from wool to polythene.

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A train is moving at a speed of 50 km/h. How many hours will it take the train to travel 600 kilometers?
EastWind [94]

Answer:

12 hours

Explanation:

600 divided by 50 is 12

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List the factors that affecting frictional force ?​
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Answer:

The frictional force between two bodies depends mainly on three factors: (I) the adhesion between body surfaces (ii) roughness of the surface (iii) deformation of bodies

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If we want to know the velocity that an object is traveling, we must know the
Serga [27]
If my memory serves me well, if we want to know the velocity that an object is traveling, we must know the <span>direction and speed. Velocity includes two these points listed in the previous sentence which means the answer is D.</span>
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3 years ago
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An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
3 years ago
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fomenos
The answer to that probably would be C excuse me if I am wrong.
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