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3 years ago

The amount of friction divided by the weight of an object forms a unit less number called the blank of friction

Physics

1 answer:

Mekhanik [1.2K]3 years ago

8 0

Answer:

Coefficient

Explanation:

For an object on level ground, the ratio of the friction force to the weight is the coefficient of friction.

F / W = μ

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2 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

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Answer:

Part a)

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Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

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Part c)

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$E_2 = 7992 N/C$

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Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

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now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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