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Pavlova-9 [17]
3 years ago
13

A force of 100N moves a body on a horizontal frictionless surface when......

Physics
1 answer:
zalisa [80]3 years ago
4 0

Answer:

I. Friction force exerted on the body is less than 100N

Explanation:

For  a body to be static, the moving force must be equal to the frictional force. Since the frictional force is a force of opposition. It tends to oppose the moving force acting on an object.

Hence if the moving force is greater than the force of friction, the Force of fiction will not be able to overcome the moving hence the body will tend to move.

Therefore, for a body to move, Fm > Ff or Ff < Ff

Fm is the moving force

Ff is the force of friction

Given

Fm = 100N

For the 100N body to move the frictional force must be less than 100N

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3 years ago
If 1.8 1016 electrons enter a light bulb in 3 milliseconds, what is the magnitude of the electron current at that point in the c
Mice21 [21]

Answer:

I = 0.96 A

Explanation:

No of electrons, n=1.8\times 10^{16}

Time, t = 3 ms = 3\times 10^{-3}\ s

We need to find the electric current. We know that electric charge per unit time is equal to the electric current.

I=\dfrac{q}{t}

q = ne (Quantization of electric charge)

I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A

So, the electric current is 0.96 A.

7 0
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A 15 kg dog jumps out a stationary sled which has a mass of 40 kg. If
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Velocity of the sled is 3.2 m/s
5 0
2 years ago
A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

q_1 = 1.11 nC

q_2 = 8.88 nC

q_3 = 71 nC

Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

E_3 = 15975 N/C

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

V = 0.0335 m^3

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

3 0
3 years ago
True or false atoms may be held together by an exchange of electrons​
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The answer is true...............
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3 years ago
Read 2 more answers
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