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Lunna [17]
3 years ago
9

Electric circuits and electric current

Physics
1 answer:
lidiya [134]3 years ago
6 0

an electric current is a flow of electric charge in electric circuits this is carried by moving electrons in a wire and an electric circuit is an electrical network of electrical components and model of interconnection consisting electrical elements

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A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a
Galina-37 [17]

Answer:

E   = (0.56 \times 10^8 ) r   \   \ N/c

Explanation:

Given that:

\rho_o = (10^{-3} ) \ c/m^3

R = (0.1) m

To find  the electric field for r < R by using Gauss Law

{\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)

For r < R

Q_{enclosed}=(\rho) ( \pi r^2 ) l

E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}

E= \dfrac{\rho ( r)}{2 \varepsilon_o}

where;

\varepsilon_o = 8.85 \times 10^{-12}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E   = (0.56 \times 10^8 ) r   \   \ N/c

4 0
3 years ago
A constant magnetic field can be used to produce an electric current. True or False?
Umnica [9.8K]

this question is true

6 0
3 years ago
An empty paper cup is the same temperature as the air in the room. A student fills the cup with cold water. Which of the followi
Vika [28.1K]

Answer:

C

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It is

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3 years ago
When y⁷ is multiplied by y⁹ the answer is?
Allushta [10]

Answer:

y^16

Explanation:

who need to add the exponents only

7 + 9 = 16

therefore, the answer is y^16

5 0
3 years ago
A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp
Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0
2 years ago
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