All categories

3 years ago

How are flow of electrons in a conductor and the direction of conventional current related??

Physics

1 answer:

lys-0071 [83]3 years ago

5 0

Answer:

flow of electron is just opposite to the flow for current. Electrons flow from the negative terminal to the positive. Conventional current or simply current, behaves as if positive charge carriers cause current flow. Conventional current flows from the positive terminal to the negative.

You might be interested in

Suppose a 49-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 585 N sits on t

Andre45 [30]

Answer:

69.74 N

Explanation:

We are given that

Weight of sled=49 N

Coefficient of kinetic friction $,\mu_k=0.11$

Weight of person=585 N

Total weight==mg=49+585=634 N

We know that

Force needed to pull the sled across the snow at constant speed,F=Kinetic friction

$F=\mu_k N$

Where N= Normal=mg

$F=0.11\times 634=69.74 N$

Hence, the force is needed to pull the sled across the snow at constant speed=69.74 N

7 0

3 years ago

State one of Kepler's laws of gravitation

jekas [21]

All planets move around the sun

3 0

3 years ago

Read 2 more answers

If you drop a silver dollar off a building and it hits the ground in 10 seconds, how fast was the coin going just before it hit?

Illusion [34]

Objects in free fall, disregarding terminal velocity, accelerate at 9.8(m/s)/s. so for every second it was falling, it gained 9.8m/s in speed. 9.8 * 10 = 98m/s

4 0

3 years ago

Read 2 more answers

A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec

kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, $\lambda=\dfrac{\lambda_ox_o}{x}$

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

$dE=\dfrac{k\ dq}{x^2}$ ..........(1)

The linear charge density is given by :

$\lambda=\dfrac{dq}{dx}$

$dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx$

Integrating equation (1) from x = x₀ to x = infinity

$E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx$

$E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}$

$E=\dfrac{k\lambda_o}{2x_o}$

Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

What will happen to the current if the voltage is reduced to one half?

stira [4]

We use v=IR and assuming the resistance doesn’t change we can also say that the voltage and current (I) are directly proportional which means the voltage also decreases by 1/2

8 0

2 years ago

How are flow of electrons in a conductor and the direction of conventional current related??​

How are flow of electrons in a conductor and the direction of conventional current related??