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3 years ago

5

Explain how a Faraday motor works, please. Help me out

1 answer:

Simora [160]3 years ago

6 0

The battery makes electrons flow through the circuit, and the electrons move from the copper wire that's hanging down are free to move through the salty water to the aluminum foil. This causes electrons to flow through the wire.

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Joe rides his bicycle in a straight line for 2 hour with an average velocity of 13 km/h south, how far has he ridden?

dmitriy555 [2]

Answer:

ans is 3.125 km

i hope this is correct

8 0

3 years ago

What is an Atwood Machine?

Lady_Fox [76]

The best answer is letter (A) a double pulley system. Atwood Machine is normally used as a measurement in balancing to object to verify the mechanical law of motion with constant acceleration.

6 0

3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

3 years ago

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.

VMariaS [17]

Answer:

A) $d=\dfrac{1}{2F}mv^2$

B) $\Delta KE'=2\times \dfrac{1}{2}mv^2$

Explanation:

Given that

Force = F

Increase in Kinetic energy = $\dfrac{1}{2}mv^2$

$\Delta KE=\dfrac{1}{2}mv^2$

we know that

Work done by all the forces =change in the kinetic energy

a)

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

$F.d=\dfrac{1}{2}mv^2$

$d=\dfrac{1}{2F}mv^2$

b)

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE' ( F.d =Δ KE)

2ΔKE = ΔKE'

$\Delta KE'=2\times \dfrac{1}{2}mv^2$

Therefore the final kinetic energy will become the twice if the force become twice.

8 0

3 years ago

a 0.4 kg block rests on a desk. the coefficient of static friction is 0.2. You push the side of the block but do not have a spri

ladessa [460]

Static friction is greater than Applied force

7 0

3 years ago

Read 2 more answers