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3 years ago

Hey nsnsnsndhdhd dudhshdhdh

Physics

2 answers:

PolarNik [594]3 years ago

5 0

Answer:

hii

Explanation:

DochEvi [55]3 years ago

4 0

Answer:

Hi!

Explanation:

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What is the easiest way to increase the magnetic force acting on the rotor in an induction motor?

Schach [20]

Answer:

Explanation:

Magnets are of two major forms namely the permanent magnet and the temporary magnets. Temporary magnets magnetizes and demagnetize easily while permanent magnets does not magnetizes and demagnetize easily.

This permanents magnets are applicable in loudspeakers, generators, induction motor etc.

To increase the

The following will tend to increase the magnetic force acting on the rotor in an induction motor.

1. Increasing the strength of the bar magnet. Increase in strength of the magnet will lead to increase in the magnetic force acting on the rotor.

2. Increase in the magnetic line of force also known as the magnetic flux around the magnet will also increase the magnetic force acting on the rotor.

6 0

3 years ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

3 years ago

Which is harder, air tight or water tight seals?

Vladimir [108]

Should be an air tight seal

4 0

3 years ago

Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various

Alexxx [7]

Answer:

$Q_{2}=1200cm^{3}/s$

Explanation:

Given data

Q₁=200cm³/s

We know that:

$F=n\frac{vA}{l}\\$

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

n₂=6.0n₁

Q=ΔP/R

$Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s$

3 0

3 years ago

An object that is slowing down is accelerating.<br> True<br> False

Art [367]

Answer:

True

Explanation

When an object slows down the Acceleration is in the other direction which “ slows it down

8 0

3 years ago