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3 years ago

If the heating curve is reversed, what would best describe the boiling point?

Physics

1 answer:

Slav-nsk [51]3 years ago

7 0

Answer:

point of condensation

Explanation:

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Charges flow from___<br> voltage to____<br> voltage.

iogann1982 [59]

Answer:

high to low

Explanation:

5 0

3 years ago

You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-

Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

$E_P=\int dE \cos$

$E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}$

$\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ$

$\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}$

If we put an electron on point P, then force on point e is :

$\vec{F}=-|e|\vec{E_P}$

$F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}$

If r >> x , then $\frac{x^2}{r^2} \approx 0$

Then, $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0

3 years ago

What is eletro magnet

Leona [35]

A soft metal core made into a magnet by the passage of electric current through a coil surrounding it.

7 0

3 years ago

The power of a bulb is 100 watt .what does it mean

natita [175]

Answer:

Explanation: 100 watts is a unit of power.

A watt is also the amount of energy being consumed. So the more watts the brighter the light bulb is lit.

3 0

3 years ago

A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when

choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

$F_t - mg = ma$

here since speed is constant or it is at rest

so we will have

$a = 0$

$F_t = mg$

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

$F_t - mg = ma$

$F_t = mg + ma$

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

$F_t - mg = -ma$

$F_t = mg - ma$

so tension force is less than the weight of crate if it is accelerating downwards

4 0

3 years ago