A free electron is one which has become detached from a covalent bond between two atoms and is able to move around from atom to atom and possibly take part in electric current flow.

3 0

3 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

8 0

3 years ago

Each second, 1250 m3 of water passes over a waterfall 150 m high. Three-fourths of the kinetic energy gained by the water in fal

mote1985 [20]

Answer:

The generator produces electrical energy at a rate of 1378125000 J per second.

Explanation:

volume of water falling each second is 1250 $m^{3}$

height through which it falls, h is 150 m

mass of 1 $m^{3}$ of water is 1000 kg

⇒mass of 1250 $m^{3}$ of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8 $\frac{m}{sec^{2} }$

in falling through 150 m in each second, by Work-Energy Theorem:

Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it

⇒KE = mgh

= 1250000×9.8×150 J

= 1837500000 J

Electrical Energy = $\frac{3}{4}$ (KE)

= $\frac{3}{4}$ ×1837500000

= <u>1378125000 J per second</u>

8 0

3 years ago

Unpolarized light passes through a combination of two ideal polarizers. The transmission axes of the first polarizer and the sec

Yuliya22 [10]

Answer:

62.5 %

Explanation:

Let the initial intensity of unpolarized light is Io.

After first polariser the intensity of light becomes I'.

So, $I' = \frac{I_{0}}{2}$

Now it passes through another polariser. The angle between the first polariser and the second polariser is given by Ф. The intensity is I''.

According to the law of Malus

$I'' = I' Cos^{2}\phi$

Here, Ф = 30 degree

$I'' = \frac{I_{0}}{2} Cos^{2}30=0.375I_{0}$

The percentage change in the intensity is given by

$\frac{I_{0}-I''}{I_{0}}\times 100=\frac{I_{0}-0.375I_{0}}{I_{0}}\times100$

= 62.5 %

7 0

3 years ago

Watt (w) is a drived unit why​

Watt (w) is a drived unit why