All categories

2 years ago

Describe about simple cell. with well label diagram.

1 answer:

8 0

A simple electric has two terminals i.e. negative and positive, And it has a top and a bottom namely Metal Cap and Metal disc.

$\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}$

<h3>Description of each part :-</h3>

Metal cap:- The part which is above the positive terminal, a cap of metal through which the positively charged electrons are passed.
Metal disc:- The part which is below the negative terminal, a disc of metal through which the negatively charged electrons are passed.
Positive terminal:- In a circuit the current flows from positive terminal to negative terminal.
These terminals are also known as cathode and anode.
Cathode is the other name for positive terminal and anode is the other name for negative terminal.

You might be interested in

What was the main view how the world worked geologically prior to the 1960s? It was generally believed that continents and ocean

attashe74 [19]

Answer:

It was generally believed that mountains were produced by vertical forces

Explanation:

The main view of the world worked geologically prior to the 1960s was that the mountains were formed by the vertical forces of nature.

The early people prior to 1960s believed in many different natural phenomenons and they give their own reasons for their occurrence. But later many researchers and geophysicists studied the formation of the earth and came with possible answers to these questions.

Thus the answer is

" It was generally believed that mountains were produced by vertical forces."

3 0

3 years ago

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ

vitfil [10]

Answer:

0.85c

Explanation:

Rest mass of Kaon $M_{0K}$ = 494 MeV/c²

Rest mass of proton $M_{0P}$ = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy $E_{0K}$ = 494c² MeV

for the proton, rest energy $E_{0P}$ = 938c² MeV

Recall that the rest energy, and the total energy are related by..

$E$ = γ $E_{0}$

which can be written in this case as

$E_{K}$ = γ $E_{0K}$ ...... equ 1

where $E$ = total energy of the kaon, and

$E_{0}$ = rest energy of the kaon

γ = relativistic factor = $\frac{1}{\sqrt{1 - \beta ^{2} } }$

where $\beta = \frac{v}{c}$

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

$E_{K}$ = $E_{0P}$ ......equ 2

where $E_{K}$ is the total energy of the kaon, and

$E_{0P}$ is the rest energy of the proton.

From $E_{K}$ = $E_{0P}$ = 938c²

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = $\frac{1}{\sqrt{1 - \beta ^{2} } }$ = 1.89

1.89 $\sqrt{1 - \beta ^{2} }$ = 1

squaring both sides, we get

3.57( 1 - $\beta^{2}$ ) = 1

3.57 - 3.57 $\beta^{2}$ = 1

2.57 = 3.57 $\beta^{2}$

$\beta^{2}$ = 2.57/3.57 = 0.72

$\beta = \sqrt{0.72}$ = 0.85

but, $\beta = \frac{v}{c}$

v/c = 0.85

v = 0.85c

7 0

3 years ago

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

2 years ago

Calculate Speed The 2-kg metal ball moving at a speed of 3 m/s strikes a 1-kg wooden ball that is at rest. After the collision,

enot [183]

Answer:

53466kg.

Explanatiokn:

5 0

1 year ago

In a nuclear reactor, the control rods are: A.made of graphite B.used to keep the reactor cool C.used to absorb free neutrons D.

algol13

I think its E. the control rods are used to control the fission rate of uranium and plutonium. Hope this helps!!

4 0

3 years ago

Read 2 more answers

Describe about simple cell. with well label diagram.​

Describe about simple cell. with well label diagram.