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3 years ago

11

What is the difference between primary cells and secondary cells?

2 answers:

Rus_ich [418]3 years ago

7 0

Answer: The main difference is that while primary cells cannot be charged, secondary cells can be recharged.

Explanation:

A primary cell is an electrochemical that can't be recharged. The chemicals will be replaced regularly e.g. Leclanche cell.

The reactions in a secondary cell can be reversed through an external electrical energy source, heve, they can be recharged. Examples of secondary cells include nickel and lead storage battery.

The main difference is that while primary cells cannot be charged, secondary cells can be recharged.

TiliK225 [7]3 years ago

3 0

Answer:

The major difference between a primary cell and the secondary cell is that primary cells are the ones that cannot be charged but secondary cells are the ones that are rechargeable

Explanation:

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In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m a

balu736 [363]

Answer:

The value is $\lambda = 214.3 \ nm$

Explanation:

From the question we are told that

The slit separation is $d = 3.00 * 10^{-5} m$

The distance of the screen is $D = 2.00\ m$

The order of fringe is n = 7

The path difference is $y = 10.0 \ cm = 0.1 \ m$

Generally the path difference is mathematically represented as

$y = \frac{n * \lambda * D}{ d}$

=> $0.1 = \frac{7 * \lambda * 2.00 }{ 3.00 * 10^{-5}}$

=> $\lambda = \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }$

=> $\lambda = \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }$

=> $\lambda = 2.143 *10^{-7} \ m$

=> $\lambda = 214.3 \ nm$

6 0

3 years ago

Two 1 MHz radio antennas emitting in-phase are separated by600

algol [13]

Answer:

1 km

Explanation:

$d_0$ = Gap between antennas = 600 m

$\nu$ = Frequency = 1 MHz

$z_1$ = Distance to receiver = 2 km

c = Speed of light = $3\times 10^8\ m/s$

Wavelength is given by

$\lambda=\dfrac{c}{\nu}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1\times 10^6}\\\Rightarrow \lambda=300\ m$

Distance to be moved is given by

$D=\dfrac{z_1\lambda_0}{d_0}\\\Rightarrow D=\dfrac{2\times 300}{600}\\\Rightarrow D=1\ km$

The distance to be moved is 1 km north.

6 0

3 years ago

If you increase the frequency of a sound wave four times, what will happen to its speed?

ozzi

The speed will increase four (4) times when the frequency of sound wave is increased four (4) times.

<h3>How to calculate the frequency of a wave.</h3>

Mathematically, the frequency of a sound wave is given by this formula:

$F = \frac{V}{\lambda}$

<u>Where:</u>

F is the frequency of a sound wave.

V is the velocity of a sound wave.

$\lambda$ is the wavelength of a sound wave.

<u>Note:</u> The frequency of the sound wave is increased four (4) times.

F = 4F

Substituting the given parameters into the formula, we have;

$4F = \frac{V}{\lambda}\\\\V=4F\lambda$

Therefore, the speed will increase four (4) times when the frequency of sound wave is increased four (4) times.

Read more on sound wave here: brainly.com/question/19250621

4 0

3 years ago

The belief that race indicates certain traits and capacities and that one's own race is superior is called

EleoNora [17]

The answer is RACISM.

7 0

4 years ago

A Jack Rabbit hops 12.8 meters per second. How long would it take for him to hop 353 m? Round to the nearest tenth place.

adoni [48]

Time = Distance/speed

353/12.8 =27.57 seconds

4 0

3 years ago