? What is the difference between the Primitive cell and convectional cell

s344n2d4d5 [400]

I would say B but I’m not 100%

5 0

2 years ago

Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? S

Zepler [3.9K]

Answer:

About two kilometers away

$\rm distance=2.401\ km$

Explanation:

Given:

The time gap between the light and sound to travel to the house, $t=7\ s$

Since the clouds are formed in the troposphere region of the atmosphere which extends from 8 kilometers to 12 kilometers above the earth-surface and the velocity of light is 300000 kilometers per second so it is visible almost instantly, hence we neglect the time taken by the light to travel to the house from the clouds.

∴Distance between the lightning-strike and the house:

$\rm distance=v\times t$

we have the speed of sound as: $v=343\ m.s^{-1}$

So,

$\rm distance=343\times 7$

$\rm distance=2401\ m$

$\rm distance=2.401\ km$

6 0

2 years ago

We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 4.2

monitta

Answer:

The number of atoms is $N = 4.37*10^{22} \ atoms$

Explanation:

From the question we are told that

The mass of coin is $m_n = 4.2g$

Number of atom in one mole = $n =6.02*10^{23} \ atoms$

Molar mass of nickel $M = 57.8g$

Now the relation to obtain the number of atom in the nickel coin is

$N = \frac{Mass \ of Nickel\ coin}{Molar \ mass\ of nickel } * No\ of\atoms \ in \ \ one\ mole\ of\ nickel$

$= \frac{4.2}{57.8}* 6.02*10^{23}$

$=4.37 *10^{22} atoms$

8 0

3 years ago

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0

3 years ago

All of the following involve waves of electromagnetic energy except _______.

r-ruslan [8.4K]

All of the following involve waves of electromagnetic energy except the rumble of thunder during a storm.

Electromagnetic waves are used to transmit long/short/FM wavelength radio waves, and TV/telephone/wireless signals or energies. They are also responsible for transmiting energy in the form of microwaves, infrared radiation (IR), visible light (VIS), ultraviolet light (UV), X-rays, and gamma rays.

The correct answer between all the choices given is the second choice or letter B. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0

3 years ago

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