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Y_Kistochka [10]

3 years ago

14

During lightning strikes from a cloud to the ground, currents as ... currents as high as 2.50×10^4 amps can occur and last for a

bout 40.0 microseconds . how much charge is transferred from the cloud to the earth during such a strike?

1 answer:

scZoUnD [109]3 years ago

7 0

Hello!

The answer is 1.00 Coulomb

Thank you!!!

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A horizontal force of 675 N is needed to overcome the force of static friction between a level floor and a 300-kg crate. What is

White raven [17]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

Monochromatic light passes through a double slit, producing interference, the distance between the slit centres is 1.2 mm and th

Alik [6]

Answer:

The wavelength of the light is $7200\ \AA$ .

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes $\beta= 0.3\ cm$

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

$\beta=\dfrac{D\lambda}{d}$

Put the value into the formula

$0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}$

$\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}$

$\lambda=7.2\times10^{-7}\ m$

$\lambda=7200\ \AA$

Hence, The wavelength of the light is $7200\ \AA$ .

8 0

2 years ago

Distance versus Displacement Worksheet

Umnica [9.8K]

when we find the distance we will add all the blocks so

distance = 6+6+4

distance = 14blocks

when we find the displacement we will add and minus too

As you can read he goes to the south 6 and to north 6 so he leave that place and back to the place again so the displacement is 0. and again he goes to the west 4 blocks so the displacement = <em><u>4blocks</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>west</u></em>

6 0

3 years ago

Read 2 more answers

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

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3 years ago

What is the disadvantage of the parallax method, especially for studying distant parts of the Galaxy?

Katena32 [7]

Answer and Explanation:

Parallax method is used for finding the distance of objects in space there are two types of parallax method that is stellar parallax and trigonometric parallax.The disadvantage of using parallax method is that it can can not reach so far in the Galaxy due to this reason parallax method is generally not used for measuring distance in galaxy.

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3 years ago