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3 years ago

14

Which of the following is an elastic collision?

2 answers:

UNO [17]3 years ago

6 0

I think its B. Two cars colliding and sticking together.or C. A wad of gum sticking to a window.

fiasKO [112]3 years ago

3 0

D is also an example

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When a loose brick is resting on a wall, it has energy. When the brick is pushed off the wall and is falling down, the amount of

OLga [1]

Potential
Potential
Kinetic

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3 years ago

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A man walks 6 hrs from his house to his office at an average speed of 2km/hr.find the difference between his house and office

laiz [17]

4 is the difference sorry if i got it wrong :( :(

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3 years ago

How do you change matter into other phases of matter?

Vera_Pavlovna [14]

Hey there!

There's many ways to do it - like melting and evaporating.

For example, we'll use water. Plain old water in a water bottle. Right now, it's in its liquid state of matter, but say you put it in the freezer for an hour. That would change its state of matter to solid, since it would be solid ice. Now, if you were to put it out in the sun on a blazing hot day for a couple of hours, it would evaporate and become water vapor, a gas. Lastly, if you can cool that water vapor it becomes a liquid again.

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3 0

3 years ago

You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength

Ray Of Light [21]

Answer:

wavelength $\lambda$ = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = $\frac{D\lambda}{d}$ ....................1

$\lambda$ is Wavelength of light and D is Distance between screen and slit and d is slit width

so put here value and we get

$\lambda$ = $\frac{2.96*0.325*10^{-6}}{2.20}$

$\lambda$ = 437.27 × $10^{-9}$ m

wavelength $\lambda$ = 437.27 nm

4 0

3 years ago

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

muminat

Answer:

$2.1\cdot 10^{21}$ electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a charged sphere, so we have:

$E_b=3.00\cdot 10^6 N/C$ is the maximum electric field strength tolerated by the air before breakdown occurs

$r=1.00 km = 1000 m$ is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C$

Assuming that the cloud is negatively charged, then

$Q=-333.3 C$

And since the charge of one electron is

$e=-1.6\cdot 10^{-19}C$

The number of excess electrons on the cloud is

$N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}$

5 0

3 years ago