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Gnoma [55]
3 years ago
9

Metals experience plastic deformation when _____.

Physics
1 answer:
SOVA2 [1]3 years ago
3 0
THE FORCE IS REMOVED BUT THE BODY DOES NOT RETURN TO ITS' ORIGINAL SHAPE (ALSO CALLED PERMANENT DEFORMATION)
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What property or properties of a wave determine it's speed?
mash [69]
The properties of the wave don't determine its speed. The properties of the medium do. You can FIND the speed by measuring the wave's frequency and wavelength.
7 0
3 years ago
Which of the following represents the greatest energy transition from a higher energy level to a lower one?
prohojiy [21]
You need to use Planck's law:
E = h·υ = (h·c)/λ

Without making all the calculations, a fraction is bigger than another when the denominator is smaller. Therefore you need to find the smallest wavelength (λ) which is 450nm.

You could also be helped by colors: in order of decreasing energy, you have blue - green - yellow - red.

In any case, the correct answer is a).
8 0
3 years ago
3. When encountering low visibility from rain or fog, you
zimovet [89]

Answer:

c. low beams and fog lights

Explanation:

When encountering low visibility from rain or fog, use your low beams and  fog lights. High beams will only increase the glare. If you can't see at least five seconds in front of you, don't drive. Pull over and put your hazards on until it clears up.

i just got the answer wrong and the drivers ed gave me this explanation !!

8 0
3 years ago
Read 2 more answers
The spectra of most galaxies show redshifts. this means that their spectral lines _________.
spin [16.1K]
Have wavelengths that are longer than normal.
3 0
2 years ago
A boy throws a ball with an initial velocity of 19.6 m/s. What maximum height does the ball reach?
Wewaii [24]
<h2>Hello!</h2>

The answer is: 19.59 m

<h2>Why?</h2>

Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.

When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:

We will use the following formula:

Vf^2=Vi^2+2*g*s

Where:

Vf= Final velocity = 0

Vi= Initial velocity = \frac{19.6m}{s}

g = Gravity Acceleration = \frac{9.81m}{s^{2} }

s = Traveled distance

0=19.6^2+2*-9.81*s\\s=\frac{19.6^2}{2*9.81}=\frac{384.16}{19.62}=19.59m

Have a nice day!

8 0
2 years ago
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