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Lina20 [59]
2 years ago
8

Name and Title:

Physics
1 answer:
Natalka [10]2 years ago
6 0

The variable that is altered by the independent variable is called the dependent variable. the spectrum given out by the hypothetical unknown celestial object.

Name : Christopher Andre

Instructor name : Albert brown

Name of the lab : ELS Laboratory

The term electromagnetic spectrum refers to the range of electromagnetic radiation's frequencies, as well as the wavelengths and photon energies connected to each frequency.

By analyzing the absorption spectra of the planets and moons, the electromagnetic spectrum experiment seeks to identify the components that make up their atmospheres.

From below one hertz to over 1025 hertz, electromagnetic waves are included in the electromagnetic spectrum.

The wavelengths that correlate to the frequency range from tens of thousands of kilometers to a small portion of the size of an atomic nucleus.

Starting at the low frequency (long wavelength) end of the spectrum, each frequency band's electromagnetic waves are referred to by a variety of names.

Hence the dependent variable is the one that changes as a result of the independent variable.

Learn more about electromagnetic spectrum here

brainly.com/question/13803241

#SPJ10

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If your prediction turns out to be false your experiment has been a failure
nikdorinn [45]

This is false. Your hypothesis, or prediction, is just that: a prediction. Saying its a failure will result in bias.

3 0
3 years ago
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According to Kepler, what do all bodies in orbit around another have in common?
dolphi86 [110]
D.
All follow an elliptical path that has two foci, rather than a circular path.

Kepler found paths are elliptical, not circular
3 0
3 years ago
A balloon filled with helium gas at 1.00 atm occupies 15.6 L. Will the volume of the balloon increase or decrease in the upper a
Mariana [72]

Answer:

The volume of the balloon increases in the upper atmosphere.

Explanation:

p1= 1 atm

p2= 0.15 atm

V1= 15.6 L

V2= ?

p1*V1= p2 * V2

V2= (p1/p2)*V1

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6 0
3 years ago
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

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65 years but anything can happen to them
I’m not really sure but I hope this helps
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