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V125BC [204]
3 years ago
7

Four of the types of electromagnetic waves are microwaves,radio waves, infreqd waves, and visible light waves which has the most

energy
Physics
1 answer:
harkovskaia [24]3 years ago
7 0

Answer:

Explanation:

You can nuke your breakfast with microwaves. That is really not possible with light waves but distance is involved here or radio waves. B and C are not true. D is a little troublesome. Infrared rays do not have more energy than microwaves but they do have energy.

A is the answer.

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Answer:

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Explanation:

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4 0
3 years ago
A positively charged particle is moving toward the South in a region of space where there is a magnetic field directed down (tow
Alexxandr [17]

Answer:

F = Q V X B

V = south

B = down

V X B = east

8 0
2 years ago
What is the wavelength of a proton traveling at a speed of 6.21 km/s? what would be the region of the spectrum for electromagnet
lukranit [14]
<span>The wavelength of a proton traveling at a speed of 6.21 km/s is wavelength = (6.63E-34 Js) / (1.67Eâ’27kg x 6.21E3m/s) = 6.393E-9 m Region of the spectrum for electromagnetic radiation of this wavelength will be Xrays.</span>
8 0
4 years ago
A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s.
ELEN [110]
Momentum = mass x velocity

Before collision
Momentum 1 = 2 kg x 20 m /s = 40 kg x m/s
Momentum 2 = 3 kg x -10m/s = -30 kg x m/s

After collision
Momentum 1 = 2 kg x -5 m/s = -10 m/s
Momentum 2 = 3 kg x V2 = 3V2

Total momentum before = total momentum after
40 + -30 = -10 + 3V2
V2 = <span>6.67 m/s

Total kinetic energy before
</span><span>= (1/2) [ 2 kg * 20 m/s * 2 + 3 kg * ( -10 m/s) *2 ]
= 550 J
</span>
<span>Total kinetic energy after
</span>= (1/2) [ 2 kg * ( - 5 m/s) * 2 + 3 kg * 6.67 m/s *2 ]
= 91.73 J

Total kinetic energy lost during collision
=<span>550 J - 91.73 J
= 458.27 J</span>

8 0
3 years ago
Read 2 more answers
A 0.17kg ball rolls at 0.75m/s to the right on a frictionless surface and collides with a 0.17kg ball rolling to the left at 0.6
mario62 [17]

Both momentum and kinetic energy are conserved in elastic collisions (assuming that this collision is perfectly elastic, meaning no net loss in kinetic energy)

To find the final velocity of the second ball you have to use the conversation of momentum:

*i is initial and f is final*

Δpi = Δpf

So the mass and velocity of each of the balls before and after the collision must be equal so

Let one ball be ball 1 and the other be ball 2

m₁ = 0.17kg

v₁i = 0.75 m/s

m₂ = 0.17kg

v₂i = 0.65 m/s

v₂f = 0.5

m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f

Since the mass of the balls are the same we can factor it out and get rid of the numbers below it so....

m(v₁i + v₂i) = m(v₁f + v₂f)

The masses now cancel because we factored them out on both sides so if we divide mass over to another side the value will cancel out so....

v₁i + v₂i = v₁f + v₂f

Now we want the final velocity of the second ball so we need v₂f

so...

(v₁i + v₂i) - v₁f = v₂f

Plug in the numbers now:

(0.75 + 0.65) - 0.5 = v₂f

v₂f = 0.9 m/s


8 0
3 years ago
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