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3 years ago

Which of the following best describes why we say that light is an electromagnetic wave?

Physics

1 answer:

Kaylis [27]3 years ago

8 0

Answer: The passage of a light wave can cause electrically charged particles to move up and down.

Explanation:

Electromagnetic waves are transversal waves, they are a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.

This means the oscillation of the wave occurs in the transversal direction to its propagation. In addition, electromagnetic waves are spread thanks to the electromagnetic fields produced by moving electric charges.

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I really need help with this, i dont know what to do first, PLEASE I NEED HELP DUE TOMORROW!!!

enot [183]

Can I tell you what makes this problem so hard ?

It's having all the data WITHOUT HAVING THE STORY !

We first have to figure out what all those things are. I mean, we don't even know what F is, what d is, what Kef or Vi is, or how W figures in to the whole thing. You really have no mercy !

If my hunch is correct, the story goes like this:

-- There's an object sailing along, minding its own business, not bothering anybody, and its speed is 7.2 meters per second.

-- Somebody jumps out in front of the object and begins to push back on it with 215 Newtons of force, trying to slow it down and stop it.

-- The object is only able to go another 13 meters, pushing the guy backwards but slowing down, and then it stops.

-- The question is: What is the mass of the object ?

Now I'll go ahead and solve the problem that I just invented:

-- Kinetic energy = (1/2) (mass) (speed²)

Before anybody touched it, the object's kinetic energy was

KE = (1/2) (mass) (7.2 m/s)²

KE = (25.92) x (mass)

-- Since that's the energy the object had, THAT's how much work the guy has to do in order to make the object stop.

Work = (force) x (distance)

Work = (215 N) x (13 meters)

Work = 2,795 N-m

-- And there you go. The work the guy did to stop the object is the amount of energy the object had before he came along.

(25.92) x (mass of the object) = 2,795 N-m

Divide each side by 25.92:

Mass of the object = (2,795 N-m) / (25.92)

<em>Mass = 107.83 kilograms</em>

5 0

3 years ago

b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

Nat2105 [25]

Answer:

b) Betelgeuse would be $\approx 1.43 \cdot 10^{6}$ times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is $\approx 1.37 \cdot 10^{-5} }$ the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

$B = \displaystyle{\frac{L}{4\pi d^2}}$ (1)

where $B$ is the brightness, $L$ is the star luminosity and $d$ , the distance from the star to the point where the brightness is calculated (measured). Thus:

b) $B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}$ and $B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}$ where $L_{Sun}$ is the Sun luminosity ( $3.9 x 10^{26} W$ ) but we don't need to know this value for solving the problem. $ly$ is light years.

Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is $1.581 \cdot 10^{-5}\ ly$ . Then

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }$

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0

3 years ago

Which forces are acting on the student and the skateboard in the instant in which they are pushing off the wall? (Select all tha

diamong [38]

E all of the answers above correlate to the student and his skateboard

7 0

2 years ago

Read 2 more answers

calculate the currrent passing acrooss a bulb if it has a resistane of 15 ohms and a potentinal difference of 6v is applied acro

lawyer [7]

V = IR

6 V = I × 15 Ohms

I = 6 V : 15 Ohms

I = 0,4 A

#LearnWithEXO

4 0

2 years ago

What is the graph of the relationship between the volume of a gas at a constant pressure

Bogdan [553]

We know that P1V1 = P2V2, if there is a constant pressure, then the P1 and P2 can cancel out, so it is V1=V2 that is whats left.

6 0

3 years ago