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3 years ago

5

If the frequency of an electromagnetic wave increases, does the number of waves passing by you increase, decrease, or stay the s

ame? a. The number of waves passing by stays the same because the speed of light stays the same.
b. The number of waves passing by increases because the speed of light increases.
c. The number of waves passing by decreases because the speed of light decreases.
d. The number of waves passing by increases because the speed of light stays the same.

1 answer:

jarptica [38.1K]3 years ago

3 0

Answer:

option(d)

Explanation:

The frequency of a wave is the property of the source of wave.

The velocity of all the electromagnetic waves is same as the speed of light. It only changes as the light passes through one medium to another.

The frequency is defined as the number of waves coming out from the source in 1 second.

As the frequency of wave increases, the number of wave coming per second increases.

So, the number of waves passing by increases but the speed remains same.

Option (d)

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Answer:

The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

Explanation:

Given;

initial temperature of the liquid, t₁ = 76.3 +/- 0.4⁰C

final temperature of the liquid, t₂ = 67.7 +/- 0.3⁰C

The change in temperature of the liquid is calculated as;

Δt = t₂ - t₁

Δt = (67.7 - 76.3) +/- (0.3 - 0.4)

Δt = (-8.6) +/- (-0.1)

Δt = 8.6 +/- 0.1 ⁰C

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Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

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r = distance between the star and the planet

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so:

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Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

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stepan [7]

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Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
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By rearranging the equation, we find d:
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Answer:

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nika2105 [10]

Answer:

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