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2 years ago

What does it mean when someone says that radio waves can be modulated?

Physics

2 answers:

Ghella [55]2 years ago

8 0

Answer:

It means that information can be added to the radio waves and can be carried through electronic or optical signals.

Explanation:

Modulation is the process of converting data into radio waves by adding information to an electronic or optical carrier signal. A carrier signal is one with a steady waveform -- constant height, or amplitude, and frequency

Feliz [49]2 years ago

5 0

It can be carried through the electronic

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Discuss potential behavioral concerns for people should they travel to Mars

Anna35 [415]

Answer:

The ability of our bodies to adapt to different levels of gravity. You would become weaker and your heart is use to zero gravity. Boredom because there isn't much to in space. When intelligent people get bored, it's not pretty all the time...

6 0

3 years ago

Grade 12 math: vectors as forces?

sashaice [31]

Larger tug force = 2*cos20 forward + 2*sin20 to the right (starboard)
<span>Smaller tug force = 1*cos15 forward + 1*sin15 to the left (port) </span>

<span>Resultant force forward = 2*cos20 + 1*cos15 = 2.8453 (4dp) </span>
<span>Resultant force to right = 2*sin20 - 1*sin15 = 0.4252 (4dp) </span>

<span>Angle of resultant force = arctan(0.4252 / 2.8453) = arctan(0.1494) = 8.5 degrees</span>

4 0

3 years ago

6. A 145-g baseball moving 30.5 m/s strikes a stationary 5.75-kg brick resting on small rollers so it moves without significant

Sindrei [870]

Answer:

Explanation:

a )

momentum of baseball before collision

mass x velocity

= .145 x 30.5

= 4.4225 kg m /s

momentum of brick after collision

= 5.75 x 1.1

= 6.325 kg m/s

Applying conservation of momentum

4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.

v = - 13.12 m / s

b )

kinetic energy of baseball before collision = 1/2 mv²

= .5 x .145 x 30.5²

= 67.44 J

Total kinetic energy before collision = 67.44 J

c )

kinetic energy of baseball after collision = 1/2 x .145 x 13.12²

= 12.48 J .

kinetic energy of brick after collision

= .5 x 5.75 x 1.1²

= 3.48 J

Total kinetic energy after collision

= 15.96 J

3 0

4 years ago

Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at

bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

The electric field produced by a single-point charge is given by

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$ .

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$

This means that we have to solve the equation

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0$

Re-arranging it,

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0\\\frac{(8-x)^2-x^2}{x^2(8-x)^2}=0$

$(8-x)^2-x^2=0\\64+x^2-16x-x^2=0\\64-16x=0\\64=16x\\x=4 m$

So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

Read 2 more answers

A p-wave arrives at 3:00:00 and the s-wave arrives at 3:07:20, what is the exact distance that the seismic station is away from

maxonik [38]

The exact distance from the seismic station to the epicenter is 6000 km.

<h3>Epicenter of earthquake</h3>

The earthquake's epicenter is the point above the fault location on the earth's surface.

Given that:

A p-wave arrives at 3:00:00 and the s-wave arrives at 3:07:20.

Difference in arrival time = 3:07:20 - 3:00:00 = 7 minutes 20 seconds

From the earthquake time travel graph, a time difference of 7 min 20 sec is at x = 6

The exact distance from the seismic station to the epicenter is 6000 km.

Find out more on Epicenter of earthquake at: brainly.com/question/1969968

4 0

2 years ago