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Lorico [155]
3 years ago
10

In order to transmit information via radio waves, the waves need to be changed somehow. For car radios this can happen in two wa

ys. For AM radios, which property of the carrier wave is modulated before it is transmitted?
Physics
1 answer:
laila [671]3 years ago
7 0

Answer: amplitude

Explanation: This describes the maximum amount of the displacement of a particle from it rest position. Usually, it is measured in metres

Since we are considering AM which is amplitude modulation, a technique used in electronic communication, most commonly for broadcasting information through a radio carrier wave. In amplitude modulation, the amplitude (signal strength) of the carrier wave is diversified in proportion to that of the message signal being broadcasted.

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Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
IRINA_888 [86]

Answer:

Time : <u>7.96 s</u>

Distance Traveled : <u>357.8 m</u>  

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s    (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s    (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s    (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

<u>t₂ = 7.96 s</u>

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

<u>S = 357.8 m</u>

7 0
3 years ago
Which of Newton's laws says that an object in motion stays in motion, and an object at rest stays at rest unless acted upon by a
Vinil7 [7]
Newtons first law states that.
6 0
3 years ago
Using the scientific definition of work, does moving an object a greater amount of distance always require a greater amount of w
tester [92]
The answer is no. If you are dealing with a conservative force and the object begins and ends at the same potential then the work is zero, regardless of the distance travelled. This can be shown using the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the object.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.

3 0
3 years ago
1. What three particles are found in an atom?
Hoochie [10]

Answer:

Protons, Electrons, and neutrons

7 0
3 years ago
Read 2 more answers
Abby throws a ball straight up and times it. she sees that the ball goes by the top of a flagpole after 0.50 s and reaches the l
Arisa [49]
The time it takes for a ball to complete the trip is given to be 4.10 s. This implies that the trip going up of the ball is equal to 2.05 s. Using this time to determine the initial speed, we have,
                                   Vf = Vo - gt
Vf is the final speed which is equal to 0 on the topmost location of the ball.
                                     0 = Vo - (9.8 m/s²)(2.05 s)
                                        Vo = 20.09 m/s

At 0.5 s,
                                    Vf = 20.09 m/s - (9.8 m/s²)(0.5 s)
                                          Vf = 15.19 m/s

Thus, the speed of the ball when the height is in level with the top of the flagpole is 15.19 m/s. 
5 0
3 years ago
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