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lara [203]
3 years ago
9

What kind of modulator is used in this scenario?

Physics
2 answers:
icang [17]3 years ago
6 0

Answer:

the answer is amplitude

Explanation:

sladkih [1.3K]3 years ago
3 0

The unmodulated carrier wave is going into the box, and when it comes out, its AMPLITUDE has been modulated.

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A watermelon is dropped from the top of a 80m tall building. We want to find the velocity of the watermelon after it falls for 1
White raven [17]

Answer:

v_f = v_i + at

v_f = 13.23 m/s

Explanation:

Height Of the watermelon when it is dropped is given as

h = 80 m

time of fall under gravity

t = 1.35 s

now if water melon start from rest then we have

v_i = 0

acceleration due to gravity for watermelon

a = 9.81 m/s^2

now we need to find the final speed of watermelon

v_f = v_i + at

so we will have

v_f = 0 + (9.81)(1.35)

v_f = 13.23 m/s

7 0
3 years ago
In science, does acceleration happen when an object is slowing down?
Irina18 [472]

Answer:

According to our principle, when an object is slowing down, the acceleration is in the opposite direction as the velocity. Thus, this object has a negative acceleration. In Example D, the object is moving in the negative direction (i.e., has a negative velocity) and is speeding up.

7 0
2 years ago
One end of a string 4.32 m long is moved up and down with simple harmonic motion at a frequency of 75 Hz . The waves reach the o
max2010maxim [7]

To solve this problem, we will apply the concepts related to the kinematic equations of linear motion, which define speed as the distance traveled per unit of time. Subsequently, the wavelength is defined as the speed of a body at the rate of change of its frequency. Our values are given as,

\text{Length of the string} = L = 4.32 m

\text{Frequency of the wave} = f = 75 Hz

\text{Time taken to reach the other end} = t = 0.5 s

Velocity of the wave,

V = \frac{L}{t}

V = \frac{4.32 m}{0.5s}

V = 8.64m/s

Wavelength of the wave,

\lambda = \frac{V}{f}

\lambda = \frac{8.64m/s}{75Hz}

\lambda = 0.1152m

\lambda = 11.52cm

Therefore the wavelength of the waves on the string is 11.53 cm

4 0
3 years ago
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xeze [42]
D) Submarine is your answer. Have a great rest of your day!
7 0
3 years ago
Read 2 more answers
A person holds a rifle horizontally and fires at a target. The bullet leaves the muzzle of the rifle with a velocity of 460 m/s.
Trava [24]

Answer:

the distance travelled from the bullet to the target  is 391m

Explanation:

Hello! To solve this exercise we must follow the following steps.

1. the bullet travels with constant speed which means that the distance traveled to the target is given by the following equation

X=(V1)(T1)

T1=\frac{X}{V1} =\frac{x}{460}

where

X=target distance

V1=bullet speed=460m/s

T1=

time it takes for the bullet to reach the target

2. The distance the sound travels is given by the following equation (it is the same as the distance from the person to the target)

X=(V2)(T2)

T2=\frac{X}{V2} =\frac{x}{340}

X=

target distance

V2= speed of sound=340m/s

T2=   time it takes the sound of the Bullet to return.

3. The total time it takes for the person to hear the bullet(T=2s) is the sum of the time it takes for the bullet to reach the target, plus the time it takes for the sound to reach the person, with the above we infer the following equation

T=T1+T2

2=T1+T2

4. Finally we use the equations found in step 1 and 2 to find the distance traveled using algebra.

2=\frac{x}{340}+\frac{x}{460} \\x(\frac{1}{340} +\frac{1}{460} )=2\\\ X= \frac{2}{(\frac{1}{340} +\frac{1}{460} )} \\\\x=391m

the distance travelled from the bullet to the target  is 391m

3 0
3 years ago
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