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Scilla [17]
3 years ago
7

AM radio signals use amplitude modulation of the radio waves to transmit a signal. A typical wavelength of an AM radio wave is 3

00. meters. What is the frequency of such a radio wave? a) 1.00 µHz b) 1.00 mHz c) 1.00 kHz d) 1.00 MHz e) 1.00 Hz
Physics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

The frequency of such a radio wave is 1 MHz.

(d) is correct option

Explanation:

Given that,

Wave length = 300 m

We know that,

Speed of light c= 3\times10^{8}\ m/s

We calculate the frequency

Using formula of frequency

c = f\times\lambda

f = \dfrac{c}{\lambda}

Put the value into the formula

f=\dfrac{3\times10^{8}}{300}

f =1\times10^{6}\ Hz

f =1\ MHz

Hence, The frequency of such a radio wave is 1 MHz.

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What are various paradox related to time travel?​
erma4kov [3.2K]

Answer:

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Explanation:

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5 0
2 years ago
Many scientists would argue that the reproductive system is the most important system in the body. Which of these would best def
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3 years ago
Read 2 more answers
A ball that has a mass of 0.25 kg spins in a circle at the end of a 1.6 m rope. the ball moves at a tangential speed of 12.2 m/s
NARA [144]

The centripetal force acting on the ball will be 23.26 N.The direction of the centripetal force is always in the path of the center of the course.

<h3>What is centripetal force?</h3>

The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.

The given data in the problem is;

m is the mass of A ball = 0.25 kg

r is the radius of circle= 1.6 m rope

v is the tangential speed = 12.2 m/s

\rm F_C is the centripetal force acting on the ball

The centripetal force is found as;

\rm F_C = \frac{mv^2}{r}  \\\\ F_C = \frac{0.25 \times (12.2)^2}{1.6}  \\\\ F_C=23.26\ N

Hence the centripetal force acting on the ball will be 23.26 N.

To learn more about the centripetal force refer to the link;

brainly.com/question/10596517

4 0
2 years ago
A body of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2
erastova [34]
You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2

Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.

-F=25(-1.8)
F=45N

Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.

45=u(25g)
45=u(25*10)

Therefore, the coefficient of friction is 0.18

Hope that helps




5 0
3 years ago
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