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luda_lava [24]
3 years ago
7

What is the wavelength of the carrier wave of a campus radio station, broadcasting at a frequency of 97.2 MHz (million cycles pe

r second or million hertz)?
Physics
1 answer:
andrey2020 [161]3 years ago
8 0

Answer:

Wavelength is 3.08 meters.

Explanation:

It is given that,

Frequency, f=97.2\ MHz=97.2\times 10^6\ Hz

We need to find the wavelength of the carrier wave of a campus radio station. It can be calculated by using following relation as :

c=f\times \lambda

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8\ m/s}{97.2\times 10^6\ Hz}

\lambda=3.08\ m

So, the wavelength of the carrier wave is 3.08 meters. Hence, this is the required solution.

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Two children ride side-by-side on a carousel. Their paths are shown in the image below.
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The child represented by a star on the outside path.

Explanation:

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The standard measure used to compare sound intensities is the ______ .
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How long does it take for a Ford Econoline van moving at 39.5 m/s to travel 600 m?
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4 years ago
A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

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