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3 years ago

7

If the frequency of an FM wave is 8.85 × 10¥ hertz, what is the period of the FM wave?

2 answers:

larisa [96]3 years ago

7 0

The value of the period of the wave is the reciprocal of its frequency in hertz (1/s). From the given above, the value of the period is calculated below,
t = 1 / f = 1 / (8.85 x 10^7 1/s)
Thus, the value of period (t) is equal to 1.13 x 10^-8 s. The answer is letter B.

natka813 [3]3 years ago

6 0

for plato users it is 1.10 X 10^-8 just took the test got 100 percent so ik its right

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Answer:

(1) 0.333 Hz

(2) 4 sec

(3) 2 sec, 0.5 Hz

Explanation:

(1) We have given time period of pendulum is 3 sec

So T = 3 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{3}=0.333Hz$

(2) Frequency of the pendulum is given f = 0.25 Hz

Time period is equal to $T=\frac{1}{f}=\frac{1}{0.25}=4sec$

(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds

So time taken to complete 1 back and forth swings = $=\frac{20}{10}=2sec$

So time period T = 2 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{2}=0.5Hz$

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Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

$\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}$

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

$\lambda = \frac{c}{f}$

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

$\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m$

Hence, the distance is:

$d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m$

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

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