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iVinArrow [24]
3 years ago
9

Answer this plsss A mobile phone operates at 900 MHz. What wavelength does it use?

Physics
1 answer:
Aleks [24]3 years ago
7 0

Answer:

The wavelength = 0.3333 meters at 900 MHz, therefore, = /4 = 0.08333 meters.

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QUESTION 9
Alik [6]

Answer:

The answer is option a.

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3 years ago
A sinusoidal wave has period 0.20 s and wavelength 2.0 m. What is the wave speed?
il63 [147K]

Answer:10m/s

Explanation:

Wave speed ,v=for

Where π= wavelength=2m

Period =1/f f=frequency of wave

F=1/period

=1/0.2=5Hz

So speed of waves,v=5×2=10m/s

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3 years ago
Check all choices below that are correct. Increasing the frequency increases the current. Changing the frequency does not affect
Annette [7]

Answer:

the experyoeyv

Explanation:

3 0
3 years ago
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A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0
3 years ago
What is the difference between static friction and kinetic friction
sergejj [24]
The Force of Static Friction<span> keeps a stationary object at rest! Once the Force of</span>Static Friction<span> is overcome, the Force of </span>Kinetic Friction<span> is what slows down a moving object.</span>
3 0
3 years ago
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