A cross country skier moves 32 m north, then 65 m south, and finally 16 m north. What is the distance? What is the displacement
1 answer:
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Answer:
The tension in the string is .
Explanation:
For a string with tension and linear density
carrying a transverse wave at speed
it is true that
solving for we get:
Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is
And it's linear density (mass per unit length) is
Therefore, the tension in the cord is
or in micro newtons
Answer:
the required minimum length of the attenuator is 3.71 cm
Explanation:
Given the data in the question;
we know that;
= c / 2a
where f is frequency, c is the speed of light in air and a is the plate separation distance.
we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s
plate separation distance a = 7.11 mm = 0.0711 cm
so we substitute
= 3 × 10¹⁰ / 2( 0.0711 )
= 3 × 10¹⁰ cm/s / 0.1422 cm
= 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }
Now, we determine the minimum wavelength required
Each non propagating mode is attenuated by at least 100 dB at 15 GHz
so
Attenuation constant TE₁ and TM₁ expression is;
∝₁ = 2πf/c × √( ( / f)² - 1 )
so we substitute
∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )
∝₁ = 3.1079 × 10⁻⁷
∝₁ = 310.79 np/m
Now, To find the minimum wavelength, lets consider the design constraint;
20log₁₀ = -100dB
we substitute
20log₁₀ = -100dB
= 3.71 cm
Therefore, the required minimum length of the attenuator is 3.71 cm