A uniform plane wave traveling in air is incident upon a flat, lossless, and infinite in extent dielectric interface with a diel
ectric constant of 8. In the air medium, a standing wave is formed. If the normalized magnitude of the incident E-field is Eo = 1, determine the maximum value of the E-field standing wave pattern in air and the shortest distance l (in λo) from the interface where the first maximum in the E- field standing wave pattern will occur (normalized to the incident field). Select one: a. Maximum value of the E-field standing wave = 2.48; shortest distance l = λo/4 b. Maximum value of the E-field standing wave = 2.48; shortest distance l = λo/2 c. Maximum value of the E-field standing wave = 1.48; shortest distance l = λo/4 d. Maximum value of the E-field standing wave = 0.52; shortest distance l = λo/2
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Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
The correct answer is "3. Both increase"
In fact, the frequency of a vibrating string is given by
where L is the length of hte string, T the tension and
the linear mass density.
We can see that string is made shorter, L decreases, so the frequency f increases. And since the pitch has the same behaviour of the frequency, it increases as well.
In fact, the frequency of a vibrating string is given by
where L is the length of hte string, T the tension and
We can see that string is made shorter, L decreases, so the frequency f increases. And since the pitch has the same behaviour of the frequency, it increases as well.