Answer:
The answer is "
".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

Potential energy shifts:


Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.



This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
Given the following in the problem:
Distances : 2.0 m and 4.0 m
Sound waves : 1700 hz
Speed of sound : 340 m/s
Get the wavelength of the sound by using the formula:
Lambda = speed of sound/sound waves
Lambda = 340 m/s / 1700 hz
Lambda = 0.2
Get the path length difference to the point from the two speakers
L1 = 4mL2 = sqrt (42+ 22) m
Delta = 4.47
x = delta / lambda
If the outcome is nearly an integer, the waves strengthen at the point. If it is nearly an integer +0.5 the waves interfere destructively at the point. If it is neither the point is somewhat in in the middle.
Solving x = (4.47 – 4) / (0.2) = 2.35 an integer +0.5 so it’s a point of destructive interference.
Answer:
the intensity of the light after passing through the two polarizing filters is 4.11 units
Explanation:
Given the data in the question;
the intensity of an unpolarized light; I₀ = 25.0 units
when the unpolarized light passes through the first polarizer, its intensity reduces to half of its initial value;
⇒ I₁ = I₀/2 = 25/2 = 12.5 units
the angle between the transmission axes of two polarizers is;
∅ = 55° - 0° = 55°
The intensity of the light after passing through two polarizing filters will be;
I₂ = I₁cos²∅
we substitute
I₂ = 12.5 × cos²(55)
I₂ = 12.5 × 0.3289899
I₂ = 4.11 units
Therefore, the intensity of the light after passing through the two polarizing filters is 4.11 units
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