Answer:
a) correct answer is C
, b) 14º from the west to the north, c) v_{1g} = 300.79 km / h
Explanation:
This is a relative speed exercise using the addition of speeds.
1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C
2 and 3) In this case we must compose the speed using the Pythagorean Theorem.
² =
² +
²
where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth
in this case let's clear the speed of the airplane with respect to the Earth
v_{1g} = √(v_{1a}² - v_{ag}²)
v_{1g} = √ (310² - 75²)
v_{1g} = 300.79 km / h
we find the direction of the airplane using trigonometry
sin θ = v_{ag} / v_{1a}
θ = sin⁻¹ (v_{ag} /v_{1a})
θ = sin⁻¹ (75/310)
θ= 14º
the pilot must direct the aircraft at an angle of 14º from the west to the north
C. when the circuit is closed
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.