Answer:
The answer is A. Which is true
Answer:
200N
Explanation:
EVERY FORCE IS OPPOSED BY AN EQUAL FORCE, REGARDLESS OF THE WEIGHT OF THE OBJECTS APPLYING THE FORCE.
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.
Answer:
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Explanation:
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Answer:
R_cm = 4.66 10⁶ m
Explanation:
The important concept of mass center defined by
R_cm = 1 / M ∑ x_i m_i
where M is the total mass, x_i and m_i are the position and masses of each body
Let's apply this expression to our case.
Let's set a reference frame where the axis points from the center of the Earth to the Moon,
R_cm = 1 / M (m_earth 0 + m_moon d)
the total mass is
M = m_earth + m_moon
the distance from the Earth is zero because all mass can be considered to be at its gravimetric center
let's calculate
M = 5.98 10²⁴ + 7.35 10²²
M = 6.0535 10₂⁴24 kg
we substitute
R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )
R_cm = 4.66 10⁶ m
