Since the lady is capable of delivering 2 strokes in 1 second, we know that it takes 0.5 seconds for one stroke of the hammer.
It is given that one more blow is heard even after she is seen as stopping hammering. This means that the sound from the last blow reached us a little later: 0.5 seconds after the last blow.
So, we know that time taken for the sound from the last blow to reach us t = 0.5 s
Speed of Sound is given as V = 340 m/s
Distance of the lady from us D = ?
Using the equation D = V.t, we get D = (340)(0.5) = 170 m
Thus, we can understand that the lady is 170 meters away from us.
Answer:
Check the explanation
Explanation:
a) in solving the first question, we will be choosing a spherical ball of radius 1 m, Therefore the width of the object is the diameter = 2m.
b)Given that and according to the question, the radius of the electron's orbit = 1x105 x radius of the object = 1x105 x1 = 1x105 m
<u>Answer</u>
To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use<em>. </em>
<u>Explanation</u>
A vernier caliper is an instrument that is used to measure the diameter of small circular objects such as diameter of a wires, thickness of an iron sheet.
The objects to be measured is place between the jaws of the calipers.
The vernier scale has two scales, the vernier scale and the main scale which is the very top scale.<em> To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use. </em>
I think it's longitudinal wave because the particles move parallel to the direction that the wave is traveling.
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.