Period is T = 1/f. The frequency, f, is 5cycles/10s = 0.5. So the period T=1/0.2 = 2.
Answer: D
Neither A nor B
Explanation:
In order to check the clearances for rod and main bearings, you need a set of micrometers and a dial-bore gauge
Measuring the inside diameter of a main or rod bearing will require a dial bore gauge. The best ones to use are accurate down to 0.0001-inch.
So, both technician A and B are incorrect
Hot, soft rock rise from the bottom of the mantle towards the top, cools, and sinks back through the mantle.
Answer:
Explanation:
The path length difference = extra distance traveled
The destructive interference condition is:
where m =0,1, 2,3........
So, ←
![\Delta d = (m+1/2)\lamb da9/tex]so [tex]\Delta d = \frac{\lambda}{2}](https://tex.z-dn.net/?f=%5CDelta%20d%20%3D%20%28m%2B1%2F2%29%5Clamb%20da9%2Ftex%5D%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3Eso%20%3C%2Fstrong%3E%5Btex%5D%5CDelta%20d%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%7D)
⇒ λ = 2Δd = 2×10 = 20
<span>Her center of mass will rise 3.7 meters.
First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so
8.5 m / 9.8 m/s^2 = 0.867346939 s
And the distance a object under constant acceleration travels is
d = 0.5 A T^2
Substituting known values, gives
d = 0.5 9.8 m/s^2 (0.867346939 s)^2
d = 4.9 m/s^2 * 0.752290712 s^2
d = 3.68622449 m
Rounded to 2 significant figures gives 3.7 meters.
Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.</span>
