D)four cells that each have half as many chromosomes as the parent cell
Answer:
f = 614.28 Hz
Explanation:
Given that, the length of the air column in the test tube is 14.0 cm. It can be assumed that the speed of sound in air is 344 m/s. The test tube is a kind of tube which has a closed end. The frequency in of standing wave in a closed end tube is given by :


f = 614.28 Hz
So, the frequency of the this standing wave is 614.28 Hz. Hence, this is the required solution.
Answer:
Vf₂ = 2 Vf₁
It shows that final speed of Joe is twice the final speed of Jim.
Explanation:
First, we analyze the final speed of Jim by using first equation of motion:
Vf₁ = Vi + at
where,
Vf₁ = final speed of Jim
Vi = initial speed of Jim = 0 m/s
a = acceleration of Jim
t = time of acceleration for Jim
Therefore,
Vf₁ = at ---------------- equation (1)
Now, we see the final speed of Joe. For Joe the parameters will become:
Vf = Vf₂
Vi = 0 m/s
a = a
t = 2t
Therefore,
Vf₂ = 2at
using equation (1):
<u>Vf₂ = 2 Vf₁</u>
<u>It shows that final speed of Joe is twice the final speed of Jim.</u>
<span>a = (v2 - v1)/t = (10 - 6)/2 = 2 m/sec/sec (average acceleration)</span>