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.
Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
First we'll calculate the energy it posesses
G.P.E = mgh = 0.2 * 10 * 100 = 200 J
Now we'll calculate the temperature rise
Q = m * c * (t2 - t1)
Q/(m * c) = t2-t1
t2 = Q/(m * c) + t1 = 200/(0.2 * 400) + 0 = <span>2.5 C</span>
