Answer:
Explanation:
Given that,
Mass of block
M = 2kg
Spring constant k = 300N/m
Velocity v = 12m/s
At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0
xo = 0
It velocity is 12m/s at t=0
Then, it initial velocity is
Vo = 12m/s
Then, amplitude is given as
A = √[xo + (Vo²/ω²)]
Where
xo is the initial amplitude =0
Vo is the initial velocity =12m/s
ω is the angular frequency and it can be determine using
ω = √(k/m)
Where
k is spring constant = 300N/m
m is the mass of object = 2kg
Then,
ω = √300/2 = √150
ω = 12.25 rad/s²
Then,
A = √[xo + (Vo²/ω²)]
A = √[0 + (12²/12.5²)]
A = √[0 + 0.96]
A = √0.96
A = 0.98m
Answer:
Use Fc centripetal force as positive and W the weight as negative
N = m v^2 / R + m g
v^2 = (N - m g) R / m
v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2
v = 18.1 m/s
Note: N - m g is the net force producing the centripetal force
1) True 2) True 3)True I hope I helped
Answer:
fb = 240.35 Hz
Explanation:
In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.
Open tube:
(1)
vs: speed of sound = 343m/s
L: length of the open tube = 0.47328m
You replace in the equation (1):
Closed tube:

L': length of the closed tube = 0.702821m

Next, you use the following formula for the beat frequency:

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz