In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis
1 answer:
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Answer:
Explanation:
Two frequencies with magnitude 150 Hz and 750 Hz are given
For Pipe open at both sides
fundamental frequency is 150 Hz as it is smaller
frequency of pipe is given by
where L=length of Pipe
v=velocity of sound
for n=1
and f=750 is for n=5
thus there are three resonance frequencies for n=2,3 and 4
For Pipe closed at one end
frequency is given by
for n=0
for n=2
Thus there is one additional resonance corresponding to n=1 , between and
Answer:
Fa=774 N
Fb=346 N
Explanation:
We will solve this problem by equating forces on each axis.
- On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
- On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative
While towing we know that car is mot moving in y-direction so net force in y-axis must be zero
⇒∑Fy=0
⇒
⇒
⇒
Given that resultant force on car is 950N in positive x-direction
⇒∑Fx=950
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Therefore approximately, Fa=774 N and Fb=346 N