Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone,
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :
v is speed of sound
Let f is the fundamental frequency. It is given by :
The relation between f and f₂ can be written as :
So, the fundamental frequency of the pipe is 409 Hz.
Answer:
No, projectile launched at 20° landed first.
Explanation:
70° projectile had a vertical velocity of v*sin(70), which is significantly bigger than v*sin(20). Since v=vx+at, we can see that
solve for t, you can check that by substituting velocity for any number.
answer: transverse and longitudinal
The velocity of the two balls after the collision is 0.73 m/s.
The velocity of the two balls after the collision can be calculated using the formula below.
<h3>Formula:</h3>
- mu+m'u' = V(m+m')............... Equation 1
<h3>Where:</h3>
- m = mass of the first ball
- m' = mass of the second ball
- u = initial velocity of the first ball
- u' = initial velocity of the second ball
- V = velocity of the two balls after the collision.
make V the subject of the equation
- V = (mu+m'u')/(m+m')................ Equation 2
From the question,
<h3>Given:</h3>
- m = 0.25 kg
- m' = 0.3 kg
- u = 1 m/s
- u' = 0.5 m/s
Substitute these values into equation 2
- V = [(0.25×1)+(0.3×0.5)](0.25+0.3)
- V = 0.4/0.55
- V = 0.73 m/s.
Hence, the velocity of the two balls after the collision is 0.73 m/s
Learn more about collision here: brainly.com/question/7694106