Answer:
(A) Fo = 72 Hz
(B) The pipe is open at both ends
(C) The length of the pipe is 2.38m
This problem involves the application of the knowledge of standing waves in pipes.
Explanation:
The full solution can be found in the attachment below.
For pipes open at both ends the frequency of the pipe is given by
F = nFo = nv/2L where n = 1, 2, 3, 4.....
For pipes closed at one end the frequency of the pipe is given by
F = nFo = nv/4L where n = 1, 3, 5, 7...
The full solution can be found in the attachment below.
Acceleration = (force) / (mass)
If there is no friction . . .
Acceleration = (200N) / (50 kg) = 4m/s²
Change in speed = (acceleration) x (time) = (4 m/s²) x (10 sec) = 40 m/s
<em>Final speed = (speed of the cart before the force began) + (40 m/s more)</em>
(That's about <em><u>89 mph faster</u></em> than the cart was moving before the force started.)
Answer:
Explanation:
100 mph = 100 x 1760 x 3 / ( 60 x 60 )
= 146.67 ft / s
70 mph = 70 x 1760 x 3 / ( 60 x 60 )
= 102.67 ft/s
Let mass of the ball = m .
The ball which was moving with a velocity of 146.67 ft/s is turned back by the action of force of bat on the ball and it moved in opposite direction with velocity of - 102.67 ft/s ( - ve sign due to reversed direction )
If m be the mass of ball
change in momentum = m x 146.67 - ( - m x 102.67 )
= m x 146.67 + m x 102.67
= 249.34 m
Change in momentum = impulse = force x time
Putting the given values
249.34 m = force x 1 x 10⁻³ ( time = 10⁻³ s )
force = 249.34 m x 10³ poundal .
Here m is mass of the ball in lb .
B,C,D i believe. Hope this help
