The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be C is 367.42 Hz.
A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.
The fundamental frequency in the tube is given by
where,
Since, T=37+273 K = 310 K
v = 331 m/s
Using this, we get:
Hence, the fundamental frequency is 367.42 Hz.
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(a) +9.30 kg m/s
The impulse exerted on an object is equal to its change in momentum:
where
m is the mass of the object
is the change in velocity of the object, with
v = final velocity
u = initial velocity
For the volleyball in this problem:
m = 0.272 kg
u = -12.6 m/s
v = +21.6 m/s
So the impulse is
(b) 155 N
The impulse can also be rewritten as
where
F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)
is the duration of the collision
In this situation, we have
So we can re-arrange the equation to find the magnitude of the average force:
Answer:
Speed of the this part is given as
Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate
Explanation:
As we know by the momentum conservation of the system
we will have
here we know that
the momentum of two parts are equal in magnitude but perpendicular to each other
so we will have
now from above equation we have
Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate