6489 for the founding product
Answer:
Explanation:
Given
Lowest four resonance frequencies are given with magnitude
50,100,150 and 200 Hz
The frequency of vibrating string is given by
where n=1,2,3 or ...n
L=Length of string
T=Tension
Mass per unit length
When string is clamped at mid-point
Effecting length becomes
Thus new Frequency becomes
i.e. New frequency is double of old
so new lowest four resonant frequencies are 100,200,300 and 400 Hz
According to the law of conservation of momentum:
m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision
Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.
m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?
Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.
The velocity of the 2nd car after the collision is
0.03m/s.
Answer:
The right solution is "24.39 per sec".
Explanation:
According to the question,
⇒
The time will be:
⇒
hence,
⇒
Answer:
Explanation:
We shall apply law of conservation of momentum during the collision of ball A and B .
Total momentum before collision of A and B = .35 x 10 = 3.5 kg m/s
Let the velocity of B after collision be v .
Total momentum after collision = .35 x 2 + .35v
According to law of conservation of momentum
.35 x 2 + .35v = 3.5
.35 v = 2.8
v = 8 m /s .
The direction of B will be same as direction of A .