Answer: 6m/s
Explanation:
Using the law of conservation of momentum, the change in momentum of the bodies before collision is equal to the change in momentum after collision.
After collision, the two objects will move at the same velocity (v).
Let mA and mB be the mass of the two objects
uA and uB be their velocities before collision.
v be their velocity after collision
Since the two objects has the same mass, mA= mB= m
Also since object A is at rest, its velocity = 0m/s
Velocity of object B = 12m/s
Mathematically,
mAuA + mBuB = (mA+mB )v
m(0) + m(12) = (m+m)v
0+12m = (2m)v
12m = 2mv
12 = 2v
v = 6m/s
Therefore the speed of the composite body (A B) after the collision is 6m/s
Its rays point away from the charge
Answer:
1/2 Hz
Explanation:
A simple harmonic motion has an equation in the form of
where A is the amplitude, 
is the angular frequency and 
is the initial phase.
Since our body has an equation of x = 5cos(π t + π/3) we can equate 
and solve for frequency f
f = 1/2 Hz
Answer:
Use the form of equation:
Q=mL
You have the specific latent heat of vaporization L = 2.260*10^{6}
And Q, the heat energy supplied, which equals 1695 KJ = 1695*10^{3} J
So you can get the mass by substitution in the formula below.
Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
