Given
m1(mass of the first object): 55 Kg
m2 (mass of the second object): 55 Kg
v1 (velocity of the first object): 4.5 m/s
v2 (velocity of the second object): ?
m3(mass of the object dropped): 2.5 Kg
The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:
Pa= Pb
Where Pa is the momentum before collision and Pb is the momentum after collision.
Now applying this law for the above problem we get
Momentum before collision= momentum after collision.
Momentum before collision = (m1+m2) x v1 =(55+5)x 4.5 = 270 Kgm/s
Momentum after collision = (m1+m2+m3) x v2 =(55+5+2.5) x v2
Now we know that Momentum before collision= momentum after collision.
Hence we get
270 = 62.5 v2
v2 = 4.32 m/s
As the plane falls the parabolic path remains directly below as the plane continues to fly over. This give more of an overview. When the package falls vertical acceleration happens as there is a vertical velocity as the package falls form high above. The downwards motion of gravity acts on the package if the approximated projectile motion ignoring air resistance.
Answer:
the blood from carrying oxygen to the tissues of the body
Explanation:
Answer:
I think its false...........
Answer:
The acceleration of Chevy Corvette is 
.
Explanation:
Given that,
Initial speed of Chevy Corvette Z06, u = 0 (it starts from rest)
Final speed of Chevy Corvette Z06, v = 60 mph
1 mph = 0.44704 m/s
60 mph = 26.82 m/s
Time, t = 3 seconds
We need to find the acceleration of Chevy Corvette. The rate of change of velocity is called acceleration of an object. It is given by :
So, the acceleration of Chevy Corvette is . Hence, this is the required solution.
