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
Answer:
SI unit of k (spring constant) = N/m
Explanation:
We have expression for force in a spring extended by x m given by
F = kx
Where k is the spring constant value.
Taking units on both sides
Unit of F = Unit of k x Unit of x
N = Unit of k x m
Unit of k = N/m
SI unit of k (spring constant) = N/m
Answer:
Approximately 
.
Assumption: air resistance on the rocket is negligible. Take 
.
Explanation:
By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.
.
Note that in this case, the uppercase letter 
in the units stands for "mega-", which is the same as 
times the unit that follows. For example, 
, while 
.
Convert the mass of the rocket and the thrust of its engines to SI standard units:
- The standard unit for mass is kilograms:
. - The standard for forces (including thrust) is Newtons:
.
At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:
- Thrust (which is supposed to go upwards), and
- Weight (downwards due to gravity.)
The thrust on the rocket is already known to be 
. Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately 
. Hence, the weight on the rocket would be approximately 
.
The magnitude of the net force on the rocket would be
.
Apply the formula to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)
.
Answer:
Matter
Pure substances Mixture
Element compound Homogenous Heterogenous
