Answer: 3 m/s
Explanation:
We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:
The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore
So, the final total momentum will also be
And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:
from which we find the velocity of the solid ball
Answer:
a) P = 44850 N
b) 
Explanation:
Given:
Cross-section area of the specimen, A = 130 mm² = 0.00013 m²
stress, σ = 345 MPa = 345 × 10⁶ Pa
Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa
Initial length, L = 76 mm = 0.076 m
a) The stress is given as:
on substituting the values, we get
or
Load, P = 44850 N
Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>
b)The deformation (
) due to an axial load is given as:
on substituting the values, we get
or
Answer:
1) Recollapsing universe
2) critical universe
3) Coasting universe
Explanation:
According to the smallest ration (ratio actual mass density to current density) to largest ration, rank of models for expansion of universe are
1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.
2) critical universe - in this, expansion of universe is very low.
3) Coasting universe - in this, expansion of universe is steady and uniform
Use Newton’s second law: F=ma
m= 250kg. a= 750ms-2
F= 250 x 750 = 187 500N
