Answer:
Explanation:
a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.
b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.
If v be the velocity of the third part along a direction making angle θ
with x axis ,
x component of v = vcosθ
So momentum along x axis after explosion of third part = mv cosθ
= 10 v cosθ
Momentum along x of first part = - 5 x 42 m/s
momentum of second part along x direction =0
total momentum along x direction before explosion = total momentum along x direction after explosion
0 = - 5 x 42 + 10 v cosθ
v cosθ = 21
Similarly
total momentum along y direction before explosion = total momentum along y direction after explosion
0 = - 5 x 38 + 10 v sinθ
v sinθ= 21
squaring and and then adding the above equation
v² cos²θ +v² sin²θ = 21² +19²
v² = 441 + 361
v = 28.31 m/s
Tanθ = 21 / 19
θ = 48°
D. When a substance reacts with another substance, it shows its chemical property
Answer:à
Explanation:waves carry energy in the direction in which they move
We are given –
- Mass of boiling ball is, m = 4 kg
- Speed is, v = 3 m/s
- Momentum, P =?
As we know –
↠Momentum = Mass × Speed(Velocity)
↠Momentum = 4 × 3 kgm/s
↠Momentum = 12 kgm/s
- Henceforth,Momentum will be 12 kgm/s.
Tuberculosis. Reason: I just took the test and got it right