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°
the control would be A. the bottle with 0% concentration, because you're not changing anything.
Rubbing both pieces cause each piece to have a negative charge.
When two parts have the same they repel each other, so holding one piece up tot he end of the other piece would push it away.
Because one piece is held in the middle by a string, it would rotate the piece in a circle.
If they held the piece to the other end of the one held by a string it would start to rotate in the opposite direction.
Answer:mile
Explanation: heres a hint think aboyt the distance between your house to school
So the initial velocity is 15 m/s, the final velocity is 0 since it's at a complete stop and time is 10 seconds. Therefore:
Therefore, the acceleration is -1.5 m/s^2. The reason it's negative is due to the fact that the vector is going against it's original movement since it's decelerating.
