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°
Carbon is the basis of all organic molecules
Answer:
Explanation:
<em>Assuming the triangle is a right triangle,</em>
the magnitude of R can be found using the Pythagorean theorem,
R = sqrt(9.7^2+6^2) = 11.41 m
The angle can be found by arctangent, which is
angle = atan(9.7/6) = 58.26 degrees.
Answer:
I = 27kg.mi/h
Explanation:
In order to calculate the impulse of the ball, you use the following formula:
(1)
m: mass of the ball = 0.3kg
v: speed of the ball after the bat hit it = 60mi/h
vo: speed of the ball before the bat hit it = 30mi/h
You replace the values of all parameters in the equation (1):
where the minus sign of the initial velocity means that the motion of the ball is opposite to the final direction of such a motion.
The imulpse of the ball is 27 kg.miles/hour
