Answer:
R = 4.24 x 10⁻⁴ m
Explanation:
given,
mass of the person = 60.3-kg
mass of the bullet = 10 gram = 0.01 Kg
velocity of bullet = 389 m/s
angle made with the horizontal = 45°
using conservation of momentum.
M v + m u = ( M + m ) V
60.3 x 0 + 0.01 x 389 = (60.3 + 0.01) V
V = 0.0645 m/s
for calculation of range
R = 4.24 x 10⁻⁴ m
the distance actor fall is R = 4.24 x 10⁻⁴ m
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
Answer:
Only the goalie is allowed inside the goal crease. The only exception when another player is allowed in the goal area is when they take off from outside the goal area, and shoots or passes the ball before landing. To avoid interference with other players, the player must then exit the goal area as soon as possible.
Explanation:
