The collapsed answer of Penchalreddy Badepalli is correct. The composition of two reflections via two mirror making an angle \alpha is equivalent to a single rotation by an angle 2\alpha, hence 2 * 60 deg = 120 deg. And turns is independent of the absolute orientation of the two mirrors in space and/or the direction of incidence of the incoming ray.
One could use elementary geometry to prove this (if you presume the direction of incidence is irrelevant imagine hitting the first mirror at 90 deg, then going retro right back along the normal to the first mirror, and follow the directions).
Given that,
Mass of trackler, m₁ = 100 kg
Speed of trackler, u₁ = 2.6 m/s
Mass of halfback, m₂ = 92 kg
Speed of halfback, u₂ = -5 m/s (direction is opposite)
To find,
Mutual speed immediately after the collision.
Solution,
The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :
So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.
Answer:
a) 588,000 N
b) 294000 N
Explanation:
Given that
Density of water = 1000kg/m3
(g) = 9.8m/s2
volume is given as (V)= 5m*4m*3m
a) force will be equal to weight of water
b) at either end
[A = wh]
F = 294000 N
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu