Answer:3.31m/s², to the right
Explanation:
According to the law of conservation of momentum of a body, change in momentum of bodies before collision is equal to the change in momentum after collision.
Momentum = mass × velocity
M1 and M2 be the masses of the first and second skaters respectively
Let u1 and u2 be the velocities of the first and second skaters respectively.
v be their common velocity after collision
M1 = 77kg M2 = 66kg u1 = 4m/s² u2 = 2.5m/s²
According to the law we have
M1u1 + M2u2 = (M1+M2)v
77(4) + 66(2.5) = (77+66)v
308 + 165 = 143v
V = 473/143
V = 3.31m/s²
Their velocity after collision will become 3.31m/s²
They will both move towards the right after collision because the mass of the body moving to the right is higher than the other mass and the mass is also moving at a higher velocity than the other.
The pressure of the water on the diver is given in an expression written as:
<span>p=15+15/33d
where p is the pressure and d is the distance of the diver </span><span>below the surface.
The pressure is calculated as follows:
</span>p=15+15/33(100) = 15.00 pounds per square feet
Therefore, the correct answer is option A.
Answer:
velocity of girl with respect to surface of ice 1.154 m/s
Explanation:
Plank Mass , M = 150 kg
Girl Mass , m = 45 kg
velocity of the girl with resect to plank, v1 = 1.50 m/s
velocity of the plank + girl = v2
from the conservation of momentum
Momentum (plank+girl) = - momentum of the girl
(M+m)v2 = -mv1
v2 = -(45 kg)/(195 kg) * 1.50
v2 = -0.346 m/s
velocity of girl with respect to surface of ice =
v1+v2 = 1.50 + (-0.346) = 1.154 m/s
There is a possibility but not extremely likely
Answer:
1.28 s
Explanation:
Given:
Δy = 8 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(8 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.28 s
