Answer:
The work done is 13.875 J.
Explanation:
Given that,
Mass of ball = 0.25 kg
Initial speed = 20 m/s
Final speed = 17 m/s
We need to calculate the work done
Using formula of work done
Work done by the friction
Negative sign shows the friction always against displacement.
Hence, The work done is 13.875 J.
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Answer:
the magnitude of the momentum of the two-ball system immediately after collision is 32.31 kg.m/s
Explanation:
Given;
mass of the first ball, m₁ = 1.0 kg
mass of the second ball, m₂ = 2.0 kg
initial velocity of the first ball, v₁ = 30 m/s due west
initial velocity of the second ball, v₂ = 6 m/s due north
From the principle of conservation of linear momentum;
the total momentum before collision = total momentum after collision
The westward momentum of the first ball, = m₁v₁ = 1 x 30 = 30 kg.m/s
The northward momentum of the second ball = m₂v₂ = 2 x 6 = 12 kg.m/s
The resultant momentum of the two balls;
R² = 30² + 12²
R² = 1044
R = √1044
R = 32.31 kg.m/s
Therefore, the magnitude of the momentum of the two-ball system immediately after collision is 32.31 kg.m/s
Answer:
v = -10⁵ m/s
Explanation:
given,
speed of asteroid,v' = 100 m/s
mass of superman = m
mass of asteroid,M = 1000 m
recoil velocity of superman,v= ?
using conservation of momentum.
m u + M u' = m v + M v'
initial velocity of asteroid and superman is equal to zero
0 + 0 = m v + 1000 m x 100
m v = -100000 m
v = -10⁵ m/s
superman's velocity after throwing the asteroid is equal to v = -10⁵ m/s