Answer:
8.75
Explanation:
First, find the force of friction.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.9 m/s)² = F (1.4 m)
F = 11.7 N
Next, find the distance at the new velocity.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d
d = 8.75 m
Answer:
Vx= 11.0865(m/s)
Vy= 6.4008(m/s)
Explanation:
Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:
V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)
The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:
Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)
And for the same theorem the speed on the Y axis will be:
Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)
Answer:
The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
Explanation:
Given that,
Mass of the cart, m = 250 g = 0.25 kg
Initial velocity of the cart, u = 0.31 m/s (due right)
Mass of another cart, m' = 500 g = 0.5 kg
Initial velocity of the another cart u' = -0.22 m/s (due left)
Let p is the total momentum of the system before the collision. It is given by :

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.