Mass of the car = 1200 kg
Mass of the truck = 2100 kg
Total mass of car and truck = 2100 + 1200 = 3300 kg
Since, the car pushes the truck. Hence, they will move together and will have same acceleration.
Let the acceleration be a.
According to Newton's second law:
F(net) = ma
F = 4500 N
4500 = 3300 × a
a = 1.36 m/s^2
Let the force applied by the car on truck be F.
F = F(net) on the truck
F = ma
F = 2100 × 1.36
F = 2856 N
Hence, the force applied by the car on the truck is 2856 N
Answer:
1.04%
Explanation:
Given that,
The power of drill = 3.5 kW = 3500 W
Transferred kinetic energy = 5000 kJ during 15 seconds of use.
We need to find the percentage efficiency of the drill. It can be given by :
Where
Po and Pi are output and input powers.
So,
So, the percentage efficiency of the drill is 1.04%.
Answer:
The minimum angular speed 57.0 rad/s
Explanation:
In this exercise we must use the relationships between the angular and linear magnitudes, for the speeds we have
v = w r
As they give us that linear velocity is constant, we will calculate the angular velocity for the two radii, external and internal
r1 = 11.75 cm (1 m / 100cm) = 0.1175 m
w1 = v / r1
w1 = 6.70 / 0.1175
w1 = 57.0 rd / s
R2 = 4.50 cm = 0.0450 m
w2 = 6.70 / 0.0450
w2 = 148 rad / s
The minimum angular speed 57.0 rad / s
Speaking that the trash can absorb the momentum. It only should go 5
