Answer:
The speed of the car when load is dropped in it is 17.19 m/s.
Explanation:
It is given that,
Mass of the railroad car, m₁ = 16000 kg
Speed of the railroad car, v₁ = 23 m/s
Mass of additional load, m₂ = 5400 kg
The additional load is dropped onto the car. Let v will be its speed. On applying the conservation of momentum as :
v = 17.19 m/s
So, the speed of the car when load is dropped in it is 17.19 m/s. Hence, this is the required solution.
When the truck's weight is added to the boat, the boat sinks 5 cm deeper,
and displaces additional water whose weight is equal to the weight of the
truck.
The volume of the additional displaced water is
(3.9 m) x (6.3 m) x (5.0 cm)
= (3.9 m) x (6.3 m) x (0.05 m) = 1.2285 m³ .
The weight of that much water is the weight of the truck.
Mass of 1 liter of water = 1 kilogram
1.2285 m³ = 1,228.5 liters = 1,228.5 kg of water.
Weight = (mass) x (gravity)
= (1,228.5 kg) x (9.8 m/s²) = 12,039 Newtons.
(about 2,708 pounds)
Power = (force) x (distance / time) = force x speed .
We know the force = 800N.
We have a speed = 30km/hr, but in order to use it in the power formula,
it has to be in meters/second, so we have some work to do first.
(30 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = 300 / 36 m/sec .
Power = (force) x (speed) = (800 N) x (300/36 m/s) = <em>6-2/3 kilowatts </em>
Work = (power) x (time) = (6,666-2/3 joule/sec) x (25sec) = <em>166,666-2/3 joules</em>.
The figure for power is slightly weird ... 746 watts = 1 horsepower,
so the truck's engine is only delivering about 8.9 horsepower.
Very fuel-efficient, but I don't think they drive trucks that way.
Is called potential energy
