Answer:
the potential energy will also change
Explanation:
kinetic energy and potential energy are inversely proportional to each other, so if kinetic energy changes, potential energy will also change..
Answer:
There is no actual question attached to this, to get a real answer be sure to include the documents/question that is provided on your work.
Explanation:
Answer: a. 198.6J b. - 198.6J
Explanation: Parameters given:
m = 15kg
g = 9.8m/s²
∅ = 12°
a. Work done by the force Fp on the cart if the ramp is 6.5m long.
Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N
Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J
b. The work done by the force mg on the cart.
Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance
= 15kg x -9.8m/s² x Sin12° x 6.5m
= - 198.6J
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.
Answer:
The time interval of acceleration for the bus is 2.20 seconds
Explanation:
Acceleration is the rate of change of velocity
→ 
where a is the acceleration, v is the final velocity, u is the initial velocity
and t is the time
The given is:
The uniform acceleration = -4.1 m/s²
The bus slows from 9 m/s to 0 m/s
We need to find the time interval of acceleration for the bus
Lets use the rule above
→ a = -4.1 m/s² , v = 0 m/s , u = 9 m/s
→ 
Multiply both sides by t
→ -4.1 t = -9
Divide both sides by -4.1
∴ t = 2.20 seconds
<em>The time interval of acceleration for the bus is 2.20 seconds</em>