Kinetic Energy = 1/2 * m * v²
1/2 * 30 * 20²
1/2 * 30 * 400
12000/2
6000 J.
Answer:
a) 20 seconds
b) No.
Explanation:
t = Time taken for jet to stop
u = Initial velocity = 100 m/s (given in the question)
v = Final velocity = 0 (because the jet will stop at the end)
s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)
a = Acceleration = -5.00 m/s² (slowing down, so it is negative)
a) Equation of motion

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.
Answer:
I believe the answer would be A. point x
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.