Answer:
The slope of a graph of position vs time
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
Acceleration = (change in speed) / (time for the change) = 9/3 = <em>3 m/s²</em> .
His mass makes no difference.
Answer:
1.02 m/s²
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 0 m/s
Final velocity (v) = 6.6 m/s
Time (t) = 6.5 s
Acceleration (a) =.?
Acceleration can simply be defined as the change of velocity with time. Mathematically, it can be expressed as:
a = (v – u) / t
Where:
a is the acceleration.
v is the final velocity.
u is the initial velocity.
t is the time.
With the above formula, we can obtain the acceleration of the car as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 6.6 m/s
Time (t) = 6.5 s
Acceleration (a) =.?
a = (v – u) / t
a = (6.6 – 0) / 6.5
a = 6.6 / 6.5
a = 1.02 m/s²
Therefore, the acceleration of the car is 1.02 m/s²
Solution :
Speed of the air craft,
= 262 m/s
Fuel burns at the rate of,
= 3.92 kg/s
Rate at which the engine takes in air,
= 85.9 kg/s
Speed of the exhaust gas that are ejected relative to the aircraft,
=921 m/s
Therefore, the upward thrust of the jet engine is given by

F = 85.9(921 - 262) + (3.92 x 921)
= 4862635.79 + 3610.32
= 
Therefore thrust of the jet engine is
.