Answer:
<em>The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.</em>
Explanation:
First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. <em>Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight</em>
Answer:
100 m/s²
Explanation:
First, convert to SI units:
50 g = 0.050 kg
Now plug into the formula:
F = ma
5 N = (0.050 kg) a
a = 100 m/s²
M = 600 kg
u = 0 (as the car was at rest initially)
v = 5 m/s
Initial kinetic energy = ½mu²
= ½×600×0
= 0
Final kinetic energy = ½mv²
= ½×600×25
= 7500 J
Answer:
difference in flight time= 0.3023 hour
Explanation:
The question is incomplete, but I found it in your textbook.
Spped of aircraft = 850 km/h
Opposing speed of wind = 90km/h
Hence, the net speed when it's travelling west = 850 - 90 = 760 km/hr
The distance covered = 1200km
time taken = distance/ time = 1200/ 760 = 1.5789 hours
When coming back, the speed of the wind is complementary to the speed of the aircraft so
net speed when it's coming back = 850 +90 = 940 km/hr
time taken in this instance = 1200/ 940 = 1.2765 hours
Hence, the difference in flight time= 1.5789 - 1.2765 = 0.3023 hour
