Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled
Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²
Answer: 1.95 x 10¹⁴ m/s²
Answer:
A.The vertical velocity is constantly increasing as the ball falls.
B.The horizontal velocity does not noticeably change as the ball falls.
G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.
H.The velocity vector of the ball changes as it travels through the air.
Explanation:
As the ball is projected horizontally so here the vertical component of the velocity is zero
So the time to reach the ground is given as

so we will have

so this is the same time as the ball is dropped from H height
Since there is no force in horizontal direction so its horizontal velocity will always remain constant while vertical velocity will change at constant rate which is equal to acceleration due to gravity.
So overall the velocity vector will change due to net acceleration g
B - A theory seems to be the closest
Explanation :
It is given that,
Mass of the car, m = 1000 kg
Force applied by the motor, 
The static and dynamic friction coefficient is, 
Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,




So, the acceleration of the car is
. Hence, this is the required solution.