Answer:
0.799 m/s if air resistance is negligible.
Explanation:
For how long is the ball in the air?
Acceleration is constant. The change in the ball's height
depends on the square of the time:
,
where
is the change in the ball's height.
is the acceleration due to gravity.
is the time for which the ball is in the air.
is the initial vertical velocity of the ball.
- The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground.
. - Gravity pulls objects toward the earth, so
is also negative.
near the surface of the earth. - Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result,
.
Solve for
.
;
;
;
.
What's the initial horizontal velocity of the ball?
- Horizontal displacement of the ball:
; - Time taken:

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.
.
Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.
1. e) None of the above is necessarily true.
2.d) Without knowing the mass of the boat and the sack, we cannot tell.
To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,
Q = V*A
Where,
A= Cross-sectional Area
V = Velocity
The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,


Our values are given as,


Re-arrange the equation to find the first ratio of rates we have:



The second ratio of rates is



Answer:
v (speed) = S / t = 4 * 400 m / (6 * 60 sec) = 4.4 m/s
The average velocity is zero because there is no net vector displacement.
The correct order they go in is "1-4-2-3" The correct answer is D.