The cheetah needs 197.8 seconds to complete the 5.50 *100 m distance.
The distance traveled divided by the time taken is the most popular formula for calculating average speed. The average speed is determined by dividing the whole distance by the total time required to complete the distance. The other calculation, provided you have the initial and final speeds, adds them together and divides by 2.20. calculating the time needed for a cheetah to travel 5.50 102 meters.
10 km/h average speed is equal to 10 / 3.6, or 2.78 m/s.
Dist. = 5.50 102
Time =?
average speed = time / distance
time=5.50 x 100/ 2.78
Time = 197.8 s
Therefore, it takes a cheetah 197.8 seconds to travel 5.50 102 meters.
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30 N can not be a possible answer for the magnitude of the resultant of these vectors.
<h3>Resultant vector</h3>
The term resultant vector refers to the result obtained when two or more vectors are combined in magnitude and direction. It is that singular vector that has the same effect in magnitude and direction as two or more vectors acting together.
The resultant vector can not be be obtained algebraically rather it must be obtained geometrically. Hence, 30 N can not be a possible answer for the magnitude of the resultant of these vectors.
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1. In the first 1.5 seconds, the lift accelerates from 0 to 3m/s. By definition, the acceleration is the ratio between the change in velocity and the time elapsed to change the velocity.
The change in velocity is 
.
The time elapsed is 1.5 seconds, so the acceleration is
meters per second squared.
2. We know, from the previous point, that the lift travelled 20m from the first floor. Since it returns to the first floor after the ascent, it must travel again those same 20m, just in reverse (descending instead of ascending). So, the total distance travelled is 
meters.
The displacement, though, is zero, because it measures the distance between the starting and ending point of a certain motion. Since the lift starts and ends its motion at the same place (the first floor), its total displacement is zero.
The boxer can just hit the tissue paper with a force as extensive as the tissue paper can apply on the boxer, and the low mass tissue can just apply a weak force.
Let a boxer hitting on heavy bag. The boxer hits the bag while the bag hits back on the boxer and stops its movement. A pair of forces is engaged with the hitting the bag. The force pair can be quite large.
Now considering boxer hitting a bit of tissue paper. The boxer can just apply as much force on the tissue paper as the tissue paper can apply on the boxer.
The boxer can`t apply any force whatsoever except if what is being hit applies a similar amount of force back. An interaction requires a pair of forces acting on two separate objects.
Answer:Remains same
Explanation:
Given
It is given that the depth is constant and we are only increasing the height of object submerged in the water
thus Force due to Pressure remains same as this force depends upon the depth of object with respect to free surface of Fluid.
Pressure at a depth h with density 
of fluid is given by
and 
