Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s
v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m
Answer:
d = 3.5*10^4 m
Explanation:
In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:
(1)
where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

hence, the displacement of the airplane is 3.45*10^4 m
Answer:
D
friction acts in the opposite direction of motion but does not affect the motion of the object
Answer:
Moment of inertia of the system is 289.088 kg.m^2
Explanation:
Given:
Mass of the platform which is a uniform disk = 129 kg
Radius of the disk rotating about vertical axis = 1.61 m
Mass of the person standing on platform = 65.7 kg
Distance from the center of platform = 1.07 m
Mass of the dog on the platform = 27.3 kg
Distance from center of platform = 1.31 m
We have to calculate the moment of inertia.
Formula:
MOI of disk = 
Moment of inertia of the person and the dog will be mr^2.
Where m and r are different for both the bodies.
So,
Moment of inertia
of the system with respect to the axis yy.
⇒ 
⇒ 
⇒ 
⇒
The moment of inertia of the system is 289.088 kg.m^2
Kinetic energy = 1/2mv^2
=1/2(25)(1^2)
= 12.5J