Answer:652.05 J
Explanation:
Given
Weight of lifter 
vertical distance move 
Work done in lifting the weight is equal change in Potential Energy of weight
Change in Potential Energy 

therefore work done is equal to
Answer:
The balance between incoming energy from the sun and outgoing energy from Earth ultimately drives our climate. This energy balance is governed by the first law of thermodynamics, also known as the law of conservation of energy.
Velocity is defined by rate of change in the position
which we can also write as

while acceleration is defined as rate of change in velocity

so acceleration and velocity both are rate of change in position and rate of change in velocity with respect to time respectively
out of all above statement the correct answer must be
<u>Acceleration equals change in velocity divided by time. </u>
(1)
Cheetah speed: 
Its position at time t is given by
(1)
Gazelle speed: 
the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by
(2)
The cheetah reaches the gazelle when
. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:



(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.
Cheetah: 
Gazelle: 
So, the gazelle should be ahead of the cheetah of at least
