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It is gaining potental energy which will then transfer to knetic energy as it falls
<span>the classification system</span>
Answer:
9.773m/s2
Explanation:
Given,
h=8848m
The value of sea level is 9.08m/s2
So,
Let g′ be the acceleration due to the gravity on the Mount Everest.
g′=g(1−h2h)
=9.8(1−640000017696)
=9.8(1−0.00276)
9.8×0.99724
=9.773m/s2
Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s2
Hope it helped!!!
Answer:
The potential energy (P.E) at the top is 392 J
The kinetic energy (K.E) at the top is 0 J
The potential energy (P.E) at the halfway point is 196 J.
The kinetic energy (K.E) at the halfway point is 196 J.
Explanation:
Given;
mass of the rock, m = 2 kg
height of the cliff, h = 20 m
speed of the rock at the halfway point, v = 14 m/s
The potential energy (P.E) and kinetic energy (K.E) when its at the top;
P.E = mgh
P.E = (2)(9.8)(20)
P.E= 392 J
K.E = ¹/₂mv²
where;
v is velocity of the rock at the top of the cliff = 0
K.E = ¹/₂(2)(0)²
K.E = 0
The potential energy (P.E) and kinetic energy (K.E) at the halfway point;
P.E = mg(¹/₂h)
P.E = (2)(9.8)(¹/₂ x 20)
P.E = 196 J
K.E = ¹/₂mv²
where;
v is velocity of the rock at the halfway point = 14 m/s
K.E = ¹/₂(2)(14)²
K.E = 196 J.
