Answer:
7.98 m
Explanation:
In the given question,
distance above surface= 2 m
Distance penny from person = 8 m
Since the swimming pool is filled with water and atmosphere has air therefore the refractive index phenomenon will occur.
The refractive index of water: air is 4/3 (1.33).
Using the formula, 4/3 = real depth, apparent depth
real depth= 4/3 x apparent depth
Now, calculating apparent depth = 8 - 2
= 6 m
therefore, real depth = 4/3 x apparent depth
= 1.33 x 6
= 7.98
thus, 7.98 m is the real depth of water.
Centripetal acceleration is (speed-squared) / (radius)
CA = (6 m/s)² / (9 m)
CA = (36 m²/s²) / (9 m)
CA = (36/9) (m²/m·s²)
<em>Centripetal acceleration = 4 m/s²</em>
V^2=u^2 +2aS
U is found first by considering that first 8 secs and using v=u+at. {different v and u though}
V=-u+gt.
Magnitude of u = magnitude of v if there is no resistance ( because the conservation of energy says the k. E. must be the same when it passes you as when it left your hand).... up is negative here, down is positive.
V+v=gt
2v= g x 8
V=4xg.= the initial velocity for the next calculation
V^2=(4g)^2+(2xgx21)
So v can be calculated.
Answer:0.253Joules
Explanation:
First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.
F = ke where;
F is the force
k is spring constant = 34N/m
e is the extension = 0.12m
F = 34× 0.12 = 4.08N
To get work done,
Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.
Work done = Force × Distance
Since F = 4.08m, distance = 0.062m
Work done = 4.08 × 0.062
Work done = 0.253Joules
Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules
