Answer:
47 m
Explanation:
Data obtained from the question include the following:
Length of dry leg 1 (L1) = 40 m
Length of dry leg 2 (L2) = 25 m
Length of swimming course (L) =..?
The length of the swimming course can be obtained by using pythagoras theory as shown below:
L² = L1² + L2²
L² = 40² + 25²
L² = 1600 + 625
L² = 2225
Take the square root of both side.
L = √2225
L = 47.1 ≈ 47 m
Therefore, the length of the swimming course is approximately 47 m.
In this item, we are given with the x-component of the velocity. The y-component is equal to 0 m/s. The time it takes for it to reach the volume can be related through the equation,
d = V₀t + 0.5gt²
Substituting the known values,
225 = (0 m/s)(t) + (0.5)(9.8)(t²)
Simplifying,
t = 6.776 s
To determine the distance of the student from the edge of the building, we multiply the x-component by the calculated time.
range = (12.5 m/s)(6.776 s)
range = 84.7 m
<em>Answer: 84.7 m</em>
To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.
Gravitational potential energy can be defined as
As M=m, then
Where,
m = Mass
G =Gravitational Universal Constant
R = Distance /Radius
PART A) As half its initial value is u'=2u, then
Therefore replacing we have that,
Re-arrange to find v,
Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s
PART B) With a final separation distance of 2r, we have that
Therefore
Therefore the velocity when they are about to collide is 
Electric pumps are not useful they clog
