Answer:

Explanation:
The kinematic parameters from the moment the two cyclists begin to accelerate until they meet are:
Initial parameters:
: Initial speed of cyclist A
: Initial speed of cyclist B
Final parameters:
: Final speed of cyclist B
distance of cyclist A = distance of cyclist B

time of cyclist A = time of cyclist B
Cyclist B Kinematics







Cyclist A Kinematics









Answer:
a) 
b) 
Explanation:
The law of conservation of mechanical energy states that total mechanical energy remains constant during oscillation. Mechanical energy is defined as the sum of kinetic energy and potential energy:

a) The position is one-third the amplitude. So, we have
. Replacing and solving for K

b) The potential energy is defined as:

Replacing:

F1x + F2x = Rx
↓
Rx = F1x + F2x
↓
Rx = F1 cos45° + F2
↓
Rx = (50N)(cos45°) + 60N
↓
Rx = 95N
Similarly, if we sum all the y components, we will get the y component of the resultant force:
F1y + F2y = Ry
↓
Ry = F1y + F2y
↓
Ry = F1 sin45° + 0
↓
Ry = F1 sin45°
↓
Ry = (50N)(sin45°)
↓
Ry = 35N
At this point, we know the x and y components of R, which we can use to find the magnitude and direction of R:
Rx = 95N
Ry = 35N
Answer:
1.503 J
Explanation:
Work done in stretching a spring = 1/2ke²
W = 1/2ke²........................... Equation 1
Where W = work done, k = spring constant, e = extension.
Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.
Substitute into equation 1
W = 1/2(26)(0.34²)
W = 13(0.1156)
W = 1.503 J.
Hence the work done to stretch it an additional 0.12 m = 1.503 J
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s