Answer:
Final velocity of the first person is 3.43m/s and that of the second person is 0.0242m/s
Explanation:
Let the momentum of the first person, the ball second person be Ma, Mb and Mc.
From the principle of the conservation of momentum, sum of the momentum before collision is equal to the sum of the momentum after collision.
Ma1 + Mb1 = Ma2 + Mb2.
The ball and the first person are both moving together with a common velocity 3.45m/s.
Let the velocity of the first person be v1
Therefore
67.5×3.45+ 0.041×3.45= 67.5v1 + 0.041×34
233.02 = 1.39+ 67.5v1
67.5v1 = 233.02 - 1.39 = 231.61
v1 = 231.61 / 67.5
v1 = 3.43m/s
The second person and the ball move together with a common velocity after catching the ball.
For the second person and the ball let their final common velocity be v
Mb2 + Mc2 = Mb3 + Mc3
0.041 × 34 + 57.5 ×0 = (57.5 + 0.041)×v
57.541v = 1.39
v = 1.39 /57.541
v = 0.0242m/s
Answer:
The desire to continue with your exercise would be the correct answer! :D
Answer:
Series circuit
Explanation:
Because the electrical components are connected one after each other in a single loop, rather than alongside each other, forming extra loops.
Given:
k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs
Solution:
F = -kx
mx" = -kx
x" + (k/m)x = 0
characteristic equation:
r^2 + k/m = 0
r = i*sqrt(k/m)
x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t)
ω = sqrt(k/m)
2π/T = sqrt(k/m)
T = 2π*sqrt(m/k)
T = 2π*sqrt(0.03106 slugs / 100 lb/ft)
T = 0.1107 s (period)
x(0) = 1/12 ft = 0.08333 ft
x'(0) = 0
1/12 = Asin(0) + Bcos(0)
B = 1/12 = 0.08333 ft
x' = Aω*cos(ωt) - Bω*sin(ωt)
0 = Aω*cos(0) - (1/12)ω*sin(0)
0 = Aω
A = 0
So B would be the amplitude. Therefore, the equation of motion would be x
= 0.08333*cos[(2π/0.1107)t]
