The answer to the set of questions are as follows:
a. 2 ( 2/w2C) = 4xc2 = Xl2 / Xc2 = 4
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The inductors reactance is greater than the capacitor
b. </span>(1/ 3w1C ) = (1/ 9w3C ) = 1/9 Xc3 = X l2/ Xc2 = 1/9
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The capacitors reactance is greater than inductor
I hope my answer has come to your help. God bless and have a nice day ahead!
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The objects that look black are all tied for first place.
Answer:
F t = m Δv impulse delivered = change in momentum
Δv = 100 * .1 / .5 = 20 m/s original speed of puck
KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J initial KE of puck
E = μ m g d energy lost by puck
Ff = μ m g = m a deceleration of puck due to friction
a = μ g = 9.8 * .2 = 1.96 m/s^2
v2 = a t + v1 = -1.96 * 4 + 20 = 12.2 m/s speed of puck on striking box
m v2 = M V conservation of momentum when puck strikes box
V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s speed of box after collision
KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J KE of box after collision
KE = μ M g d energy lost by box in sliding distance d
d = 23.3 / (.3 * .8 * 9.8) = 9.91 m distance box slides
Explanation:
Average acceleration is change in velocity over time.
a = Δv / Δt
a = (22.0 m/s − (-25.0 m/s)) / 0.00350 s
a = 13,400 m/s²
