Answer:
A) 3.48m/s
B) 3.92m
C) 2.32m
D 23.33m/s
Explanation:
ma(t)=mg-At
a(t)= g - (At/m)
V(t)= integrala(T)dT = gt- (At^2/2m)
Initial x coordinator of the box is zero
X(t)= integralV(t)dt= 1/2gt^2-(At^3/6m)
a) V =( 9.81×1) -(38×1^2/2×3)
V= 9.81-6.33= 3.48m/s
b)-AT^2/2m + gT= 0
T=2mg/A= (2×3×9.81)/38
T= 1.549m
X(T)= (1/2×9.81×1.549^3)- (38×1.549^3/6×3)
X(T)= 11.768- (141.23/18) = 11.768 - 7.85= 3.92m
C) 1/2gT''^2 - AT''^3/6m =0
The only non trivial solution is T''= 3mg/A
T=(3×3×9.81)/38 = 2.32m
D) V = 9.81×3) - (38×3^2/6)
V= 29 - 5.667= 23.33m/s
Answer:
a. 2 Hz b. 0.5 cycles c . 0 V
Explanation:
a. What is period of armature?
Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz
b. How many cycles are completed in T/2 sec?
The period, T = 1/f = 1/2 Hz = 0.5 s.
So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,
Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.
c. What is the maximum emf produced when the armature completes 180° rotation?
Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0
E = E₀ × 0 = 0
E = 0
So, at 180° rotation, the maximum emf produced is 0 V.
