Answer:
6.77 m/s
Explanation:
First, in the x direction:
Given:
Δx = 3.17 m
v₀ = v cos 30.8° = 0.859 v
a = 0 m/s²
Δx = v₀ t + ½ at²
(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²
3.17 = 0.859 v t
3.69 = v t
Next, in the y direction:
Given:
Δy = 0.432 m
v₀ = v sin 30.8° = 0.512 v
a = -9.81 m/s²
Δy = v₀ t + ½ at²
(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²
0.432 = 0.512 v t − 4.905 t²
Two equations, two variables. Solve for t in the first equation and substitute into the second equation:
t = 3.69 / v
0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²
0.432 = 1.89 − 66.8 / v²
66.8 / v² = 1.458
v² = 45.8
v = 6.77
Answer:
Explanation:
Given an LC circuit
Frequency of oscillation
f = 299 kHz = 299,000 Hz
AT t = 0 , the plate A has maximum positive charge
A. At t > 0, the plate again positive charge, the required time is
t =
t = 1 / f
t = 1 / 299,000
t = 0.00000334448 seconds
t = 3.34 × 10^-6 seconds
t = 3.34 μs
it will be maximum after integral cycle t' = 3.34•n μs
Where n = 1,2,3,4....
B. After every odd multiples of n, other plate will be maximum positive charge, at time equals
t" = ½(2n—1)•t
t'' = ½(2n—1) 3.34 μs
t" = (2n —1) 1.67 μs
where n = 1,2,3...
C. After every half of t,inductor have maximum magnetic field at time
t'' = ½ × t'
t''' = ½(2n—1) 1.67μs
t"' = (2n —1) 0.836 μs
where n = 1,2,3...
