Answer:
B) - 5.0 m
Explanation:
B is located on a positive location, 15m from the starting point A. Hence, since E is located a positive distance 10m from A, the difference becomes 10 - 15 = - 5.0 m
Kinetic Energy = 1/2mv^2
m= 1200kg
v= 24 m/s
KE = 1/2 (1200kg)(24m/s)^2 = 345,600 N
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...
Answer:
What is observed that connects radio emissions in the galactic nucleus with the emissions in the halo or radio lobes? There is a jet of matter coming out of the nucleus, which often points toward the lobes.
