Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m
Vf = 480.83 m/s
a=v/t
t=v/a
t=480.83/34
t=14.142 s
Answer:
D. only briefly while being connected or disconnected.
Explanation:
As we know that transformer works on the principle of mutual inductance
here we know that as per the principle of mutual inductance when flux linked with the primary coil charges then it will induce EMF in secondary coil
So here when AC source is connected with primary coil then it will give output across secondary coil because AC source will have change in flux with time.
Now when we connect DC source across primary coil then it will not induce any EMF across secondary coil because DC source is a constant voltage source in which flux will remain constant always
So here in DC source the EMF will only induce at the time of connection or disconnection when flux will change in it while rest of the time it will give ZERO output
so correct answer will be
D. only briefly while being connected or disconnected.
Parallel has more than one circuit or form of energy
series has only one form of energy circuit
Finding acceleration= final velocity-initial velocity/ time taken (or A= V-U/T)
Final speed= 2m
Initial speed= 0m
Time taken= 2 seconds
2-0/2 so it’ll be 1m/s
2-0=0
2/2=
