Answer:
When a wire that carries electrical current is placed in a magnetic field the wire experiences a force.
Explanation:
The electric motor is a device which convert electrical energy into mechanical energy ie when current carrying conductor is placed in magnetic field it experience a force. Flemings left hand rule explains the direction of the current.
The electric motor works by attraction and repulsion of magnetic field.
So the option d explains basic concept of simple motor ie when when current carrying wire is placed in the magnetic field it experience magnetic repulsive force.
Answer:
15 Joules
Explanation:
work = charge x potential difference
= 10 x 1.5
= 15
Answer:
C) only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results.
Explanation:
Selectivity is the ability of a receiver to respond only to a specific signal on a wanted frequency and reject other signals nearby in frequency.
If a receiver is overly selective, only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. Whereas, if a receiver is underselective, the receiver can pick different signals on different frequencies at the same time.
The solution to the problem is as follows:
<span>First, I'd convert 188 mi/hr to ft/s. You should end up with about ~275.7 ft/s.
So now write down all the values you know:
Vfinal = 275.7 ft/s
Vinitial = 0 ft/s
distance = 299ft
</span>
<span>Now just plug in Vf, Vi and d to solve
</span>
<span>Vf^2 = Vi^2 + 2 a d
</span><span>BTW: That will give you the acceleration in ft/s^2. You can convert that to "g"s by dividing it by 32 since 1 g is 32 ft/s^2.</span>
As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.
The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.
The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.
A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.
Learn more about capacitor here:-brainly.com/question/14883923
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