Answer;
-The white light shines forward and to both sides (225 Degrees) and is required on all power-driven vessels. A masthead light must be displayed by all vessels when under engine power. The absence of this light indicates a sailboat under sail.
Explanation;
-Masthead light; Means a white light placed over the fore and aft center-line of the vessel showing an unbroken light over an arc of the horizon of 225 degrees and so fixed to show the light from right ahead to 22.5 degrees abaft of the beam on either side of the vessel, except that on a vessel of less than 12 meters (39'4") in length the masthead light shall be placed as nearly as practical to the fore and aft center-line of the vessel.
Answer:
0.28802
2.57162 W
14.28 W
53.55 W
6.07142 W
Explanation:
R = 280Ω
L = 100 mH
C = 0.800 μF
V = 50 V
ω = 10500rad/s
For RLC circuit impedance is given by
Power factor is given by
The power factor is 0.28802
The average power to the circuit is given by
The average power to the circuit is 2.57162 W
Power to resistor
Power to resistor is 14.28 W
Power to inductor
Power to the inductor is 53.55 W
Power to the capacitor
The power to the capacitor is 6.07142 W
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass kg, charge +e = C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed m/s. The proton comes momentarily to rest at a distance m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are m apart?
Explanation:
The given data is as follows.
Mass of proton = kg
Charge of proton =
Speed of proton =
Distance traveled =
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
=
where,
U =
Putting the given values into the above formula as follows.
U =
=
=
Therefore, we can conclude that the electric potential energy of the proton and nucleus is .