Answer:
Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere
Explanation:
The formula to be used here is
Q = It
where Q is the quantity of electricity and it is measured coulombs (C); 2.8 × 10⁻⁸ C or 0.000000028 C
I is current and it is measured in ampere (amps or A); unknown
t is time and it is measured in seconds (s); 0.05 s
Since, average current is what is unknown
I =Q/t
I = 0.000000028/0.05
I = 5.6 × 10⁻⁷ A
Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere
Let car A's starting position be the origin, so that its position at time <em>t</em> is
A: <em>x</em> = (40 m/s) <em>t</em>
and car B has position at time <em>t</em> of
B: <em>x</em> = 100 m - (60 m/s) <em>t</em>
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They meet when their positions are equal:
(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>
(100 m/s) <em>t</em> = 100 m
<em>t</em> = (100 m) / (100 m/s) = 1 s
so the cars meet 1 second after they start moving.
They are 100 m apart when the difference in their positions is equal to 100 m:
(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m
(subtract car B's position from car A's position because we take car A's direction to be positive)
(100 m/s) <em>t</em> = 200 m
<em>t</em> = (200 m) / (100 m/s) = 2 s
so the cars are 100 m apart after 2 seconds.
Recreational boaters have a role in keeping the waterways safe and secure. Violators of the restrictions can expect a quick and severe response. The operator should not approach within yards and slow to minimum speed within 500 yards of any of united states naval vessel. If the vessel need to pass within 100 yards of a united states naval vessel for safe passage, must contact the united states naval vessel or the united states coast guard escort vessel on VHF-FM channel 16. In addition,
• The operator must perceive and avoid all security areas.
• Avoid commercial port operation areas particularly those that include military cruise line or petroleum amenities.
• The operator must perceive and avoid other controlled areas near dams, power plants, etc.
• The operator of the vessel must not discontinue or anchor underneath bridges or in the channel.
Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,
Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,
Therefore, the resistance of the bulb is 484 Ω
Answer:
2649600 Joules
Explanation:
Efficiency = 40%
m = Mass of air = 92000 kg
v = Velocity of wind = 12 m/s
Kinetic energy is given by
The kinetic energy of the wind is 6624000 Joules
The wind turbine extracts 40% of the kinetic energy of the wind
The energy extracted by the turbine every second is 2649600 Joules
