To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)
To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)
If a problem says the acceleration is some positive value than solve using that value, a negative acceleration is said to be deceleration. E.g. a car decelerating at 10 m/sec can be said to be accelerating at -10 m/sec.
If a problem states decelerates at A, then use -A for acceleration in the classic equations which are for acceleration. If a problem says accelerates at a negative value like -A the use -A as the value for acceleration, it can also be said to be decelerating at A.
Answer:
Its new current will be 240.0milliAmps
Explanation:
According to ohms law,
V = IR
V is the supply voltage in a circuit
I is the current
R is the resistance
If the current in the wires of a circuit is 120.0 milliAmps, the voltage will be:
V1 = 120R1
R1 = V1/120...(1)
If the voltage impressed across the ends of the circuit were doubled (with no change in its resistance),
V2 = 2V1
V2 = IR2
2V1 = IR2
R2 = 2V1/I ... (2)
Since there are no changes in resistance
R1 = R2
Equating 1 and 2, we have:
V1/120 = 2V1/I
1/120 = 2/I
Cross multiplying
I = 2×120
I = 240milliAmps
Answer:
Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;
The closest (minimum) distance of Neptune from the Earth is 29 AU
The farthest (maximum) distance of Neptune fro the Earth is 31 AU
Explanation:
The following parameters are given;
The distance from the Earth to the Sun = 1 AU
The distance of Neptune from the Earth = 30 AU
We have;
When the Sun is between the Earth and Neptune, the distance is found by the relation;
Distance from the Earth to Neptune = 30 + 1 = 31 AU
When the Earth is between the Sun and Neptune, the distance is found by the relation;
Distance from the Earth to Neptune = 30 - 1 = 29 AU
Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU
The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.