A few different ways to do this:
Way #1:
The current in the series loop is (12 V) / (total resistance) .
(Turns out to be 2 Amperes, but the question isn't asking for that.)
In a series loop, the current is the same at every point, so it's
the same current through each resistor.
The power dissipated by a resistor is (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power. That's R1 .
And by the way, it's not "drawing" the most power. It's dissipating it.
Way #2:
Another expression for the power dissipated by a resistance is
(voltage across the resistance)² / (resistance) .
In a series loop, the voltage across each resistor is
[ (individual resistance) / (total resistance ] x battery voltage.
So the power dissipated by each resistor is
(individual resistance)² x [(battery voltage) / (total resistance)²]
This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .
Way #3: (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.
===> When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).
Answer: 37.5 kg in 3 s.f.
Explanation:
Answer:
Gamma Rays
Explanation:
Gamma rays have the highest energies the shortest wavelengths and the highest frequencies. They have the highest energies because they have the highest frequencies, the higher the frequency the more energy. ( I think, maybe double check with other answers )
Good Luck :)
The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
Answer:
Distance, d = 778.05 m
Explanation:
Given that,
Force acting on the car, F = 981 N
Mass of the car, m = 1550 kg
Initial speed of the car, v = 25 mi/h = 11.17 m/s
We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:
Let d is the distance covered by car. Using second equation of motion as :
So, the car will cover a distance of 778.05 meters.
