Answer:
i am not a point taker i help people when they need help
Explanation:
3
i think it is right
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:
i think its 1.5 m/s to the right?
Explanation:
To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are
The radius then would be
And
PART A ) For the Surface Area we have that,
Deriving we have that the change in the Area is equivalent to the maximum error, therefore
Maximum error:
The relative error is that between the value of the Area and the maximum error, therefore:
PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so
Therefore the Maximum Error would be,
Replacing the value for the radius
And the relative Error
