Answer:
dC = 72m < dA = 80m < dB = 90m
Explanation:
In order to calculate the distance traveled by each mobile, you use the following formula:
(1)
d: distance traveled
v: speed of the car
t: time of the motion
You replace the values of v and t for each mobile:
mobile A:
mobile B:
mobile C:
Then, you obtain that he distances traveled by the mobiles show:
dC = 72m < dA = 80m < dB = 90m
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:
0.7 mJ
Explanation:
<u>Identify the unknown: </u>
The work required to turn the dial from 180° to 0°
<u>List the Knowns: </u>
Capacitance when the dial is set at 180°: C = 350 pF = 350 x 10^-12 F Capacitance when the dial is set at 0°: C = 100 pF = 100 x 10^-12 F
Voltage of the battery: V = 130 V
<u>Set Up the Problem:</u>
<em><u>Energy stored in a capacitor: </u></em>
U_c=1/2*V^2*C
=1/2*Q^2/C
<em><u>When the dial is set at 180°:</u></em><em> </em>
U_c=1/2*(130)^2*350*10^-12=10^-4
Q=√2*U_c*C=4*10^-7
<u><em>When the dial is set at 0°:</em></u>
U_c=1/2*(4*10^-7)^2/100*10^-12
=8*10^-4 J
<u><em>Solve the Problem: </em></u>
ΔU_c=7*10^-4 J
=0.7 mJ
note:
there maybe error in calculation but method is correct
Gravitational force equals GMm/r^2, where G is constant, M and m are the masses, and r is distance.
For I, if both masses double, the formula becomes G2M2m/r^2, or 4GMm/r^2. Therefore, the gravitational force will quadruple or 4x.
For II, if the distance between the object doubles, the formula becomes GMm/(2r)^2 or GMm/4r^2. In this case, the gravitational force is 1/4x the initial force.
the correct answer is unbalanced because if it was balanced it would be floating because it would have no push or pull force.
