Answer: E = 0.85
Therefore the efficiency is: E = 0.85 or 85%
Explanation:
The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH
E = W/QH.
W=QH – QC,
Where Qc is the output heat.
That is,
E=1 - Qc/QH
E =1 - Tc/TH
where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir.
Note: The unit of temperature must be in Kelvin.
Tc = 300K
TH = 2000K
Substituting the values of E, we have;
E = 1 - 300K/2000K
E = 1 - 0.15
E = 0.85
Therefore the efficiency is: E = 0.85 or 85%
<span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is:
Ff = UsN
where Us is the coefficient of static friction and N is the normal force.
In order to get the crate moving you must first apply enough force to overcome the static friction:
Fapplied = Ff
Since Fapplied = 43 Newtons:
Fapplied = Ff = 43 = UsN
and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11
43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>
The answer is B. Nonrenewable
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:
80.33F
Explanation:
(300-273.15)*9/5+32=80.33
