Explanation:
Formula for maximum efficiency of a Carnot refrigerator is as follows.
..... (1)
And, formula for maximum efficiency of Carnot refrigerator is as follows.
...... (2)
Now, equating both equations (1) and (2) as follows.
=
=
=
= 2.5
Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.
Answer:
- Newton's first law applies. An object at rest will stay that way until a force is applied.
- Any amount of effort can be applied to any amount of mass (in the ideal case). The question is not sufficiently specific.
Explanation:
A force is required to move an object because the object will stay at rest until a force is applied.
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The effort required to lift or push two masses instead of one depends on the desired effect. For the same kinetic energy, no more effort is required. For the same momentum, half the effort is required for two masses. For the same velocity, double the effort is required.
Answer:
its 1
Explanation:
Several factors can increase the rate of a chemical reaction. In general, anything that increases the number of collisions between particles will increase the reaction rate, and anything that decreases the number of collisions between particles will decrease the chemical reaction rate.
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Answer:
70713
Explanation:
Because you need to multiply the amount of water lost (2430) by the time (29.1) which will equal 70713J/g needed to counter the loss.
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Answer:
Explanation:
Given that,
First Capacitor is 10 µF
C_1 = 10 µF
Potential difference is
V_1 = 10 V.
The charge on the plate is
q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC
q_1 = 100 µC
A second capacitor is 5 µF
C_2 = 5 µF
Potential difference is
V_2 = 5V.
Then, the charge on the capacitor 2 is.
q_2 = C_2 × V_2
q_2 = 5µF × 5 = 25 µC
Then, the average capacitance is
q = (q_1 + q_2) / 2
q = (25 + 100) / 2
q = 62.5µC
B. The two capacitor are connected together, then the equivalent capacitance is
Ceq = C_1 + C_2.
Ceq = 10 µF + 5 µF.
Ceq = 15 µF.
The average voltage is
V = (V_1 + V_2) / 2
V = (10 + 5)/2
V = 15 / 2 = 7.5V
Energy dissipated is
U = ½Ceq•V²
U = ½ × 15 × 10^-6 × 7.5²
U = 4.22 × 10^-4 J
U = 422 × 10^-6
U = 422 µJ