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Olin [163]
3 years ago
7

A resistor R and a capacitor C are connected in series to a battery of terminal voltage V0. Which of the following equations rel

ating the current I in the circuit and the charge Q on the capacitor describes this circuit?
(A) V0 + QC – I2R = 0
(B) V0 – Q/C – IR = 0
(C) V02 – Q2/2C – I2R = 0
(D) V0 – CI – I2R = 0
(E) Q/C – IR = 0
Physics
1 answer:
Lesechka [4]3 years ago
5 0
Answer-C VO2-Q2/2C-I2R=0
Q1/2C currents upon the charge I2R on the capacitor. Voltage V02 goes best
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Which sound waves in the electromagnetic spectrum are considered low energy waves
Mariulka [41]
Radio waves are the waves with the lowest energy in the electromagnetic spectrum. X-rays and gamma rays are the highest. Sound is not part of the electromagnetic spectrum.
7 0
3 years ago
I need help thanks :)))))))))
dsp73

°C = (5/9) · (°F-32)

The "wet" thermometer is the upper one ... you can see the wet cloth wrapped around the bulb at the end.  It's reading 70° F.

°C = (5/9) · (38) = 21.1° C

The "dry" thermometer is the lower one.  It's reading 80° F.

°C = (5/9) · (48) = 26.7° C

So it looks like choice-A is your answer.

6 0
2 years ago
______ can occur when water-saturated soil turns from a solid to a liquid as a result of an earthquake.
jek_recluse [69]
The answer is liquefaction
6 0
2 years ago
xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi
slega [8]

Answer:

Approximately \rm 90\; kJ.

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (E^{\circ}(\text{cell})) is equal to

E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode}).

There are two half-reactions in this question. \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu and \rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of E^{\circ}(\text{cell}) should be positive.

In this case, E^{\circ}(\text{cell}) is positive only if \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu is the reaction takes place at the cathode. The net reaction would be

\rm Cu^{2+} + Pb \to Cu + Pb^{2+}.

Its cell potential would be equal to 0.339 - (-0.130) = \rm 0.469\; V.

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell}),

where

  • n is the number moles of electrons transferred for each mole of the reaction. In this case the value of n is 2 as in the half-reactions.
  • F is Faraday's Constant (approximately 96485.33212\; \rm C \cdot mol^{-1}.)

\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}.

5 0
2 years ago
Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3
RideAnS [48]

Answer:

Expression of work done is

W = Fd cos\theta

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

W = Fd cos\theta

here we know that

F = 12.6 N

d = 15.4 m

\theta = 15.3 degree

so we will have

W = 12.6 \times 15.4 cos15.3

W = 187.2 J

5 0
3 years ago
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