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san4es73 [151]
3 years ago
13

Does anyone know how to work this out?

Physics
1 answer:
Vladimir [108]3 years ago
8 0

Answer:

\frac{17}{14}*5 = Approximately 6

Explanation:

<u>What we need to know:</u>

1. Given that this is a parallel circuit, each pathway will receive the full voltage from the source. Therefore, each pathway will receive 17V.

2. The current (I) stays the same for all loads in series. Therefore, I2 and I3 are equal since they are connected in series.

3. Ohm's law states that Voltage = Current × Resistance (V=IR)(I=V/R)(R=V/I)

<u>1) Calculate the current for the path containing R2 and R3</u>

I_2_,_3=\frac{V_2_,_3}{R_2_,_3}\\I_2_,_3=\frac{17}{5+9}\\I_2_,_3=\frac{17}{14}

Because the current running through this path is \frac{17}{14} A, then I2 is \frac{17}{14} A.

<u>2) Use Ohm's Law by plugging in I2 and R2</u>

V_2=I_2*R_2\\V_2=\frac{17}{14} *5

V_2=6 (approximate)

Therefore, V2 is approximately 6V.

I hope this helps!

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4 years ago
In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_1-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_3)}{Q_1}}

also,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}

or

\frac{T_3}{T_1}=\frac{Q_3}{Q_1}

thus,

\eta=1-\frac{(T_3)}{T_1}}

thus,

\eta=1-\frac{(240\ K)}{500\ K}}

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Efficiency = 52%

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A charge of 2.00 μC flows onto the plates of a capacitor when it is connected to a 12.0-V potential source. What is the minimum
uysha [10]

To develop this problem we will apply the concept of energy conservation. For which the work carried out must be equivalent to the potential energy stored on the capacitor. We will start by finding the capacitance to later be able to calculate the energy and therefore the work in the capacitor

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Here,

C = Capacitance

V = Potential difference between the plates

Q = Charge between the capacitor plates

At the same time the energy stored in the capacitor can be defined as,

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We will start by finding the value of the capacitance, so we will have to,

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Finally using the expression for the energy we have that,

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8 0
4 years ago
Read 2 more answers
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