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3 years ago

10

Consider the following cyclic process carried out in two steps on a gas. Step 1: 44 J of heat is added to the gas, and 20. J of

expansion work is performed. Step 2: 61 J of heat is removed from the gas as the gas is compressed back to the initial state. Calculate the work for the gas compression in Step 2.

1 answer:

notka56 [123]3 years ago

3 0

Answer:37 J

Explanation:

Given

Step :1

Heat added Q=44 J

Work done=-20 J

$\Delta E_1=Q+W=44-20=24 J$

Step :2

Heat added Q=-61 J

work done $W_2$

$\Delta E_2=Q+W_2$

$\Delta E_2=61+W_2$

$\Delta E_1+\Delta E_2=0$

as the process is cyclic

$44-20-61+W_2=0$

$W_2=37 J$

work done in compression is 37 J

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The circuit you should use to find the open-circuit voltage is

fiasKO [112]

Answer:

Incomplete questions check attachment for circuit diagram.

Explanation:

We are going to use superposition

So, we will first open circuit the current source and find the voltage Voc.

So, check attachment for open circuit diagram.

From the diagram

We notice that R3 is in series with R4, so its equivalent is given below

Req(3-4) = R3 + R4

R(34) = 20+40 = 60 kΩ

Notice that R2 is parallel to the equivalent of R3 and R4, then, the equivalent of all this three resistor is

Req(2-3-4) = R2•R(34)/(R2+R(34))

R(234) = (100×60)/(100+60)

R(234) = 37.5 kΩ

We notice that R1 and R(234) are in series, then, we can apply voltage divider rule to find voltage in R(234)

Therefore

V(234) = R(234) / [R1 + R(234)] × V

V(234) = 37.5/(25+37.5) × 100

V(234) = 37.5/62.5 × 100

V(234) = 60V.

Note, this is the voltage in resistor R2, R3 and R4.

Note that, R2 is parallel to R3 and R4. Parallel resistor have the same voltage, then voltage across R2 equals voltage across R34

V(34) = 60V.

Now, we also know that R3 and R4 are in series,

So we can know the voltage across R4 which is the Voc we are looking for.

Using voltage divider

V4 = Voc = R4/(R4 + R(34)) × V(34)

Voc = 40/(40+60) × 60

Voc = 24V

This is the open circuit Voltage

Now, finding the short circuit voltage when we short circuit the voltage source

Check attachment for circuit diagram.

From the circuit we notice that R1 and R2 are in parallel, so it's equivalent becomes

Req(1-2) = R1•R2/(R1+R2)

R(12) = 25×100/(25+100)

R(12) = 20 kΩ

We also notice that the equivalent of Resistor R1 and R2 is in series to R3. Then, the equivalent resistance of the three resistor is

Req(1-2-3) = R(12) + R(3)

R(123) = 20 + 20

R(123) = 40 kΩ

We notice that, the equivalent resistance of the resistor R1, R2, and R3 is in series to resistor R4.

So using current divider rule to find the current in resistor R4.

I(4) = R(123) / [R4+R(123)] × I

I(4) = 40/(40+40) × 8

I(4) = 4mA

Then, using ohms law, we can find the voltage across the resistor 4 and the voltage is the required Voc

V = IR

V4 = Voc = I4 × R4

Voc = 4×10^-3 × 40×10^3

Voc = 160V

Then, the sum of the short circuit voltage and the open circuit voltage will give the required Voc

Voc = Voc(open circuit) + Voc(short circuit)

Voc = 24 + 160

Voc = 184V.

3 0

3 years ago

A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele

Fantom [35]

The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, $\Delta t$ :
$I= \frac{Q}{\Delta t}$
First we need to find the net charge flowing at a certain point of the wire in one second, $\Delta t=1.0 s$ . Using I=0.92 A and re-arranging the previous equation, we find
$Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C$

Now we know that each electron carries a charge of $e=1.6 \cdot 10^{-19} C$ , so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
$N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}$

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3 years ago

A solid sphere of weight 42.0 N rolls up an incline at an angle of 36.0°. At the bottom of the incline the center of mass of the

Alecsey [184]

Answer:

Part a)

$KE = 77.95 J$

Part b)

$L = 3.16 m$

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

$KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2$

$KE = \frac{7}{10} mv^2$

$KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)$

$KE = 77.95 J$

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

$mgLsin\theta = KE$

$\frac{42}{9.81}(9.81)L sin36 = 77.95$

$L = 3.16 m$

Part c)

by equation of energy conservation we know that

$\frac{7}{10}mv^2 = mgL sin\theta$

so here we can see that distance L is independent of the mass of the sphere

7 0

3 years ago

What is the y axis on a graph?

viktelen [127]

^^^^^^^^^^^^^^^^^^^^^^^^^^^ is correct

5 0

3 years ago

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

irakobra [83]

Answer:

1.2 amps :)

Explanation:

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

Known:

R = 10.0 Ω

V = 12.0 V

Unknown:

I = ???

I = V/R

= 12.0 V / 10.0 Ω

= 1.2 amps

6 0

3 years ago