What kind of thermal transfer 1.conduction 2.convective 3.radiation from the sun?

A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the ope

MariettaO [177]

Answer:

Xc = (0.467 - 0.427j)R

Explanation:

Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is

Z = √[R² + (XL - XC)²]

Since the inductive reactance XL equals the resistance R, we have that

Z = √[R² + (XL - XC)²]

Z = √[R² + (R - XC)²]

Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]

Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus

Z' = √[R² + (R - XC')²]

Z' = √[R² + (R - 2XC)²]

The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]

Since the current doubles, I' = 2I.

V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]

1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]

√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]

squaring both sides, we have

[R² + (R - XC)²] = 4[R² + (R - 2XC)²]

expanding the brackets, we have

[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]

[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]

2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²

collecting like terms, we have

16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²

14RXC - 15XC² = 6R²

15XC² - 14RXC + 6R² = 0

Using the quadratic formula to find XC, we have

$XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri$

Since it is capacitive, we take the negative part.

So, Xc = (0.467 - 0.427j)R

4 0

3 years ago

Two moles of helium are initially at a temperature of 21.0 ∘Cand occupy a volume of 3.30×10−2 m3 . The helium first expands at c

Anettt [7]

Answer:

(B) The total internal energy of the helium is 4888.6 Joules

(C) The total work done by the helium is 2959.25 Joules

(D) The final volume of the helium is 0.066 cubic meter

Explanation:

(B) ∆U = P(V2 - V1)

From ideal gas equation, PV = nRT

T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3

P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal

∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules

(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal

Assuming a closed system

(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules

(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter

6 0

4 years ago

A +2.00nc point charge is at the origin, and a second -5.00nc point charge is on the x-axis at x = 0.800m find the magnitude of

Damm [24]

You have to reduce 2.00 an5.00 I order to use the×that=0.800

3 0

3 years ago

A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

IrinaK [193]

Answer : The specific heat capacity of the alloy $1.422J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$C_1$ = specific heat of alloy = ?

$C_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of alloy = 21.6 g

$m_2$ = mass of water = 50.0 g

$T_f$ = final temperature of system = $31.10^oC$

$T_1$ = initial temperature of alloy = $93.00^oC$

$T_2$ = initial temperature of water = $22.00^oC$

Now put all the given values in the above formula, we get

$21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC$

$c_1=1.422J/g^oC$

Therefore, the specific heat capacity of the alloy $1.422J/g^oC$

6 0

3 years ago

An uncharged series RC circuit is to be connected across a battery. For each of the following changes, determine whether the tim

slavikrds [6]

a) Increase

b) Unchanged

c) Increase

Explanation:

a)

The charge on a capacitor charging in a RC circuit connected to a battery follows the exponential equation:

$Q(t)=Q_0 (1-e^{-\frac{t}{RC}})$

where

$Q_0 = CV$ is the final charge stored in the capacitor, where C is the capacitance and V is the voltage of the battery

t is the time

R is the resistance of the circuit

The capacitor reaches 90% of its final charge when

$Q(t)=0.90Q_0$

Substituting and re-arranging the equation, we find:

$0.90Q_0 = Q_0(1-e^{-\frac{t}{RC}})\\0.90=1-e^{-\frac{t}{RC}}\\e^{-\frac{t}{RC}}=0.10\\-\frac{t}{RC}=ln(0.10)\\t=-RCln(0.10)=2.30RC$

We see that if we double the RC constant, then $(RC)'=2(RC)$

So the time taken will double as well:

$t'=2.30(RC)'=2.30(2RC)=2(2.30RC)=2t$

So, the answer is "increase"

b)

In this second part, the battery voltage is doubled.

According to the equation written in part a),

$Q_0 =CV$

this means also that the final charge stored on the capacitor will also double.

However, the equation that gives us the time needed for the capacitor to reach 90% of its full charge is

$t=2.30 RC$

We see that this equation does not depend at all on the voltage of the battery.

Therefore, if the battery voltage is doubled, the final charge on the capacitor will double as well, but the time needed for the capacitor to reach 90% of its charge will not change.

So the correct answer is

"unchanged"

c)

In this case, a second resistor is added in series with the original resistor of the circuit.

We know that for two resistors in series, the total resistance of the circuit is given by the sum of the individual resistances:

$R=R_1+R_2$

Since each resistance is a positive value, this means that as we add new resistors, the total resistance of the circuit increases.

Therefore in this problem, if we add a resistor in series to the original circuit, this means that the total resistance of the circuit will increase.

The time taken for the capacitor to reach 90% of its final charge is still

$t=2.30 RC$

As we can see, this time is directly proportional to the resistance of the circuit, R: therefore, if we add a resistor in series, the resistance of the circuit will increase, and therefore this time will increase as well.

So the correct answer is

"increase"

8 0

3 years ago