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Questions 1 - 10

You have 4 resistors, , , , and , set up like this:

4resistorcircuit

Their resistance are as follows:

$\begin{align*} A&=1\Omega \ B&=3\Omega \ C&=4\Omega \ D&= 2\Omega \end{align*}$

If the battery has 8V, what is the total power dissipated through the resistors?

Explanation

The equation for power is

In order to get the power, we need the current. To find the current, we need to get the total resistance, and use Ohm's Law ().

To find the total resistance, remember the equations for adding resistors is this:

$(Series)\ R_{tot}&=R_1+R_2+...+R_n$

$(Parallel)\ \frac{1}{R_{tot}}&=\frac{1}{R_1}+\frac{1}{R_2}=...=\frac{1}{R_n}$

Resistors and are in series, resistors and are in parallel, and resistors and are in series.

$R_{AB}&=1+3$

$R_{AB}= 4$

$R_{ABC}&=(\frac{1}{R_{AB}}+\frac{1}{R_C})^{-1}$

$R_{ABC}= (\frac{1}{4}+\frac{1}{4})^{-1}$

$R_{ABC}= 2$

$R_{total}&=R_{ABC}+R_{D}$

$R_{total}= 2+2$

$R_{total}= 4$

Now, we can find the current.

$I&=\frac{V}{R}$

$I=\frac{8}{4}$

Finally, we can find the power.

Therefore, the power is 16W (watts).

Which of the following is not an acceptable unit for specific heat?

$\frac{J}{N\cdot K}$

$\frac{J}{g\cdot K}$

$\frac{cal}{g\cdot K}$

$\frac{Btu}{^oF\cdot lb}$

$\frac{J}{mol\cdot K}$

Explanation

Specific heat is most commonly applied in the equation to determine the heat required/released during a temperature change:

$Q=mc\Delta T$

Rearranging this equal, we can see that this units of specific heat are units of heat per units of mass times units of temperature.

$c=\frac{Q}{m\Delta T}$

Heat energy can have the units of Joules or calories. Mass can have units of grams or kilograms. Temperature can have units of Kelvin, degrees Celsius, or degrees Fahrenheit. British thermal units (Btu) is a less common unit of heat, defined by the amount of heat required to raise one pound of water by one degree Fahrenheit.

Newtons are a unit of force, and cannot be used in the units for specific heat.

Two parallel conducting plates separated by a distance are connected to a battery with voltage . If the distance between them is doubled and the battery stays connected, which of the following statements are correct?

The electric charge on the plates is halved

The potential difference between the plates is doubled

The capacitance is unchanged

The potential difference between the plates is halved

The electric charge on the plates is doubled

Explanation

The equation for capacitance of parallel conducting plates is:

$C=\frac{\epsilon_0A}{d}$

When the distance is doubled, the capacitance changes to:

$C'=\frac{\epsilon_0A}{d'}=\frac{\epsilon_0A}{2d}=\frac{1}{2}C$

The battery is still connected to the plates, the potential difference is unchanged. Because , and the capacitance is halved while the voltage stays the same, must necessarily drop to half to account for the change.

Lazy capacitor

Consider the given diagram. If , each plate of the capacitor has surface area , and the plates at apart, determine the excess charge on the positive plate.

$3.54*10^{-10}C$

$9.26*10^{-10}C$

$6.11*10^{-10}C$

$1.11*10^{-10}C$

$4.44*10^{-10}C$

Explanation

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$

Combine equations and solve for the electric field:

$\frac{V_1}{d}=E$

Convert mm to m and plug in values:

$\frac{V_1}{d}=E$

$4000\frac{N}{C}=E$

Using the electric field in a capacitor equation:

$E=\frac{q}{A*\epsilon_0}$

Rearrange to solve for the charge:

$E*\epsilon_0*A=q$

Convert to and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

$3.54*10^{-10}C=q$

A circuit contains a battery and a $20\Omega$ resistor in series. Determine the magnitude of the magnetic field outside of the loop away from the wire.

None of these

Explanation

Using

$B=\frac{\mu_0*I}{2\pi*r}$

Converting to and plugging in values

$B=\frac{4\pi*10^{-7}*I}{2\pi*.002}$

Determining current:

$B=\frac{4\pi*10^{-7}*3}{2\pi*.002}$

Lazy capacitor

Consider the given diagram. If , each plate of the capacitor has surface area , and the plates are 0.1mm apart, determine the number of excess electrons on the negative plate.

$e=1.6*10^{-19}C$

$n=2.21*10^{9}$

$n=5.85*10^{9}$

$n=5.15*10^{9}$

$n=3.33*10^{9}$

None of these

Explanation

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$

Combine equations and find the electric field:

$\frac{V_1}{d}=E$

Convert mm to m and plugg in values:

$\frac{V_1}{d}=E$

$4000\frac{N}{C}=E$

Use the electric field in a capacitor equation:

$E=\frac{q}{A*\epsilon_0}$

$E*\epsilon_0*A=q$

Convert to and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

$3.54*10^{-10}=q$

The magnitude of total charge on the positive plate is equal to the total charge on the negative plate, so to find the number of excess elections:

$\frac{3.54*10^{-10}}{1.6*10^{-19}}=n$

$n=2.21*10^{9}$

The half-life of carbon-14 is 5730 years.

Rex the dog died in 1750. What percentage of his original carbon-14 remained in 1975 when he was found by scientists?

Explanation

225 years have passed since Rex died. Find the number of half-lives that have elapsed.

$\frac{225}{5730}=0.039267$

To find the proportion of a substance that remains after a certain number of half-lives, use the following equation:

$\left(\frac{1}{2} \right )^n$

Here, is the number of half lives that have elapsed.

$\left(\frac{1}{2} \right )^{0.039267}=0.973\cdot 100%=97.3%$

Suppose that the half-life of substance A is known to be . When substance A decays, it becomes substance B. If this is the only way that substance B can be produced, and a sample is found that contains of substance A and of substance B, what is the age of this sample?

Explanation

To find the approximate age of the sample, we need to consider the half-life of the starting material, substance A. We also need to consider the relative amounts of substance A and substance B.

Since we are told that the sample we are looking at contains substance A and substance B, we know that the sample must have started out with of substance A. Furthermore, in order to go from to , we know that a total of two half-lives must have passed. And since we know that one half-life is equal to , we can conclude that a total of must have passed.