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

\(\displaystyle Q=mc\Delta T\)

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

\(\displaystyle c=\frac{Q}{m\Delta T}\)

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.

Our equation for temperature change using specific heat is:

\(\displaystyle Q = mc\Delta t\)

Since Celsius and Kelvin use the same interval scale, we do not need to convert to find the change in temperature. Using the initial and final temperatures, the energy input, and the specific heat of copper, we can calculate the mass of copper present.

\(\displaystyle 40000J=m(0.386\frac{J}{g\cdot K})(62^oC-54^oC)\)

To simplify, we will solve for the temperature change separately.

\(\displaystyle 62^oC-54^oC=8^oC\)

Since the scale for Celsius and Kelvin is the same, this change is the same in either scale.

\(\displaystyle \Delta 8^oC=\Delta 8K\)

Use this value back in the original calculation.

\(\displaystyle 40000J=m(0.386\frac{J}{g\cdot K})(8K)\)

\(\displaystyle m=\frac{40000J}{3.088\frac{J}{g}}=12953.4g\)

Convert to kilograms to simplify.

\(\displaystyle 12953.4g\approx13kg\)

Since the ice melts, this action occurs in three stages.

1) The heating of the ice to its melting point

2) The melting of the ice into water

3) The heating of the water after it is melted

When a temperature change is involved, the relationship of heat input to temperature change is given by:

\(\displaystyle Q = mc \Delta T\)

In this formula, \(\displaystyle c\) is the specific heat (capacity) for the substance. For the phase change involved in melting, there is no temperature change (the energy goes into breaking bonds), and the amount of heat required to completely melt the substance is:

\(\displaystyle Q=m\Delta H^o_f\)

So this situation is encompassed by three expressions, representing the three parts of the problem, each of which deals with the same mass of material, \(\displaystyle m\). Keep in mind that the melting point for ice is \(\displaystyle 0^oC\), or \(\displaystyle 273K\).

1)  \(\displaystyle Q_{ice}=mc_{ice}\Delta T_{ice}\)

\(\displaystyle Q_{ice}=(2kg)(2.050\frac{kJ}{kg\cdot K})(273K-271K)\)

\(\displaystyle Q_{ice}=8.2kJ\)

2)  \(\displaystyle Q_{melt}=m\Delta H^o_{f\ H_2O}\)  (for the ice completely melting)

\(\displaystyle Q_{melt}=(2kg)(334\frac{kJ}{kg})\)

\(\displaystyle Q_{melt}=668kJ\)

3)  \(\displaystyle Q_{water}=mc_{water} \Delta T_{water}\)  (for the water warming up to its final temperature)

\(\displaystyle Q_{water}=(2kg)(4.816\frac{kJ}{kg\cdot K})(T_f-273K)\)

The values for \(\displaystyle c_{ice}\), \(\displaystyle L_{water}\), and \(\displaystyle c_{water}\) are given, and it is known that \(\displaystyle Q_{ice} + Q_{melt} + Q_{water} = 700 kJ\).

\(\displaystyle 8.2kJ+668kJ+(2kg)(4.816\frac{kJ}{kg\cdot K})(T_f-273K)=700kJ\)

Using this equation we can solve for the final temperature.

\(\displaystyle 676.2kJ+(9.632\frac{kJ}{K})(T_f-273K)=700kJ\)

\(\displaystyle T_f-273K=\frac{23.8kJ}{9.632\frac{kJ}{K}}=2.47K\)

\(\displaystyle T_f=275.47K\)

You can simplify this question tremendously by using the definition of a dielectric constant. Dielectric constant is defined as the ratio of the electrostatic force in vacuum to the electrostatic force in the medium (in this case glass).

\(\displaystyle k=\frac{F_{e,vac}}{F_{e,glass}}=5\)

The question is asking for the reciprocal of this value: the ratio of the force in glass to the force in the vacuum. Our answer is calculated by taking the reciprocal of the dielectric constant of glass.

\(\displaystyle \frac{1}{k}=\frac{F_{e,glass}}{F_{e,vac}}=\frac{1}{5}=0.2\)

Moving a charge along an equipotential surface will involve no work done by the electric field. The potential is constant throughout an equipotential surface; therefore, the potential difference experienced by the electron will be zero. Remember that energy is dependent on the potential difference.

\(\displaystyle W=q\Delta V=q(E*d)\)

If the potential difference is zero then the energy (and work) will be zero.

This question is very simple if you realize that the force experienced by both charges is equal.

The definition of the two electrostatic forces are given by Coulomb's law:

\(\displaystyle F=-k\frac{q_1q_2}{r^2}\)

In this question, we can rewrite this equation in terms of our given system.

\(\displaystyle F_a = -k\frac{q_a q_b}{r^2}=-k\frac{(0.003C)(0.006C)}{(1.5m)^2}=-k(8*10^{-6}\frac{C^2}{m^2})\)

\(\displaystyle F_b = -k\frac{q_b q_a}{r^2}=-k\frac{(0.006C)(0.003C)}{(1.5m)^2}=-k(8*10^{-6}\frac{C^2}{m^2})\)

It doesn’t matter if the charges of the two particles are different; both particles experience the same force because the charges of both particles are accounted for in the electrostatic force equation (Coulomb's law). This conclusion can also be made by considering Newton's third law: the force of the first particle on the second will be equal and opposite the force of the second particle on the first.

Since the forces are equal, their ratio will be \(\displaystyle \small 1:1\).

Two principal realizations help with solving this problem, both derived from Gauss’ law for electricity:

1) The excess charge on an ideal conducting sphere is uniformly distributed over its surface

2) A uniform shell of charge acts, in terms of electric force, as if all the charge were contained in a point charge at the sphere’s center

With these realizations, an application of Coulomb’s law answers the question. If \(\displaystyle q_{2}\) is the point charge outside the sphere, then the force \(\displaystyle F_{q2}\) on \(\displaystyle q_{2}\) is:

\(\displaystyle F_{q2} = \frac{kq_{sphere}q_{2}}{r^{2}}\)

In this equation, \(\displaystyle k\) is Coulomb’s constant, _{\(\displaystyle q_{sphere}\)}is the excess charge on the spherical conductor, and \(\displaystyle r\) is total distance in meters of \(\displaystyle q_{2}\) from the center of the conducting sphere.

\(\displaystyle r=0.20m+0.30m=0.50m\)

Using the given values in this equation, we can calculate the generated force:

\(\displaystyle F_{q2}=\frac{(8.99*10^9\frac{m^2N}{C^2})(10*10^{-6}C)(5*10^{-6}C)}{(0.5m)^2}\)

\(\displaystyle F_{q2}=\frac{0.4495m^2N}{0.25m^2}=1.798N\)

A charged particle moving through a perpendicular magnetic field feels a Lorentz force equal to the formula:

\(\displaystyle F_{B} = qvB\)

\(\displaystyle q\) is the charge, \(\displaystyle v\) is the particle speed, and \(\displaystyle B\) is the magnetic field strength. This force is always directed perpendicular to the particle’s direction of travel at that moment, and thus acts as a centripetal force. This force is also given by the equation:

\(\displaystyle F_c=\frac{mv^2}{r}\)

We can set these two equations equal to one another, allowing us to solve for the radius of the arc.

\(\displaystyle qvB = \frac{mv^{2}}{r}\)

\(\displaystyle r = \frac{mv}{qB}\)

Once the particle travels through a semicircle, it is laterally one diameter in distance from where is started (i.e. twice the radius of the circle).

\(\displaystyle r = \frac{mv}{qB} = \frac{(1.67 \times 10^{-27} kg)(1.0 \times 10^{7} \frac{m}{s})}{(1.6 \times 10^{-19} C)(3 T)}= 34.8 mm\)

Twice this value is the lateral offset of its crash point from the entrance:

\(\displaystyle 2(34.8mm)=69.6mm\)

In basic consideration, the electrical transformer works directly according to Faraday’s law, following the picture given below.

If the coils are wound tightly enough that the magnetic flux generated in any of them effectively passes undiminished through all of them, and the iron core directs all the magnetic flux from the primary to secondary coil with insignificant losses, then Faraday’s law points a direct path from \(\displaystyle V_p\) to \(\displaystyle V_s\). Faraday’s law relates the time of rate of change of magnetic flux to the voltage (emf) around a closed loop: 

\(\displaystyle V = \frac{-d \Phi_B}{dt}\)

If there are \(\displaystyle N\) loops stacked together, then the total voltage, \(\displaystyle V_t\), is related to the flux through all loops by:

\(\displaystyle V_t = -N\frac{d\Phi_B}{dt}\)

From the figure, it is seen that the voltages in each line are dropped over all loops together. Hence, for the primary winding, by Faraday’s law: 

\(\displaystyle V_p = -N_p \frac{d \Phi_B}{dt}\)

For the secondary winding:

\(\displaystyle V_s = -N_s \frac{d \Phi_B}{dt}\)

Here, \(\displaystyle N_p\) and \(\displaystyle N_s\) are the number of coils in the primary and secondary windings. The quantity \(\displaystyle \frac{d \Phi_B}{dt}\) is the same in both windings, being transmitted fully (and effectively instantly) by the iron core. This leads to the equation:

\(\displaystyle \frac{V_p}{N_p} = \frac{V_s}{N_s}\)

This can be used to solve the problem above:

\(\displaystyle N_s = \frac{V_s N_p}{V_p} = \frac{(120 V)(1000)}{10,000 V} = 12\)

NOTE:  This only works under alternating current.  Otherwise, there is no change in current to create the changing magnetic flux on which the transformer depends.

The definition of a time constant for an RC circuit is the product of resistance and charge:

\(\displaystyle R* C\)

This is defined as the time it takes the capacitor to reach a charge that is around 63% of the maximum charge. It is also the time it takes to discharge around 37% of the charge. This value is commonly used to determine the amount of charge a capacitor has or the amount of current flowing through the capacitor at any given time point.