Alan

Total energy used is 1.2kJ/s*125s = 150 kJ.  So the latent heat is 150kJ/0.4kg = 375kJ/kg

Yes.  Just type in your quadratic equation and hit the = sign.  For example:

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} = \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{3}}\\ {\mathtt{x}} = {\mathtt{2}}\\ \end{array} \right\}$$

tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x) so tan(x)/sec(x) = tan(x)*cos(x) = sin(x)

We assume the pendulum bob travels along an arc of a circle of radius 15.7cm.  The angle (in radians) is simply the distance travelled along the arc divided by the radius.  Angle = 10.5/15.7 radians 

$${\mathtt{Angle}} = {\frac{{\mathtt{10.5}}}{{\mathtt{15.7}}}} = {\mathtt{Angle}} = {\mathtt{0.668\: \!789\: \!808\: \!917\: \!197\: \!5}}$$

To get the angle in degrees:

$${\mathtt{Angleindegrees}} = \left({\frac{{\mathtt{10.5}}}{{\mathtt{15.7}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{180}}}{{\mathtt{\pi}}}}\right) = {\mathtt{Angleindegrees}} = {\mathtt{38.318\: \!833\: \!432\: \!316\: \!201\: \!9}}$$

Assuming the sum of kinetic and potential energy is conserved, that is (1/2)mv₁² + mgh₁ = (1/2)mv₂² + mgh₂ then (1/2)v₂² = (1/2)v₁² +g*(h₁-h₂)  or v₂ = sqrt(v₁²+2*g*(h₁-h₂))

$${\sqrt{{{\mathtt{18.4}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.81}}{\mathtt{\,\times\,}}\left({\mathtt{35.1}}{\mathtt{\,-\,}}{\mathtt{22.5}}\right)}} = {\mathtt{24.202\: \!727\: \!119\: \!066\: \!561\: \!9}}$$

or approximately 24.2 m/s

1.224 = e^k*20 Take logs of both sides: ln(1.224) = k*20;  rearrange: k = ln(1.224)/20

$${\mathtt{k}} = {\frac{{ln}{\left({\mathtt{1.224}}\right)}}{{\mathtt{20}}}} = {\mathtt{k}} = {\mathtt{0.010\: \!106\: \!209\: \!204\: \!506\: \!7}}$$

 so k is approximately equal to 0.01 per second. 

3.7*10⁵*9.5*10⁴ = 3.515810¹*10⁵*10⁴ = 3.515*10¹⁺⁵⁺⁴ = 3.515*10¹⁰

inverse of ln(x) is just e^x

inverse of log₁₀(x) is just 10^x

 inverse of log_b(x) is just b^x

Coulomb's law says (1/4*pi*eps)*q*Q/r^2=F  where eps is the permittivity of free space (8.854*10^-12 farad/m) q is the charge on one source, Q is the charge on the other, r is the distance between them and F is the force.  This gives (1/4*pi*8.854*10^-12)(-6*10^-6)*Q/0.05^2=65

$${\mathtt{Q}} = {\frac{{\mathtt{65}}{\mathtt{\,\times\,}}{{\mathtt{0.05}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{8.854}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{12}}\right)}}{{\mathtt{\,-\,}}\left({\mathtt{6}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{6}}}\right)}} = {\mathtt{Q}} = -{\mathtt{0.000\: \!003\: \!013\: \!363\: \!313\: \!4}}$$ Coulombs or about -3*10-6C

The potential energy is given by mass*g*height where g is the acceleration of gravity (9.81m/s^2), so energy used to heat silver is (1/3)*3.25*9.81*75 Joules

This energy equals mass*specific heat*temperature rise in the silver, so (1/3)*3.25*9.81*75 = 3.25*234*TempRise  which gives a temperature rise of

  $${\frac{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{\mathtt{9.81}}{\mathtt{\,\times\,}}{\mathtt{75}}}{{\mathtt{234}}}} = {\mathtt{1.048\: \!076\: \!923\: \!076\: \!923\: \!1}}$$  degrees C