Electrodynamics

Solve the following exercises

Problem 1

What is the potential difference needed to stop an electron that has an initial velocity of \(5.0x10_{\frac{m}{s}}^5\)
if we know that the potential energy is equal to the electric power and the electric power is equal to the kinetic energy
Δv=q Δ v= - Δ K
and to get the potential difference we must the following formula
-   Δ K= \(K_I-K_F\)
and how is multiplying happens to dividing
Δ V=\(\frac{K_I}{q}\ =\ \frac{mu^2}{^2q}\)
and the values of 'm' and 'q'
\(m=9.11x10-31kg\)
\(q=-1.6x10^{-19}c\)
we substitute the formula
\(\frac{\left(9.11x10^{-31}\right)\left(5.0x10_{\frac{m}{s}}^5\right)}{2\left[-1.6x10^{-19}c\right]}=-0.71v\)

Problem 2

Acurrent of 1.30A in a wire how many electrons go through any point in a second
knowing that I is ΔQ / Δt which is equal to 1.30A 
the formula is replaced
\(I=\frac{ΔQ}{Δt}=1.30A\) substituting \(\frac{1.30}{s}=\left(\frac{1c}{1.6x10^{-19}c}\right)\)
\(=8.125x10^{-18}c\)

Problem 3

An electrical device drives a current of 6.50A at a voltage of 240v.
a) if the voltage is dropped to 15% which will be the current assuming nothing else change
b) if the resistance of the device is reduced to 15% which will be the current at a voltage of 240v
and the values of
\(I=6.50A\)
\(V=240v\)
and the formula is dividing happens to multiplying 
\(I=\frac{v}{R}\)
Clause a:
the formula to use is
\(V=IR\) 
and 
\(V_F=6.50A\)
remains
\(VI=0.85\left(6.50A\right)R\)
\(=5.525A\)
 
Clause b:
the formula to use is 
\(R=\frac{V}{I}\)
\(RI=\frac{V}{I_I}\)
\(RF=\frac{V}{I_F}\)
which is equal to
\(\frac{R_F}{R_I}=0.85\)
The formula remains
\(I_F=\frac{I_I}{0.85}=\frac{6.50A}{0.85}=7.64A\)

Problem 4

what is the diameter of a tungsten wire of one meter that has a recistensia of .32 Ω
for this exercise we will use the following formula
\(R=\rho\frac{l}{A}\)
and  
\(\rho\)\(=5.6x10^{-8}Ω.m\) 
\(A=\pi r^2\)
\(d=2r\)
The formula will be taken as follows
\(r=\frac{d}{2}\)
\(A=\pi\left(\frac{d^2}{4}\right)\)
\(A=\frac{\rho l}{R}\)
\(\frac{\pi d^2}{4}=\frac{\rho l}{R}\)
\(d^2=\frac{4l}{\pi R}\)
\(d=\sqrt{\frac{4\rho l}{\pi R}}=\sqrt{\frac{4\left(5.6x10^{-8}\right)\left(1\right)}{\pi\left(o.32\right)}}=0.00047m\)

Problem 5

Which is the resistance of a 4.5 copper wire that has a diameter of 1.5mm
With the following data 
\(l=4.5\)
\(d=1.5mm\)
\(\rho=1.68x10^{-8}Ω.m\)
The formula that will be used
\(R=\rho\frac{l}{A}\)
What is left of the following form
\(R=\frac{4\rho l}{\pi d^2}=\frac{\left(4x1.68x10^{-8}\right)\left(4.5\right)}{\pi\left(0.0015^2\right)}\ =0.042=4.2x10^{-2}Ω\)

Problem 6

Which is the maximum power consumption of a 3v disman that knows a current of 270mA.
The formula is very simple 
\(P=IV\)
Multiply the mA already converted only to A by the V that are
\(=\left(0.27A\right)\left(3.0v\right)=0.81w\)

Problem 7

A microwave oven is designed to introduce 3.3KW of heat when you connect it to a 240V source which is its resistance.
with the following formula:
\(R=\frac{V^2}{\rho}\)
\(R=\frac{240^2}{3.3x10^{-3}}=17.45Ω\)

Problem 8

You buy a 75v bulb where the electricity is at 240v if you visit this forum in Mexico where the voltage is 120v like changing the brightness of the focus with respect to europe.
With the following formula:
\(\frac{P_{usa}}{P_{EU}}=\ \frac{Vusa}{\frac{R}{V_{\frac{EU}{R}}}}=\left(\frac{1}{2}\right)=\frac{1}{4}=0.25\)

Problem 9

At a cost of $ 0.095 / kwh, how much does it cost you to leave your yard focus of 25w if you leave it on 24 hours a day.
\(Costo=25w\left(\frac{0.095}{kwh}\right)\left(\frac{1kw}{1000w}\right)\left(\frac{365dias}{ }\right)\left(\frac{24hrs}{1\ dia}\right)\)
the result is =$\(\text{20}\) Pesos .