Enable contrast version

# Tutor profile: Kenny B.

Inactive
Kenny B.
Application Engineer with a BS in Electrical Engineering and a minor in Computer Science
Tutor Satisfaction Guarantee

## Questions

### Subject:Microsoft Excel

TutorMe
Question:

How would you clear all the formatting without removing the cell contents?

Inactive
Kenny B.

Find the "Home" tab, click on the arrow to the right of the "Clear" option under the heading labeled "Editing," and select "clear formats."

### Subject:Electrical Engineering

TutorMe
Question:

An electrical circuit has a current source, I, where I = 10sin(10t - 27) Amps, a 100 Ohm resistor, R1, in series with another 50 Ohm resistor, R2, and two capacitors in parallel with each other C1 and C2, with capacitances equal to 25 microfarads and 75 microfarads, respectively. What is the total resistance, reactance, and admittance of the circuit?

Inactive
Kenny B.

The total resistance of the circuit is the sum of all the resistive (voltage drop) components of the circuit. In this case, the only resistive components are the two resistors R1 and R2. Since R1 and R2 are in series with each other their resistances are summed up with their equivalent resistance or Req = R1 + R2. Therefore, Req = 100 Ohms + 50 Ohms = 150 Ohms. The total reactance of the circuit is the sum of all the reactive (phase shift) components of the circuit. In this case, the only reactive components are the two capacitors C1 and C2. Since C1 and C2 are in parallel to each other, we need to sum them up to calculate the equivalent capacitance or Ceq = C1 + C2. Therefore, Ceq = 25 microfarads + 75 microfarads = 100 microfarads (100 x 10^-6 F). The total reactance of the circuit is calculated using the formula Zeq = -j/(frequency*Ceq). Therefore, Zeq = -j/(10 Hz * 100 microfarads) = -j1000 Ohms. The total admittance of the circuit is the inverse of the total impedance or the sum of the resistive and reactive components of the circuit. The total impedance is equal to 150 Ohms - j1000 Ohms. Therefore, the total admittance is the inverse of that or (150 Ohms - j1000 Ohms)^-1 = 0.000147 + j0.000978 S.

### Subject:ACT

TutorMe
Question:

A car averages 25 miles per gallon. If gas costs \$3.39 per gallon, which of the following is closest to how much the gas would cost for this car to travel 2,725 typical miles? A. \$41.14 B. \$108.09 C. \$178.40 D. \$369.51 E. \$432.10

Inactive
Kenny B.

In order to solve the problem, we need to convert the price of gas per gallon into the price of gas per mile (using the car's average gas mileage in miles per gallon) and multiply that by the total number of miles to find the total cost. First, we put the unit that we want to keep on top of the first ratio. We want to find the price of gas per mile, so we start with the price of gas on top (\$3.39/1gal). Then we want to multiply that ratio by something with gallons on top to have miles be the new unit on the bottom. We have to flip the ratio of 25 miles per gallon to be 1 gallon per 25 miles (1gal/25 miles). Now let's multiply: (\$3.39/1gal) * (1gal/25 miles) = (\$3.39/25 miles) = \$0.1356/mile. Finally, we multiply that ratio by the number of miles to get the approximate cost to travel that far. \$0.1356/mile * 2,725 miles = \$369.51, which is answer D.

## Contact tutor

Send a message explaining your
needs and Kenny will reply soon.
Contact Kenny

Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage