Math Physics HomeWork Tutor: Sample Problems

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Linear Equations Standard Form and Perpendicular:

Determine the equation in standard form of the line that passes through (5,5) and is perpendicular to the line 5x + 2y =10?

Solution:
If two lines are perpendicular their slopes are negative reciprocals. First solve for y to get the slope of your line: 5x + 2y = 10 --> 2y = -5x + 10 --> y = -(5/2)x + 5 The slope of the new line is going to be 2/5 and it goes through (5, 5). We have y = (2/5)x + b
Now plug in your point (5, 5)
5 = (2/5)5 + b --> b = 3
So the equation of your new line is going to be:
y = (2/5)x + 3
now standard form:
-(2/5)x + y = 3
2x - 5y = -15 I multiplied both sides by -5

Systems of Equations:

How many minutes after 3:00 will the hands of a clock overlap?

Solution:
I prefer to look at it as a position / velocity question. It's simpler and more elegant that way. That is let the minute hands position be x = 0 and velocity be 60. Let the hour hands position be 15 and velocity 5. Why? It goes 5 minutes every hour and the minute hand goes 60 minutes every hour. The units are arbitrary and we can imagine the clock laid out on a number line. Were do the two intersect? Now it's an easy problem.
Let t = number of hours
Position minute hand:
0 + 60(t)
Position hour hand:
15 + 5(t)
now when they intersect the positions are the same:
0 + 60t = 15 + 5t
55t = 15
t = 15 / 55 = 3 / 11 = .2727... hours or 16.3636... minutes.

Physics Notes

Sample Physics Solutions

Kilowatt Hour Problems:

A nuclear power plant has an electrical power output of 1500 Megawatts and operates with an efficiency of 35%. If excess energy is carried away from the plant by a river with a flow rate of 106,000 kg/s, what is the rise in temperature of the flowing water?

Solution:
This is an interesting problem. First, how much water is 106,000 Kg/s? Since on cubic meter weighs 1,000 Kg it would be 106 cubic meters / second. That might be say a river 50 m across and 2 meters deep with a flow velocity of about 1 meter per second. We are also assuming all the heat not used to power the plant goes into the water. This is an incorrect assumption. So now for the solution:
.65 * 1500 MW = 975 MW (HEAT)
1 Megawatt = 10^6 Watts = 10^6 J/s
975 x 10^6 Joules per Second Heat
Now find the Joules imparted into each KG of Water:
(9.75 x 10^8 J/s) / (1.06 x 10^5 Kg/S) =>
9.20 x 10^3 J/Kg
Now get the specific heat of water:
4186 Joules required to heat 1 Kg 1 degree K
Now (9.20 x 10^3 J/Kg ) / 4190 J/Kg K
2.20 K Hotter
To extend this research how much of the heat is dissipated before heating the river.

An LED light uses 6 Watts and has the same lighting capacity as an incandescent that uses 60 Watts. Compare energy costs after 30,000 hours. Assume .10 per KWH

Solution:
First, lets look at the units. A KWH or Kilowatt hour is 1,000 Watts for one hour. A Watt is 1 Joule / 1 second. Therefore, a Kilowatt hour is 1,000 Watts x 3,600 S or 3,600,000 Joules!

Incandescent
30,000 hrs x 60 Watts = 1,800,000 Watt Hours or 1,800 KWH. 1,800 KWH x .10 /KWH = $180 dollars!
LED
30,000 hrs x 6 Watts = 180,000 Watt Hours or 180 KWH. 180 KWH x .10 /KWH = $18 dollars
For the difference subtract $180 - $18 = $162. $162 represents the energy savings of using LED lights. Of course, you also save on all those incandescent replacements.

Pressure Problems:

A container is filled to a depth of 20.0 cm with water. On top of water floats a 30.0 cm thick layer of oil with a specific gravity of 0.700. What is the absolute pressure at the bottom of the container?

Solution:
First lets just get the equivalent of water in meters. The oil is .7 x .3m = .21m or equivalent to .21 meters of water. So the total is:
.2 m + .21 m = .41 m.
Now we need the pressure at the bottom of a .41 meter column of water. Now 1 Pascal = 1 Newton / m^2 So lets get the Newtons at the bottom of cubic meter of water. You could memorize it but I would figure it. A cubic meter of water weighs 1,000 Kg. F = mg = 9.8 x 1,000 = 9,800 Newtons and it's 1 meter squared at the bottom. So a 1 meter column of water produces 9,800 Pascals pressure. .41 x 9,800 = 4,018 round to 4,000 Pa. Now finally don't forget the atmospheric pressure equivalent to about 100,000 PA. So the total is 104,000 PA.

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