Math questions

Question

1. The power available in a jet liquid is directly proportional to the cross sectional area of the jet and to the cube of the velocity. By what factor will the power increase if the area and velocity is increased by 50%?

 

2. The intensity of illumination at a given point is directly proportional to the intensity of the light source and inversely proportional to the square of the distance from the light source. If a desk is properly illuminated by a 75.0-W lamp  8.00 ft from the desk, What size lamp will be needed to provide the same lighting at a distance of 12.0 ft?

 

3. The time needed to empty a vertical cylindrical tank varies directly as the square of the radius of the tank and the square root of the height. If a tank 2m high and with radius 35 cm requires 1H 10 min to empty, how long will it take to empty a tank if it is 3m high with radius 25 cm?

 

4. The number of oscillations made by a pendulum in a given time is inversely proportional to the square root of the length of the pendulum. A certain clock with a 30.00 in long pendulum makes 50 oscillations/min.  If the pendulum is shorten to 28 in, how many oscillations per minute would the pendulum now make?

Answer 1

Haroon,

 

This is all about carefully reading the wording of the question.

 

1) Power is directly proportional to the area. Increasing the area by 50% means multiplying by a factor of

150/100=3/2=1.5. Power is directly proportional to the cube of the velocity. So increasing the velocity by 50% means multiplying by 1.5 cubed. All in all the total effect is 1.5^4 times the original power.

 

2) We want the lamp to give off the same light at 12 feet as it does at 8 feet. Intensity of illumination is inversely proportional to the square of the distance, so the original change from 8 to 12 divides (inversely remember) the overall intensity of illumination by 12/8=1.5 squared. To combat the effect of the greater distance we need to multiply the intensity of light by 1.5 squared.

 

3) Let's take this one measure at a time. The tank went from 2 to 3m high, with a multiplication factor of 3/2=1.5 The radius went from 35 to 25cm or 25/35=.714 (yuck)! The empty rate varies directly with the square of the radius producing a factor of .714^2 and the square root of the height (Sqrt of 1.5) producing .624. Applying that factor to an empty rate of an hour and ten minutes (70 minutes) means 43.7 minutes.

 

4) If the pendulum is shortened from 30/28 we're talking a factor of 28/30=.93. The oscillations are inversely proportional to the square root of the length so we divide by the square root of .93, creating a factor of 1.15. (Punch this into the calculator carefully!) So the original oscillations of 50/min is now 50*1.15=57.6 oscillations a minute.

 

Hope that helps,

Deanna