Five identical light bulbs, labeled A through E, are connected to a voltage source in the series/parallel circuit shown below. Any or all of the bulbs can be placed into the circuit by turning ON the switches that are wired in series with the particular bulbs. In the photograph none of the switches have been closed so all of the bulbs are off.
The question this week involves how brightly each of the light bulbs will glow when ALL of them are inserted into the circuit by turning their switches on.
For the "answer" this week, write down a list of the bulbs in order of their brightness, from brightest to dimmest, for example:
E > D > B > A > C;
or A > B = C > D > E;
or perhaps C > D = E > A = B;
or any number of other possible combinations.
What will the sequence of brightness be?
For the answer, click Read More after April 17, 2015.
The apparatus for this week's question consists of a rod at the bottom of which there is a fixed circular mass. The rod pivots above this mass at a fixed point and is free to oscillate from side to side. Above the pivot point there is moveable trapezoidal mass, which can be adjusted either higher or lower. Catch a video of it in action by clicking your mouse on the photograph below.
Notice how the adjustable mass is very close to the pivot point.
And now the question: Suppose the mass were adjusted further upwards, away from the pivot point (pictured below). How would the of frequency of oscillation change?
There should be two parts to a good answer: What will happen, and why?
- (a) The frequency will not change; the ticking machine will oscillate the same.
- (b) The frequency will increase (i.e. the machine will tick faster). perhaps the torque exerted by the moveable mass will be greater, the effect would be much like moving a person further down a teeter-totter; the further the person is away, the more quickly rod will accelerate.
- (c) The frequency will decrease (i.e. tick slower). Perhaps the torque exerted by the moveable mass will be less, and therefore the effect would be like moving a person closer in on a teeter-totter. Less torque, less acceleration.
- (d) Other (you must explain).
Click Read More for the answer after April 10, 2015
A steel bar, the narrower of the two seen in the photograph below, is struck near its center with a rubber mallet to produce a musical tone as shown in an mpeg video when you click your mouse on the photograph.
A second steel bar, with the same length and thickness, but twice the width, as seen in the photograph, also produces a tone when struck. The question this week is about the frequency relation between the two tones.
When the two steel bars are struck, relative to the frequency of the narrower bar, the frequency of the wider bar will be:
- (a) higher.
- (b) lower.
- (c) the same.
After April 3, click Read More for the answer.
A small amount of water (a few millilitres) of water is heated in a 12-ounce soda can (photograph at left below) until the water boils, producing steam as evidenced by the condensation coming out of the can (photograph at the center below). The steam-filled can is then grabbed by a pair of tongs and quickly placed upside down on a dish of room-temperature water, as seen in the photograph at the right below, taken just before the top of the can reaches the water. You can see how the demonstration is set up in an mpeg video by clicking your mouse on the photographs, or on this link.
The question this week involves exactly what happens after the can hits the water.
When the can reaches the surface of the water:
- (a) The can will RAPIDLY implode.
- (b) The can will SLOWLY implode.
- (c) Water will be pulled up into the can, filling the can with water.
- (d) Nothing will happen.
On or after 3/27, click Read More for the answer.