Recent content by mplayer

May want to double-check that conversion. 0.02 s is an awfully quick kilometer :P

Yes. Well, yes to the equation, but not sure where the 0.02s is coming from.

Those are the answers I got for 1a. Just use a similar approach for 1b.

Ok, ask yourself this: If you were standing on the train, how fast would you perceive the car to be moving? If you take the perceived speed and use it to figure out how long it takes for the car to travel 1km (the length of the train), what number do you get? It's just as simple as t=d/v. So...

For the first one, I would try thinking about the difference in speed between the two moving objects. Ask yourself how fast the car is moving relative to the train. Then, given that information, how long does it take to travel that 1 km length of the train. You will then have a time to use to...

I think we all know what is REALLY going on here...

For the voltage between t=1 and t=3, the function describing the voltage is just a constant, or y=10.

For the second part, try drawing out a 'power triangle' and thinking about what each of the three sides represent.

One way you could think about this problem is by taking discrete intervals on the current graph and voltage graph and multiplying these values together. This will result in a graph that describes the power dissipation in the device over the given time interval. Then it will just be a matter of...

The reaction at point D has only a y-component because it is on a roller and free to move along the x-axis. Point D can only resist forces in the direction perpendicular to the beam. Point A is hinged and will have both x and y reaction components.

When you shorted the voltage source, I think you assumed you could combine the two voltage sources together like you did in the first part of the calculation, but you cant. Try calculating the current through R1 just one source at a time - the current source, V1, then V2.

You could just try converting the 2 lb. mass into kg. and the two velocities given into m/s so that solving for the change in kinetic energy would be in Joules. Should be pretty easy to find those three conversion factors.

You are right about having to find the total area under the curve to solve for the total energy expenditure. One way you could deal with the calculation is to draw out the graph in full with time on the x-axis (in seconds) and power on the y-axis (in watts). When the power is increasing...

Re=(1.5x10^11 m): radius of Earth's orbit about the Sun. It is the distance between the Earth and the Sun. Rm=(2.3x10^11 m): radius of Mars' orbit about the Sun. It is the distance between Mars and the Sun. Re<Rm: Earth is the third planet from the Sun, while Mars is the fourth. If the...

First, notice that the distance from the Earth to the sun is the same as the radius of its orbit around the sun, so don't let that confuse you. Its just given in different units. I would suggest drawing out a simple diagram on paper - a top down view of the solar system focusing on the Sun...