Recent content by smsport

Thanks again, Simon. I am however, pondering something else now. I calculated the frequency of the signal emitted into the air using equation for a moving source and a stationary observer. However, would the remainder of the problem use the original f of source (ie: 440 Hz). I'm not sure, but...

Oh, of course! The way you explain it makes it much more clear. It is actually pretty obvious, so I am a little embarrassed I missed this. I will have no problem going ahead with the problem. Thank you for your time and help, Simon!

Homework Statement A car and a bicycle are traveling directly toward each other at different speeds. The car horn has a frequency of 440Hz, and the car is traveling at 15.0m/s. The cyclist hears a sound of frequency 467Hz. The velocity of sound in air is 343m/s. i) Calculate the frequency of...

Thank you for the graph. I have to admit I'm a bit unsure of how to work this then, however. I have had zero explanation on how to rework these problems to use an appropriate function (equation) or on aspects like the phase constant, etc. So, now I just feel utterly stupid on these problems. I...

Yes, I guess that doesn't make sense using the sine. I must have gotten confused along the way as there was just no explanation given for this section except a list of these formulas and I obviously applied it incorrectly. Am I on the right path that Vmax is 1.00 m/s at t=0 if at t=0 the...

Homework Statement A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 0 with a velocity of 1.00 m/s to the right. A time t later, the particle is observed to move to the left with a velocity of 0.50 m/s. (Note the change in...

Okay, if in the W= F x cos theta x s equation, the F and s are the magnitudes of F and s then if I am solving for F using that equation the F I am solving for will just be the magnitude of F therefore positive or negative is not an issue if I am understanding that correctly. I understand that...

I just want to make sure I'm understanding negative work correctly. An example problem where there is a bullet with an initial velocity of x m/s and a final velocity of 0 m/s. It is stopped by a wood block and the question asks to determine the amount of work done on the bullet by the block. If...

Got it! Thanks again. I also think I figured out what the issue was, but I will be ever so grateful to look at your file also.

Yes, I do have/know that Normal force is perpendicular to the slope and not the weight force. I know that one very well. I can start the free body diagram similarly to a box on an incline. I must have seen something somewhere with friction force vertically that threw me off, but the way you...

I do consider the normal force perpendicular to the slope. When you note that the only actual forces are mg and normal reaction force, there is no friction force vertically down along with the weight force? For some reason I thought that was the case as I've seen other banked curve problems...

ƩHomework Statement A car of mass 2,000 kg is moving round a curve on a banked track (see diagram-We were actually only given the top picture "a" and not the bottom one, but both are relevant) at a constant speed. The coefficient of static friction between the car's tires and the track is...

I got it! I wasn't being careful enough with the directions designated as positive/negative. I forgot to denote that if the direction up the incline is positive then this also means the force downward on the object hanging off the edge is positive also as the object moving up the incline would...

Now, that confuses me more. The mass on the incline is indeed larger than the other mass. However, because they specifically indicate the coefficient of kinetic friction and the mass is moving up the incline, I thought the kinetic friction would certainly be in the direction opposite, or down...

If my intuition is correct, then I certainly did the problem wrong. Hence, my red flag when I got the answer I did. I do have the frictional force opposing the tension, however, but my calculations are not working out so I must have a sign wrong somewhere. If the Sum of Fx is the two negative...