Recent content by omc1

wait, i don't see how that equation applies here? i don't see how sin(theta) works here?

Homework Statement A scientist decides to do the double-slit experiment in a vertical tube, with the slits at the top and a viewing screen at the bottom as shown. With red light from a HeNe laser (632 nm) a nice pattern forms on the viewing screen, with the bright spots 1.00 cm apart. Now the...

yes, that is the answer, why is the mass combined because the bullet bounces off.

but i don't know either final velocity...

i thought i was using the formula for elastic collision, .03*190=.03*V2+31*V2, but I though the man did not move so I am confused about that, how would I apply the cm to this?

Homework Statement A gangster fires a bullet (m1 = 30 g, v1 = 190 m/s) at Stupendous Man (m2 = 31 kg), but it simply bounces away elastically. If Stupendous man was standing on a frictionless surface, what is the change in speed of the bullet after striking stupendous man? Be careful with...

ok that makes sense, thanks!

well at first i thought d was constant and since slope relates to the velocity then a is zero but iam just really confused when trying to read this graph

well, the slope is increasing at a, and maybe at e, its constant at c, and decreasing at d, then at b it is slowing down

Homework Statement A plot of Calvin's Mom's position vs. time is shown. Rank the acceleration of the five regions (e.g., A, B, etc.) from most negative to most positive See attachment! Homework Equations The Attempt at a Solution I think that b is the most negative and c is zero, a...

yes i mean to write that, I got 1.7 from online source, but someone else also used 1.5 but neither of those works?

Homework Statement In an electrically heated home, the temperature of the ground in contact with a concrete basement wall is 13.8oC. The temperature at the inside surface of the wall is 20.8oC. The wall is 0.118m thick and has an area of 9.43m2. Assume one kilowatt*hour of electrical energy...

yup we did, computer wanted more sig figs! thanks for your help!

ok, double check me on this because it is saying that it is wrong, (2.218)*(.099)/(pi*.0083^2)*(77.4875-5)=13.997 W/mC

ok, Q through the lower part of the bar is the same as through the top par...the temperature at the lower end is 77.4875 and 5C. so Q=K2*A*deltaT/L2 L2=9.90cm, Q=2.22W, deltaT=77.4875-5