I didnt go to school today, and so missed day of class, now i all lost someone help.
White light is incident normal to the surface of the film as shown below. It is observed that at a point where the light is incident on the film, light reflected from the surface appears green (lamda = 525...
While exploring a sunken ocean liner, the principal researcher found hte absolute pressure on the robot observatino submarine at the level of the ship to be about 413 atmospheres. The density of seawater is 1025-kg/m^3.
Calculate the gauge pressure p_g on the sunken ocean liner...
For #3, im assuming asteroid is at rest. So
Hero
m_{H}= m
v_{iH}= 0 m/s
Asteroid
m_{A}= 1000m
v_{iA}= 0 m/s
v_{fA}= 100 m/s
So momentum before = momentum after
0 = m_{A}v_{fA} + m_{H}v_{fH}
thxs man, that really helped me out. :). How about the star one:
In the five-sided star shown, the letters A, B, C, D and E are replaced by the numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the numbers at the ends of the line segments...
how about this one:
Three circles of radius s are drawn in the first quadrant of the xy-plane. The first circle is tangent to both axes, the second is tangent to the first circle and the x-axis, and the third is tangent to the first circle and the y-axis. A circle of radius r > s is...
This is not exatly homework, but it seems the best place to post it.
Let A, M, and C be digits with:
(100A+10M+C)(A+M+C) = 2005
A. 1 B. 2 C. 3 D. 4 E. 5
maybe some one can help me out on this, was one of the test questions i did not know how to do. :blushing:
Hmm, does not seem I learned those laws, onless my book and teacher are calling them by different names. O, and I asked to get this started, for example finding pressure on state 2. From what I see, is that the 150 kg weight does work on the gas? I might be wrong. But if so it does 1470 J, so...
Okay, i found this problem to be pretty tough, maybe someone can lend me a hand.
Theres a container with a piston, and theres a gas on the bottom of the piston.
State 1
A=1.2 x 10^{-2}m^{2}
T_1=0^{\circ{}}C
V_1=1.5 x 10^{-3}m^{3}
P_1=1.02 x 10^{5}Pa
State 2
Theres a 150kg weight...
Im assuming they are not interested in friction, etc.
Does not seem your awnser is right, because how could he throw it, and bounce back at a faster rate? hmmm, remember what Curious3141 said.
Momentum initial = Momentum final
Oops, I didnt even look at the graph, or read what she posted after your comment :rolleyes: . Point taken. :smile:
Well seems she didnt even need our help, she had the awnser. Right after my post.
"By the way, I got V1 = 4.727 m/s (to the right) and V2 = 21.27 m/s (to the left)."
Srry, I was gone.
I think it was an elastic collision. Because if it was not, then the awnser could not be determined.
"Most collisions between objects involve the loss of some kinetic energy and are said to be inelastic. In the general case, the final velocities are not determinable from...
with the information given, I have, using momentum:
(26)(.5) + (5)(0) = .5V_1 + 5V_2
momentum before = momentum after
dont think one can do it another way, maybe if we see the graph it would change.
I c how, 1 cal/g*C= 4.184 J/kg*C, correct?
Oops, 4.184 is already in KJ, I just noticed, that 1 cal/g*C=4184 J/kg*C. Thats whats started this whole question.
Well thxs for all the help, and again I didn't mean to be rude. :smile:
apchem, I posted that before, because, dex said i got A, and told me to post my work. I said I did post it, and I didnt really care what the letter awnser was.
Anyways thxs for the help, didnt mean to be rude.