Recent content by bs15benny

electromagnetic attraction between the two different poles (positive and negative).

the mass 2g I converted it into 0.002 kg for Q=mc∆T but for mole, I still used the mass in gram. While P2 also 100 kPa because the question said that the pressure is the same (constant) in this case is isobaric. For U the other type of energies cannot be included because it is only the sum of...

where is the image?

For part c) by using the equation gave by DrClaude U=3/2(nRT) U=3/2(0.5 x 8.314 x 373) U=2326 J then, by using Q=∆U+W Q=2326 + 1550 Q=3876 J For part d) the change in thermal energy of the gas is the same as ∆U which is 2326 J *the difference between 2329 J and 2326 J maybe because...

I just try it again by convert the 100 degrees (found that it is in C) into Kelvin which is 373 K, convert the 100000 Pa into 100 kPa. part a), using the PV/T(initial)=PV/T(final): 100(V1)/373 = 100(2V1)/Tf (100 and V1 cancels out)...

Btw DrClaude and hbk69, did you change the temperature into Kelvin? Because I'm not convert the 100 degrees because there is no unit provided.

I got part a) by using the other ideal gas equation PV/T(initial)=PV/T(final), while for V1 I got it from the other PV=nRT equation. After I got the V1, then I used the W=P∆V to find the W. Btw the ∆U is the internal energy, the sum of all K.E of the molecule, this is used at Q=∆U+W (the first...

I do agree with jbriggs444 and A.T :)

I already try found it that for, a) the final temperature is 200 degree (don't know whether it is in K or C) b)W=415.7 J (another equation, W=P∆V) c)Q=0.416 J for d) I'm not really understand about the question. is it correct? do u have the answer key to check it or not?

Here is some equations maybe useful for your problem. To find: 1. the Q, just use the Q=mc∆T 2. the final temp, use P1.V1/T1=P2.V2/T2 3. the work done, use Q=∆U+w 4. the mole(n), use n=mass/Molarmass or PV=nRT just apply it :)

There is always a torque because the wheel is rotates but the difference between going straight and left or right is only depends on the steering

I think there is a missing data which is the height of the ball after it was shot (the point where K.E=0), because it stated that the cannon is tilted up 30 degrees and the result should be about projectile motion. The height for P.E should be more than 10m and at the end the result of the final...

I think there is a missing data which is the height of the ball after it was shot (the point where K.E=0), because it stated that the cannon is tilted up 30 degrees and the result should be about projectile motion. The height for P.E should be more than 10m and at the end the result of the final...

How about using 'gaussian gun', it is about metal balls accelerates due to the attractions of magnetic forces, maybe it is the same as Vodkacannon's idea. http://www.miniscience.com/kits/Gauss_Rifle/ <- maybe this link can help you to get the idea.

I think it is because the wheel rotates counter-clockwise which is to the left.. and if its rotates clock-wise it rotates to the right. The direction of the wheel rotates proportional to the direction of force applied.