<p>yup it’s abab</p>
<p>An ideal gas may be taken from one state to another state with a different pressure, volume, and temperature along several different paths. Quantities that will always be the same for this process, regardless of which path is taken, include which of the following? I. The change in internal energy of the gas II. The heat exchanged between the gas and its surroundings III. The work done by the gas a. I only b. II only c. I and III only d. II and III only e. I, II, and III</p>
<p>II only (10 char)</p>
<p>is it II only?</p>
<p>I only (10 char)</p>
<p>explain please</p>
<p>always</p>
<p>I’m going to BOMB this test. I have never felt like I’m going to bomb an AP Test. Not Even Chem, which I had to learn by myself because the teacher sucked. Ugh. This is going to be horrible.</p>
<p>dragontoe- I completely agree. My teacher barely taught half the course and I don’t feel confident whatsoever. Good luck tomorrow!</p>
<p>Internal energy will remain the same whenever you go from one point to another, regardless of the path taken…</p>
<p>Is the answer II only?</p>
<p>
</p>
<p>I don’t think that’s true. Internal energy is based on temperature, and temperature most certainly will not “remain the same whenever you go from one point to another, regardless of the path taken.”</p>
<p>I is the answer
this question was in princeton review
if it starts at a and ends at b, the change in internal energy will ALWAYS be the same
regardless of what path it takes from a to b </p>
<p>*and in an isothermal process the change in internal energy is ALWAYS zero</p>
<p>I is right…if you have trouble conceptualizing it you can always look at the equations…PV = nRT and U = (3/2)nRT</p>
<p>A new planet is discovered that has twice the Earth’s mass and twice the Earth’s radius. On the surface
of this new planet, a person who weighs 500 N on Earth would experience a gravitational force of
a. 125 N b. 250 N c. 500 N d. 1000 N e. 2000 N</p>
<p>can someone please explain this to me</p>
<p>Sure the equation is Fg= (G(just a constant))(M1)(M2)/(R^2)</p>
<p>Therefore if Fg= 500N and you double M1 and then divide by a 4 (2 squared) you get an answer of 250 N</p>
<p>^ Think I explained this one earlier but anyway:</p>
<p>Force of a planet on an object = GMplanetMobject/rplanet^2
So if that F is 500N on earth, and the new planet has twice Earth’s mass and radius
On that planet F = G2MplanetMobject/(2rplanet)^2 = 2GMplanetMobject/4rplanet = 1/2 of the original force = 250N</p>
<p>the answer is b</p>
<p>can any1 explain this equation for me.
(E) = B(Area^2) –> B = IR/(Area^2)</p>
<p>^huh?? what is THAT from??</p>
<p>I dunno…I have it in my notes, but I totally forgot. thats why I am was asking.</p>