Hi- this is actually part of my physics homework, but all I need help with deals strictly with algebra.
This is my equation:
(0.97)(1 + Beta gas (-Ti)) = (1 + Beta steel (-Ti))
The Attempt at a Solution
How do I solve for Ti...
A child holding a helium filled balloon @ sea level (T= 20 C) lets go of the string. The balloon rises freely several thousand meters, where T = 5 C and P = 0.70 atm. Find the percentage change in the balloon's volume.
PV = nRT
What is the density of radon gas at 0 degrees C and 1 atmosphere?
0 C = 273 K
density = mass/volume g/m^3
The Attempt at a Solution
I want density = g/m^3
Below are the units of PV = nRT. I thought that breaking it down...
I don't have a homework question exactly, but I need help with an equation please.
Angular frequency: W= 2 pi/T = 2(pi)(f) f= frequency
And- W = square root of (k/m) k = spring constant m= mass
So, wouldn't T = 2 pi / square root of (k/m) ??
Why 10 degrees? I know all of this relates back to the angle that the path swings out somehow. Would it be the same if I did say, a 45 degree angle (easier to meausre) and then counted 45 cycles and divided by 45? What does the angle have to do with it??
Suppose you were kidnapped and held prisoner by space invaders in a completely isolated room, with nothing but a watch and a pair of shoes (with shoelaces of known length). Explain how you might determine whether this room is on earth or on the moon.
Ok- I guess that makes sense. However, in other contexts I would probably still be confused with the cycles, revolutions, etc... I just can't wrap my head around it- silly I know :shy: I have some more problems to practice with.
Thank you both, chrisk and LP- :smile:
Ok one last question- how exactly do the units work themselves out?
This is what I have:
1.89 x 10^5 rad = 2 pi rad/cycle
To isolate cycle don't I have to multiply by 2 pi on each side? Even then I would be left with 1/cycle? You said to divide by 2 pi rad/ cycle, but then what's left...
Ok, so when I plug numbers into W = 2pi (f) - or do it the way I stated in my previous post (same thing) I end up with this numerical value: 1.89 x 10^5 rad = 2 pi rad Is this correct? Can you please help me understand what this means? Do I need to divide through on the left by 2 pi? If...
So, would it make 1 revolution (2 pi) in 2.5 seconds? I'm really confused I guess- what I just said doesn't make sense, becuase if it did, then the answer would just be 1, which didn't take the ang. frequency into account.
What about this:
W = 2 pi/T so 7.54 x 10^4 rad/sec (2.5 sec)...
A loudspeaker diaphragm is producing a sound for 2.5 s by moving back and forth in simple harmonic motion. The angular frequency of the motion is 7.54 x 10^4 rad/s. How many times does the diaphragm move back and forth?
angular freq. (W for...
Ok :) Now another issue I am having. I took the force the raft can support, which is in newtons, as is the weight of the raft. So, how do I figure how many 60 kg persons this thing will hold? I thought it would be a simple conversion, but Im missing something somewhere. I did this:
Alright- so, I found the bouyant force, and then the weight of the raft, so the rest of the weight can go to people. So, what does the whole, "and not get their feet wet" bit have to do with this? Does it just mean that if we go over our "account balance" that the raft will sink? So, the raft...
I can't quite wrap my mind around what your saying. However, if it's of any help to get me started, I think I found the weight of water displaced by the raft. I used: m = roh x V then used W = mg. Is this correct?
A raft is made of 11 logs lashed together. They are 38.0 cm in diameter and 6.10 m long and has a density of 700 kg/m^3. How many 60 kg persons can the raft hold in freshwater while keeping everybody's feet dry?
Volume immersed/Volume total =...
This question is continued from a problem that I have already solved. This is what I know from the previous problem that will be helpful:
r= 0.20 m
roh of solid sphere= 850 kg/m^3
mass sphere= 3.94 x 10 ^-5 kg
I also have 2 apparent weights that I solved for when...
With everything I knew I was able to form two triangles and in each I knew two sides. A: 8.5 and 25 and hypotenuse unknown. B: 58.5 and 75 and hypotenuse unknown. Using the tangent inverse, I found theta for each.
Tree A is 10 m tall. Tree B is 60 m tall. You're standing on level ground at a position that is 25 m from tree A and 75 m from tree B. You're eye height is 1.5 m above the ground. Find the ratio of theta B/ theta A, of the angles (measured from the horizontal) at which you...