The 8cm mark was given to us in the problem. After the 0.6kg mass it went down to the 6cm mark. So based on that I assumed 0.9kg would put it at 4cm and 1.2kg at 2cm. It turned out to be the correct answer. But as more pressure builds up it would take more mass to push the plunger down, at least...
m1=0.3kg
Area=10cm^2 or 0.001m^2
P=F/A = 2.94 / 0.001 = 2940Pa
F=0.3 * 9.8 = 2.94N
Volume = Bh = 10x8 (currently at the 8cm mark) = 80cm^3
Now the mass increases to 0.6kg
F= 5.88N
That's as far as I got. I'm trying to figure out howto get this in P1V1=P2V2.
No this is just a cylinder with air in it and something pushing down with a force. I think the problem has to be solved for the initial pressure P1 and using P1V1=P2V2 solve for the second pressure then extract the force out and solve for mass. But I'm getting some goofy answers and I'm sure...
Homework Statement
A tube with air in it is marked off by 1cm marks up to 10cm. 0.3kg mass is pushing a plunger down to compress the air. The area of the tube is 10cm^2.
a. What is the pressure inside the sealed tube? 0.295Pa
b. What is the volume of air in the tube? 20cm^3
c. If the mass is...
Oh snap! So I got 30sec at the max time, double it because the rise and fall are symmetrical. That gives 60sec. Then using x=vt or x=600cos30(60) I got 31176.91 meters! Is that right?
Okay. So I get x=300t-0.5t^2 (For this problem we assume g=10m/s^2). At 30 seconds it will attain a height of 4500. I got this by putting y=300t-5t^2 in my calculator and finding the maximum. Now I know how high it goes and how long it is in the air. Now what do I do to find the range of it?
Homework Statement
A bullet is shot at initial velocity of 600m/s at an angle of 30 degrees. How far will it travel before landing? (Assume no air resistance)
v=600m/s
a=9.8m/s^2
t=?
distance=?
Homework Equations
x=x0+v0t+1/2at^2
The Attempt at a Solution
I'm stuck on this...