Molecules, The movemnet of molecules
-- If velocity is constant, then there is no net force
on the chair.
-- If there is no net force on the chair, then friction
must exactly balance out your push.
-- The force of friction is exactly equal in magnitude
to your push, and in exactly the opposite direction.
Answer:
3.49 seconds
3.75 seconds
-43200 ft/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
![s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%2050%3D0t%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%209.81%5Ctimes%20t%5E2%5C%5C%5CRightarrow%20t%3D%5Csqrt%7B%5Cfrac%7B50%5Ctimes%202%7D%7B9.81%7D%7D%5C%5C%5CRightarrow%20t%3D3.19%5C%20s)
Time the parachutist falls without friction is 3.19 seconds
![v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2as%2Bu%5E2%7D%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2%5Ctimes%209.81%5Ctimes%2050%2B0%5E2%7D%5C%5C%5CRightarrow%20v%3D31.32%5C%20m%2Fs)
Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity
![v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%2011%3D31.32%2B9.81t%5C%5C%5CRightarrow%20t%3D%5Cfrac%7B11-31.32%7D%7B-67%7D%3D0.3%5C%20s)
So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds
![s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%20%5Cfrac%7Bs%7D%7B2%7D%3D0t%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%20t%5E2%5C%5C%5CRightarrow%20%5Cfrac%7Bs%7D%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%20%5Cfrac%7Bs%7D%7B2%7D%3Du1.1%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%201.1%5E2)
Now the initial velocity of the last half height will be the final velocity of the first half height.
![v=u+at\\\Rightarrow v=at](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20v%3Dat)
Since the height are equal
![\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dat%5E2%3Du1.1%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%201.1%5E2%5C%5C%5CRightarrow%20%5Cfrac%7B1%7D%7B2%7Dat%5E2%3Dat1.1%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%201.1%5E2%5C%5C%5CRightarrow%200.5t%5E2-1.1t-0.605%3D0%5C%5C%5CRightarrow%20500t%5E2-1100t-605%3D0)
![t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B11%5Cleft%281%2B%5Csqrt%7B2%7D%5Cright%29%7D%7B10%7D%2C%5C%3At%3D%5Cfrac%7B11%5Cleft%281-%5Csqrt%7B2%7D%5Cright%29%7D%7B10%7D%5C%5C%5CRightarrow%20t%3D2.65%2C%20-0.45)
Time taken to fall the first half is 2.65 seconds
Total time taken to fall is 2.65+1.1 = 3.75 seconds.
When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.
![v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20a%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2s%7D%5C%5C%5CRightarrow%20a%3D%5Cfrac%7B0%5E2-240%5E2%7D%7B2%5Ctimes%20%5Cfrac%7B8%7D%7B12%7D%7D%5C%5C%5CRightarrow%20a%3D-43200%5C%20ft%2Fs%5E2)
Magnitude of acceleration is -43200 ft/s²
Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
![W=-P\Delta V=-P(V_2-V_1)](https://tex.z-dn.net/?f=W%3D-P%5CDelta%20V%3D-P%28V_2-V_1%29)
W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
= initial volume = 5.68 L
= final volume = 2.35 L
Putting values in above equation, we get:
![W=-0.96atm\times (2.35-5.68)L=3.20L.atm](https://tex.z-dn.net/?f=W%3D-0.96atm%5Ctimes%20%282.35-5.68%29L%3D3.20L.atm)
To convert this into joules, we use the conversion factor:
![1L.atm=101.33J](https://tex.z-dn.net/?f=1L.atm%3D101.33J)
So, ![3.20L.atm=3.20\times 101.3=324J](https://tex.z-dn.net/?f=3.20L.atm%3D3.20%5Ctimes%20101.3%3D324J)
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J