Density + mass / volume = 42 / 22 = 1.909 kg / m^3 ( to the nearest thousandth)
Answer:
a) For P: ![v=0.938\frac{m}{s}](https://tex.z-dn.net/?f=%20v%3D0.938%5Cfrac%7Bm%7D%7Bs%7D)
For Q: ![v = 1.876\frac{m}{s}](https://tex.z-dn.net/?f=%20v%20%3D%201.876%5Cfrac%7Bm%7D%7Bs%7D)
b) For P:
![a_{rad}=8.80\frac{m}{s^{2}}](https://tex.z-dn.net/?f=%20a_%7Brad%7D%3D8.80%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
for Q:
![a_{rad}=17.60\frac{m}{s^{2}}](https://tex.z-dn.net/?f=%20a_%7Brad%7D%3D17.60%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
c) As the distance from the axis increases then speed increases too.
Explanation:
a) Assuming constant angular acceleration we can find the angular speed of the wheel dividing the angular displacement θ between time of rotation:
![\omega =\frac{\theta}{t}](https://tex.z-dn.net/?f=%20%5Comega%20%3D%5Cfrac%7B%5Ctheta%7D%7Bt%7D%20)
One rotation is 360 degrees or 2π radians, so θ=2π
![\omega =\frac{2\pi}{0.670} =9.38\frac{rad}{s}](https://tex.z-dn.net/?f=%20%5Comega%20%3D%5Cfrac%7B2%5Cpi%7D%7B0.670%7D%20%3D9.38%5Cfrac%7Brad%7D%7Bs%7D%20)
Angular acceleration is at every point on the wheel, but speed (tangential speed) is different and depends on the position (R) respect the rotation axis, the equation that relates angular speed and speed is:
![v = \omega R](https://tex.z-dn.net/?f=%20v%20%3D%20%5Comega%20R%20)
for P:
![v = 9.38\frac{rad}{s}*0.1m=0.938\frac{m}{s}](https://tex.z-dn.net/?f=%20v%20%3D%209.38%5Cfrac%7Brad%7D%7Bs%7D%2A0.1m%3D0.938%5Cfrac%7Bm%7D%7Bs%7D)
for Q:
![v = 9.38\frac{rad}{s}*0.2m=1.876\frac{m}{s}](https://tex.z-dn.net/?f=%20v%20%3D%209.38%5Cfrac%7Brad%7D%7Bs%7D%2A0.2m%3D1.876%5Cfrac%7Bm%7D%7Bs%7D)
b) Centripetal acceleration is:
![a_{rad}= \frac{v^2}{R}](https://tex.z-dn.net/?f=%20a_%7Brad%7D%3D%20%5Cfrac%7Bv%5E2%7D%7BR%7D)
for P:
![a_{rad}= \frac{(0.938)^2}{0.1}=8.80\frac{m}{s^{2}}](https://tex.z-dn.net/?f=%20a_%7Brad%7D%3D%20%5Cfrac%7B%280.938%29%5E2%7D%7B0.1%7D%3D8.80%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
for Q:
![a_{rad}= \frac{(1.876)^2}{0.2}=17.60\frac{m}{s^{2}}](https://tex.z-dn.net/?f=%20a_%7Brad%7D%3D%20%5Cfrac%7B%281.876%29%5E2%7D%7B0.2%7D%3D17.60%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
c) As seen on a) speed and distance from axis is
because ω is constant the if R increases then v increases too.
Answer:
The temperature change per compression stroke is 32.48°.
Explanation:
Given that,
Angular frequency = 150 rpm
Stroke = 2.00 mol
Initial temperature = 390 K
Supplied power = -7.9 kW
Rate of heat = -1.1 kW
We need to calculate the time for compressor
Using formula of compression
![\terxt{time for compression}=\text{time for half revolution}](https://tex.z-dn.net/?f=%5Cterxt%7Btime%20for%20compression%7D%3D%5Ctext%7Btime%20for%20half%20revolution%7D)
![\terxt{time for compression}=\dfrac{1}{2}\times T](https://tex.z-dn.net/?f=%5Cterxt%7Btime%20for%20compression%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20T)
![\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}](https://tex.z-dn.net/?f=%5Cterxt%7Btime%20for%20compression%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cdfrac%7B1%7D%7Bf%7D)
Put the value into the formula
![\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60](https://tex.z-dn.net/?f=%5Cterxt%7Btime%20for%20compression%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cdfrac%7B1%7D%7B150%7D%5Ctimes60)
![\terxt{time for compression}=0.2\ sec](https://tex.z-dn.net/?f=%5Cterxt%7Btime%20for%20compression%7D%3D0.2%5C%20sec)
We need to calculate the rate of internal energy
Using first law of thermodynamics
![U=Q-W](https://tex.z-dn.net/?f=U%3DQ-W)
![\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20U%7D%7B%5CDelta%20t%7D%3D%5Cdfrac%7B%5CDelta%20Q%7D%7B%5CDelta%20t%7D-%5Cdfrac%7B%5CDelta%20W%7D%7B%5CDelta%20t%7D)
Put the value into the formula
![\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20U%7D%7B%5CDelta%20t%7D%3D%28-1.1%29-%287.9%29)
![\dfrac{\Delta U}{\Delta t}=6.8\ kW](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20U%7D%7B%5CDelta%20t%7D%3D6.8%5C%20kW)
We need to calculate the temperature change per compression stroke
Using formula of rate of internal energy
![\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20U%7D%7B%5CDelta%20t%7D%3D%5Cdfrac%7Bnc_%7Bv%7D%5CDelta%20%5Ctheta%7D%7B%5CDelta%20t%7D)
![\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}](https://tex.z-dn.net/?f=%5CDelta%5Ctheta%3D%5Cdfrac%7B%5CDelta%20U%7D%7B%5CDelta%20t%7D%5Ctimes%5Cdfrac%7B%5CDelta%20t%7D%7Bn%5Ctimes%20c_%7Bc%7D%7D)
Put the value into the formula
![\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D6.8%5Ctimes10%5E%7B3%7D%5Cdfrac%7B0.2%7D%7B2.0%5Ctimes20.93%7D)
![\Delta\theta=32.48^{\circ}](https://tex.z-dn.net/?f=%5CDelta%5Ctheta%3D32.48%5E%7B%5Ccirc%7D)
Hence, The temperature change per compression stroke is 32.48°.
The correct choice is Buffers.
Buffers contains both acids and bases in equilibrium.when base is added to buffers, they adjust the pH value by releasing hydrogen ions.when acid is added to buffers, they adjust the pH value by consuming hydrogen ions. hence these substances resist any change in their pH value by releasing or absorbing hydrogen ions.This process keeps the pH value constant.
To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,
![\text{Initial Energy}=\text{Final Energy}](https://tex.z-dn.net/?f=%5Ctext%7BInitial%20Energy%7D%3D%5Ctext%7BFinal%20Energy%7D)
![\frac{1}{2} mv^2=\frac{1}{2} kx^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%3D%5Cfrac%7B1%7D%7B2%7D%20kx%5E2)
Here,
m = mass
k = Spring constant
x = Displacement
v = Velocity
Rearranging to find the velocity,
![mv^2 = kx^2](https://tex.z-dn.net/?f=mv%5E2%20%3D%20kx%5E2)
![v^2 = \frac{kx^2}{m}](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7Bkx%5E2%7D%7Bm%7D)
![v = \sqrt{\frac{kx^2}{m}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7Bkx%5E2%7D%7Bm%7D%7D)
Our values are,
![m = 5.22*10^4kg](https://tex.z-dn.net/?f=m%20%3D%205.22%2A10%5E4kg)
![k = 4.58*10^5N/m](https://tex.z-dn.net/?f=k%20%3D%204.58%2A10%5E5N%2Fm)
![x = 32cm = 0.32m](https://tex.z-dn.net/?f=x%20%3D%2032cm%20%3D%200.32m)
Replacing our values we have,
![v = \sqrt{\frac{(4.58*10^5)(5.22*10^4)}{0.32}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B%284.58%2A10%5E5%29%285.22%2A10%5E4%29%7D%7B0.32%7D%7D)
![v = 2.733*10^5m/s](https://tex.z-dn.net/?f=v%20%3D%202.733%2A10%5E5m%2Fs)
Therefore the velocity is ![2.733*10^5m/s](https://tex.z-dn.net/?f=2.733%2A10%5E5m%2Fs)