Answer:
0.2932 rad/s
Explanation:
r = Radius = 2 m
= Initial angular momentum = ![275\ kgm^2](https://tex.z-dn.net/?f=275%5C%20kgm%5E2)
= Initial angular velocity = 14 rev/min
= Final angular momentum
= Final angular velocity
Here the angular momentum of the system is conserved
![I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s](https://tex.z-dn.net/?f=I_i%5Comega_i%3DI_f%5Comega_f%5C%5C%5CRightarrow%20275%5Ctimes%2014%5Ctimes%20%5Cdfrac%7B2%5Cpi%7D%7B60%7D%3D%28275%2B275%282%29%5E2%29%5Comega_f%5C%5C%5CRightarrow%20%5Comega_f%3D%5Cdfrac%7B275%5Ctimes%2014%5Ctimes%20%5Cdfrac%7B2%5Cpi%7D%7B60%7D%7D%7B275%2B275%282%29%5E2%7D%5C%5C%5CRightarrow%20%5Comega_f%3D0.2932%5C%20rad%2Fs)
The final angular velocity is 0.2932 rad/s
Answer:
diverging light rays of the bulb are collected by the reflector.
Explanation:
Answer : The final temperature of gas is 266.12 K
Explanation :
According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.
The formula will be:
![\mu_{J,T}=(\frac{dT}{dP})_H](https://tex.z-dn.net/?f=%5Cmu_%7BJ%2CT%7D%3D%28%5Cfrac%7BdT%7D%7BdP%7D%29_H)
or,
![\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}](https://tex.z-dn.net/?f=%5Cmu_%7BJ%2CT%7D%3D%28%5Cfrac%7BdT%7D%7BdP%7D%29_H%5Capprox%20%5Cfrac%7B%5CDelta%20T%7D%7B%5CDelta%20P%7D)
As per question the formula will be:
.........(1)
where,
= Joule-Thomson coefficient of the gas = ![0.13K/atm](https://tex.z-dn.net/?f=0.13K%2Fatm)
= initial temperature = ![19.0^oC=273+19.0=292.0K](https://tex.z-dn.net/?f=19.0%5EoC%3D273%2B19.0%3D292.0K)
= final temperature = ?
= initial pressure = 200.0 atm
= final pressure = 0.95 atm
Now put all the given values in the above equation 1, we get:
![0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}](https://tex.z-dn.net/?f=0.13K%2Fatm%3D%5Cfrac%7BT_2-292.0K%7D%7B%280.95-200.0%29atm%7D)
![T_2=266.12K](https://tex.z-dn.net/?f=T_2%3D266.12K)
Therefore, the final temperature of gas is 266.12 K
A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies. Frequency beat is equal to,
![f_{beat} =| f_2\pm f_1 |](https://tex.z-dn.net/?f=f_%7Bbeat%7D%20%3D%7C%20f_2%5Cpm%20f_1%20%7C)
The reference frequency in our case would be 392Hz, and since there is the possibility of the upper and lower range for the amount of beats per second that the two possible frequencies are heard would be
![f_{beat} =|392+4|= 396Hz](https://tex.z-dn.net/?f=f_%7Bbeat%7D%20%3D%7C392%2B4%7C%3D%20396Hz)
![f_{beat} =|392-4|=388Hz](https://tex.z-dn.net/?f=f_%7Bbeat%7D%20%3D%7C392-4%7C%3D388Hz)
Therefore the two possible frequencies the piano wire is vibrating at, would be 396Hz and 388Hz
Answer:
∆h = 0.071 m
Explanation:
I rename angle (θ) = angle(α)
First we are going to write two important equations to solve this problem :
Vy(t) and y(t)
We start by decomposing the speed in the direction ''y''
![sin(\alpha) = \frac{Vyi}{Vi}](https://tex.z-dn.net/?f=sin%28%5Calpha%29%20%3D%20%5Cfrac%7BVyi%7D%7BVi%7D)
![Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}](https://tex.z-dn.net/?f=Vyi%20%3D%20Vi.sin%28%5Calpha%20%29%20%3D%209.2%20%5Cfrac%7Bm%7D%7Bs%7D%20.sin%2846%29%20%3D%206.62%20%5Cfrac%7Bm%7D%7Bs%7D)
Vy in this problem will follow this equation =
![Vy(t) = Vyi -g.t](https://tex.z-dn.net/?f=Vy%28t%29%20%3D%20Vyi%20-g.t)
where g is the gravity acceleration
![Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t](https://tex.z-dn.net/?f=Vy%28t%29%20%3D%20Vyi%20-%20g.t%3D%206.62%20%5Cfrac%7Bm%7D%7Bs%7D%20-%20%289.8%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%29%20.t)
This is equation (1)
For Y(t) :
![Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}](https://tex.z-dn.net/?f=Y%28t%29%3DYi%2BVyi.t-%5Cfrac%7Bg.t%5E%7B2%7D%20%7D%7B2%7D)
We suppose yi = 0
![Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}](https://tex.z-dn.net/?f=Y%28t%29%20%3D%20Yi%20%2BVyi.t-%5Cfrac%7Bg.t%5E%7B2%7D%20%7D%7B2%7D%20%3D%206.62%20%5Cfrac%7Bm%7D%7Bs%7D%20.t-%204.9%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20.t%5E%7B2%7D)
This is equation (2)
We need the time in which Vy = 0 m/s so we use (1)
![Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s](https://tex.z-dn.net/?f=Vy%20%28t%29%20%3D%200%5C%5C0%3D6.62%20%5Cfrac%7Bm%7D%7Bs%7D%20-%209.8%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20.t%5C%5Ct%3D%200.675%20s)
So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)
![Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m](https://tex.z-dn.net/?f=Y%280.675s%29%20%3D%206.62%5Cfrac%7Bm%7D%7Bs%7D.%280.675s%29-4.9%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%20.%280.675s%29%5E%7B2%7D%20%5C%5CY%280.675s%29%20%3D2.236%20m)
2.236 m is the maximum height from the shell (in which Vy=0 m/s)
Let's calculate now the height for t = 0.555 s
![Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m](https://tex.z-dn.net/?f=Y%280.555s%29%3D%206.62%20%5Cfrac%7Bm%7D%7Bs%7D%20.%280.555s%29-4.9%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20.%280.555s%29%5E%7B2%7D%20%5C%5CY%280.555s%29%20%3D%202.165m)
The height asked is
∆h = 2.236 m - 2.165 m = 0.071 m