Answer:
0.00417 kW/K or 4.17 W/K
Second law is satisfied.
Explanation:
Parameters given:
Rate of heat transfer, Q = 2kW
Temperature of hot reservoir, Th = 800K
Temperature of cold reservoir, Tc = 300K
The rate of entropy change is given as:
ΔS = Q * [(1/Tc) - (1/Th)]
ΔS = 2 * (1/300 - 1/800)
ΔS = 2 * 0.002085
ΔS = 0.00417 kW/K or 4.17 W/K
Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.
Answer:
The dependant variable is obvrioulsy going to be the temperature of the watch, and the independan variable is going to be the amount of water ebing poured into the cups.
Explanation:
The temperature is the dependant variable by how it is the thing being observed, or recorded. Your independant variables could be a few things from wha information I have but it could be either, the amount of water being poured into the cups, the temperature of the water being poured, or the amount of time between each new temperature of wather bing poured into the cups.
Answer:
V = 5.4 m/s
Explanation:
It is given that,
Let mass of the block, m = 10 kg
Spring constant of the spring, k = 2 kN/m = 2000 N/m
Speed of the block, v = 6 m/s
Compression in the spring, x = 15 cm = 0.15 m
Let V is the speed of the block moving at the instant the spring has been compressed 15 cm. It can be calculated using the conservation of energy of spring mass system.
![\dfrac{1}{2}m^2=\dfrac{1}{2}kx^2+\dfrac{1}{2}mV^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dm%5E2%3D%5Cdfrac%7B1%7D%7B2%7Dkx%5E2%2B%5Cdfrac%7B1%7D%7B2%7DmV%5E2)
![mv^2=kx^2+mV^2](https://tex.z-dn.net/?f=mv%5E2%3Dkx%5E2%2BmV%5E2)
![V^2=\dfrac{mv^2-kx^2}{m}](https://tex.z-dn.net/?f=V%5E2%3D%5Cdfrac%7Bmv%5E2-kx%5E2%7D%7Bm%7D)
![V^2=\dfrac{10\times 6^2-2000\times (0.15)^2}{10}](https://tex.z-dn.net/?f=V%5E2%3D%5Cdfrac%7B10%5Ctimes%206%5E2-2000%5Ctimes%20%280.15%29%5E2%7D%7B10%7D)
V = 5.61 m/s
From the given options,
V = 5.4 m/s
Hence, this is the required solution.
We have a problem with three different state of the ratio of flow velocity to speed of sound.
That is,
a) Mach number to evaluate is 0.2, that mean we have a subsonic state.
The equation here for lift coefficient is,
![c_1 = 2\pi \alpha](https://tex.z-dn.net/?f=c_1%20%3D%202%5Cpi%20%5Calpha)
where
should be expressed in Rad.
![\alpha = \frac{5}{57.3}= 0.087](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B5%7D%7B57.3%7D%3D%200.087)
So replacing in equation for subsonic state,
![c_1 = 2\pi (0.087)=0.548](https://tex.z-dn.net/?f=c_1%20%3D%202%5Cpi%20%280.087%29%3D0.548)
b) In this situation we have a transonic state, so we need to use the Prandtl-Glauert rule,
![c_{t}=\frac{\c{t_0}}{\sqrt{1-M^2_{\infty}}} = \frac{0.548}{\sqrt{1-0.7^2}}=0.767](https://tex.z-dn.net/?f=c_%7Bt%7D%3D%5Cfrac%7B%5Cc%7Bt_0%7D%7D%7B%5Csqrt%7B1-M%5E2_%7B%5Cinfty%7D%7D%7D%20%3D%20%5Cfrac%7B0.548%7D%7B%5Csqrt%7B1-0.7%5E2%7D%7D%3D0.767)
c) For this case we have a supersonic state, so we use that equation,
![c_s = \frac{4\alpha}{\sqrt{M^2_{\infty}-1}}=\frac{4(0.087)}{\sqrt{2^2-1}}=0.2](https://tex.z-dn.net/?f=c_s%20%3D%20%5Cfrac%7B4%5Calpha%7D%7B%5Csqrt%7BM%5E2_%7B%5Cinfty%7D-1%7D%7D%3D%5Cfrac%7B4%280.087%29%7D%7B%5Csqrt%7B2%5E2-1%7D%7D%3D0.2)
Ha it’s not spring break yet for me but if it is for you I hope you have a good spring break as well :)