<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:
</span>
A pulley is another sort of basic machine in the lever family. We may have utilized a pulley to lift things, for example, a banner on a flagpole.
<u>Explanation:</u>
The point in a fixed pulley resembles the support of a lever. The remainder of the pulley behaves like the fixed arm of a first-class lever, since it rotates around a point. The distance from the fulcrum is the equivalent on the two sides of a fixed pulley. A fixed pulley has a mechanical advantage of one. Hence, a fixed pulley doesn't increase the force.
It essentially alters the direction of the force. A moveable pulley or a mix of pulleys can deliver a mechanical advantage of more than one. Moveable pulleys are appended to the item being moved. Fixed and moveable pulleys can be consolidated into a solitary unit to create a greater mechanical advantage.
Answer:L=109.16 m
Explanation:
Given
initial temperature ![=20^{\circ}C](https://tex.z-dn.net/?f=%3D20%5E%7B%5Ccirc%7DC)
Final Temperature ![=80^{\circ}C](https://tex.z-dn.net/?f=%3D80%5E%7B%5Ccirc%7DC)
mass flow rate of cold fluid ![\dot{m_c}=1.2 kg/s](https://tex.z-dn.net/?f=%5Cdot%7Bm_c%7D%3D1.2%20kg%2Fs)
Initial Geothermal water temperature ![T_h_i=160^{\circ}C](https://tex.z-dn.net/?f=T_h_i%3D160%5E%7B%5Ccirc%7DC)
Let final Temperature be T
mass flow rate of geothermal water ![\dot{m_h}=2 kg/s](https://tex.z-dn.net/?f=%5Cdot%7Bm_h%7D%3D2%20kg%2Fs)
diameter of inner wall ![d_i=1.5 cm](https://tex.z-dn.net/?f=d_i%3D1.5%20cm)
![U_{overall}=640 W/m^2K](https://tex.z-dn.net/?f=U_%7Boverall%7D%3D640%20W%2Fm%5E2K)
specific heat of water ![c=4.18 kJ/kg-K](https://tex.z-dn.net/?f=c%3D4.18%20kJ%2Fkg-K)
balancing energy
Heat lost by hot fluid=heat gained by cold Fluid
![\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)](https://tex.z-dn.net/?f=%5Cdot%7Bm_c%7Dc%28T_h_i-T_h_e%29%3D%20%5Cdot%7Bm_h%7Dc%2880-20%29)
![2\times (160-T)=1.2\times (80-20)](https://tex.z-dn.net/?f=2%5Ctimes%20%28160-T%29%3D1.2%5Ctimes%20%2880-20%29)
![160-T=36](https://tex.z-dn.net/?f=160-T%3D36)
![T=124^{\circ}C](https://tex.z-dn.net/?f=T%3D124%5E%7B%5Ccirc%7DC)
As heat exchanger is counter flow therefore
![\Delta T_1=160-80=80^{\circ}C](https://tex.z-dn.net/?f=%5CDelta%20T_1%3D160-80%3D80%5E%7B%5Ccirc%7DC)
![\Delta T_2=124-20=104^{\circ}C](https://tex.z-dn.net/?f=%5CDelta%20T_2%3D124-20%3D104%5E%7B%5Ccirc%7DC)
![LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}](https://tex.z-dn.net/?f=LMTD%3D%5Cfrac%7B%5CDelta%20T_1-%5CDelta%20T_2%7D%7B%5Cln%20%28%5Cfrac%7B%5CDelta%20T_1%7D%7B%5CDelta%20T_2%7D%29%7D)
![LMTD=\frac{80-104}{\ln \frac{80}{104}}](https://tex.z-dn.net/?f=LMTD%3D%5Cfrac%7B80-104%7D%7B%5Cln%20%5Cfrac%7B80%7D%7B104%7D%7D)
![LMTD=91.49^{\circ}C](https://tex.z-dn.net/?f=LMTD%3D91.49%5E%7B%5Ccirc%7DC)
heat lost or gain by Fluid is equal to heat transfer in the heat exchanger
(LMTD)
![A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1.2%5Ctimes%204.184%5Ctimes%201000%5Ctimes%2060%7D%7B640%5Ctimes%2091.49%7D%3D5.144%20m%5E2)
![A=\pi DL=5.144](https://tex.z-dn.net/?f=A%3D%5Cpi%20DL%3D5.144)
![L=\frac{5.144}{\pi \times 0.015}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B5.144%7D%7B%5Cpi%20%5Ctimes%200.015%7D)
![L=109.16 m](https://tex.z-dn.net/?f=L%3D109.16%20m)
Answer: im not sire
Explanation: very sorry im not sure
Answer:
The mass percentage of calcium nitrate is 31.23%.
Explanation:
Let the the mass of calcium nitrate be x and mass of potassium chloride be y.
Total mass of mixture = 19.12 g
x + y = 19.12 g..(1)
Mass of solvent = 149 g = 0.149 kg
Freezing point of the solution,
= -5.77 °C
Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)
The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:
i = 3
i' = 2
Freezing point of water = T = 0°C
![\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC](https://tex.z-dn.net/?f=%5CDelta%20T_f%3DT-T_f%3D0%5EoC-%28-5.77%5EoC%29%3D5.77%5EoC)
![\Delta T_f=i\times K_f\times m](https://tex.z-dn.net/?f=%5CDelta%20T_f%3Di%5Ctimes%20K_f%5Ctimes%20m)
![Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}](https://tex.z-dn.net/?f=Molality%3Dm%28mol%2Fkg%29%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7Bmass%20of%20solvent%20in%20kg%7D%7D)
![5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})](https://tex.z-dn.net/?f=5.77%5EoC%3D1.86%20%5EoC%2F%28mol%2Fkg%29%5Ctimes%20%28%5Cfrac%7B%20i%5Ctimes%20x%7D%7B164%20g%2Fmol%5Ctimes%200.149%20kg%7D%2B%5Cfrac%7Bi%27%5Ctimes%20y%7D%7B74.5%20g%2Fmol0.149%20kg%7D%29)
On solving we get:
....(2)
Solving equation (1)(2) for x and y:
x =5.973 g
y = 13.147 g
Mass percent of
in the mixture:
![\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B19.21%20g%7D%5Ctimes%20100%3D%5Cfrac%7B5.973%20g%7D%7B19.12%20g%7D%3D31.23%5C%25)
The mass percentage of calcium nitrate is 31.23%.