Answer:
I=V/R
so larger bsttery or small resistor can increase the current value.
Answer:
The heat loss per unit length is ![\frac{Q}{L} = 2981 W/m](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%202981%20W%2Fm)
Explanation:
From the question we are told that
The outer diameter of the pipe is ![d = 104mm = \frac{104}{1000} = 0.104 m](https://tex.z-dn.net/?f=d%20%3D%20104mm%20%3D%20%5Cfrac%7B104%7D%7B1000%7D%20%3D%200.104%20m)
The thickness is
The temperature of water is
The outside air temperature is ![T_a = -10^oC = -10 +273 = 263K](https://tex.z-dn.net/?f=T_a%20%3D%20-10%5EoC%20%3D%20-10%20%2B273%20%3D%20263K)
The water side heat transfer coefficient is ![z_1 = 300 W/ m^2 \cdot K](https://tex.z-dn.net/?f=z_1%20%3D%20300%20W%2F%20m%5E2%20%5Ccdot%20K)
The heat transfer coefficient is ![z_2 = 20 W/m^2 \cdot K](https://tex.z-dn.net/?f=z_2%20%3D%2020%20W%2Fm%5E2%20%5Ccdot%20K)
The heat lost per unit length is mathematically represented as
![\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%5Cpi%20%28T%20-%20Ta%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_1%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_2%7D%7D)
Substituting values
![\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%2A%203.142%20%28363%20-%20263%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B300%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B20%7D%7D)
![\frac{Q}{L} = \frac{628}{0.2107}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B628%7D%7B0.2107%7D)
![\frac{Q}{L} = 2981 W/m](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%202981%20W%2Fm)
Answer:(a)9.685 mm
(b)4.184 mm
Explanation:
Given
Wavelength of light ![(\lambda )=565nm \approx 565\times 10^{-9}m](https://tex.z-dn.net/?f=%28%5Clambda%20%29%3D565nm%20%5Capprox%20565%5Ctimes%2010%5E%7B-9%7Dm)
Width of slit(b)=0.210
(a)Width of central maximum located 1.80m from slit
![=\frac{2\lambda L}{b}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%5Clambda%20L%7D%7Bb%7D)
![=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%5Ctimes%20565%5Ctimes%2010%5E%7B-9%7D%5Ctimes%201.8%7D%7B0.210%5Ctimes%2010%5E%7B-3%7D%7D)
=9.685 mm
(b)Width of the first order bright fringe
![Y_1=\frac{\lambda \times L}{b}](https://tex.z-dn.net/?f=Y_1%3D%5Cfrac%7B%5Clambda%20%5Ctimes%20L%7D%7Bb%7D)
![Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}](https://tex.z-dn.net/?f=Y_1%3D%5Cfrac%7B565%5Ctimes%2010%5E%7B-9%7D%5Ctimes%201.8%7D%7B0.210%5Ctimes%2010%5E%7B-3%7D%7D)
![Y_1=4.84mm](https://tex.z-dn.net/?f=Y_1%3D4.84mm)
<span>In order to
change power, current or voltage should also be changed. Voltage is an
electromotive force, and also the quantitative expression that shows the
potential difference of the two points charged in an electrical field. So, before power will take place, it would
always be best to change also the voltage.</span>