Answer:
A certain battery can power a 100 Ohm-resistance device for 4 hours. How long can the battery power a 200 Ohm device? Ans: 8 ohms
Part B)
If you put two of these batteries in series with each other - still powering a device with the original 100 Ohm resistance - how long would the batteries last? (Note: you have two batteries now, so twice the energy to begin with than with one battery.)
t₁=8hr
t₂=2hr
Explanation:
Given that,
R₁ =100Ω
R₂ =200Ω
t₁ =4hr
Part A
The expression for energy dessipated across a resistor when it is connected with a single resistor (R) is
![E = \frac{V^2}{R} t](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BV%5E2%7D%7BR%7D%20t)
Here V is the voltage across the source, t is time
energy dissipated across 200ohms is the same as energy dissipated across 100ohms
![E = \frac{V^2}{R} t](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BV%5E2%7D%7BR%7D%20t)
![\frac{V^2t_1}{R_1} = \frac{V^2t_2}{R_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV%5E2t_1%7D%7BR_1%7D%20%3D%20%5Cfrac%7BV%5E2t_2%7D%7BR_2%7D)
rearrange the above equation t₂
![t_2 = \frac{R_2t_1}{R_1} \\\\= \frac{(200)(4)}{100} \\= 8hr](https://tex.z-dn.net/?f=t_2%20%3D%20%5Cfrac%7BR_2t_1%7D%7BR_1%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B%28200%29%284%29%7D%7B100%7D%20%5C%5C%3D%208hr)
![t_2 = \frac{R_2t_1}{R_1} \\\\= \frac{(200)(4)}{100} \\= 8hr](https://tex.z-dn.net/?f=t_2%20%3D%20%5Cfrac%7BR_2t_1%7D%7BR_1%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B%28200%29%284%29%7D%7B100%7D%20%5C%5C%3D%208hr)
Part b
Now two batteries are in series
total emf =2V
![\frac{(2V)^2t_1}{R_1} = \frac{V^2t_2}{R_2} \\\\\frac{4t_1}{R_1} = \frac{t_2}{4R_2} \\\\t_2=\frac{2(4)(100)}{4(100)} \\\\=2hr](https://tex.z-dn.net/?f=%5Cfrac%7B%282V%29%5E2t_1%7D%7BR_1%7D%20%3D%20%5Cfrac%7BV%5E2t_2%7D%7BR_2%7D%20%5C%5C%5C%5C%5Cfrac%7B4t_1%7D%7BR_1%7D%20%3D%20%5Cfrac%7Bt_2%7D%7B4R_2%7D%20%5C%5C%5C%5Ct_2%3D%5Cfrac%7B2%284%29%28100%29%7D%7B4%28100%29%7D%20%5C%5C%5C%5C%3D2hr)
use the equation below
Answer:
Δx is the path difference between the two waves.
...
Phase Difference And Path Difference Equation.
Formula Unit
Phase Difference \Delta \phi=\frac{2\pi\Delta x}{\lambda } Radian or degree
Path Difference \Delta x=\frac{\lambda }{2\pi }\Delta \phi meter
Answer:
and ![\Delta \theta = 495.944\,rev](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20495.944%5C%2Crev)
Explanation:
The tub rotates at constant speed and the kinematic formula to describe the change in angular displacement (
), measured in radians, is:
![\Delta \theta = \omega \cdot \Delta t](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20%5Comega%20%5Ccdot%20%5CDelta%20t)
Where:
- Steady angular speed, measured in radians per second.
- Time, measured in seconds.
If
and
, then:
![\Delta \theta = \left(31.7\,\frac{rad}{s} \right)\cdot (98.3\,s)](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20%5Cleft%2831.7%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%5Cright%29%5Ccdot%20%2898.3%5C%2Cs%29)
![\Delta \theta = 3116.11\,rad](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%203116.11%5C%2Crad)
The change in angular displacement, measured in revolutions, is given by the following expression:
![\Delta \theta = (3116.11\,rad)\cdot \left(\frac{1}{2\pi} \frac{rev}{rad} \right)](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20%283116.11%5C%2Crad%29%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B2%5Cpi%7D%20%5Cfrac%7Brev%7D%7Brad%7D%20%5Cright%29)
![\Delta \theta = 495.944\,rev](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20495.944%5C%2Crev)
The answer for this would be B!!
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