Explanation:
https://educationalghana.news.blog/2021/08/09/geography-human-physical-and-practical-for-wassce-novdec-candidates/
Answer:
The factor of the diameter is 0.95.
Explanation:
Given that,
Power of old light bulb = 54.3 W
Power = 60 W
We know that,
The resistance is inversely proportional to the diameter.
![R\propto\dfrac{1}{D}](https://tex.z-dn.net/?f=R%5Cpropto%5Cdfrac%7B1%7D%7BD%7D)
The power is inversely proportional to the resistance.
![P\propto\dfrac{1}{R}](https://tex.z-dn.net/?f=P%5Cpropto%5Cdfrac%7B1%7D%7BR%7D)
![P\propto D^2](https://tex.z-dn.net/?f=P%5Cpropto%20D%5E2)
We need to calculate the factor of the diameter of the filament reduced
Using relation of power and diameter
![\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_%7Bi%7D%7D%7BP_%7Bf%7D%7D%3D%5Cdfrac%7BD_%7Bi%7D%5E2%7D%7BD_%7Bf%7D%5E2%7D)
Put the value into the formula
![\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}](https://tex.z-dn.net/?f=%5Cdfrac%7BD_%7Bi%7D%5E2%7D%7BD_%7Bf%7D%5E2%7D%3D%5Cdfrac%7B54.3%7D%7B60%7D)
![\dfrac{D_{i}}{D_{f}}=0.95](https://tex.z-dn.net/?f=%5Cdfrac%7BD_%7Bi%7D%7D%7BD_%7Bf%7D%7D%3D0.95)
![D_{i}=0.95 D_{f}](https://tex.z-dn.net/?f=D_%7Bi%7D%3D0.95%20D_%7Bf%7D)
Hence, The factor of the diameter is 0.95.
Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
![R = \frac{6}{2.5} - 2.4 ohm](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7B6%7D%7B2.5%7D%20-%202.4%20ohm)
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
![\frac{3R}{R+3} \times 2.5 = 6](https://tex.z-dn.net/?f=%5Cfrac%7B3R%7D%7BR%2B3%7D%20%5Ctimes%202.5%20%3D%206)
R= 12 ohm
2) circuit Q
![I\times (R+0.1) =V](https://tex.z-dn.net/?f=I%5Ctimes%20%28R%2B0.1%29%20%3DV)
![R+0.1 =\frac{6}{2.5}](https://tex.z-dn.net/?f=R%2B0.1%20%3D%5Cfrac%7B6%7D%7B2.5%7D)
R = 2.3 ohm
Um you should putting it in a object that it can fill then go from there