Answer:
Explanation:
the light ray leaving a medium in contrast to the entering or incident ray.
![g = \frac{GMm}{r^2}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7BGMm%7D%7Br%5E2%7D)
where
is the gravitational constant and is about ![6.67 * 10^{-11}](https://tex.z-dn.net/?f=6.67%20%2A%2010%5E%7B-11%7D)
![g = \frac{6.67 * 10^{-11} * 25 * 0.55}{35^2}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7B6.67%20%2A%2010%5E%7B-11%7D%20%2A%2025%20%2A%200.55%7D%7B35%5E2%7D)
:D
Answer:
i. 6.923 V
ii. The e.m.f. = 22.5 V
Explanation:
i. The given parameters are;
Length of potentiometer = 1 m
The resistance of the potentiometer = 10 Ω
The e. m. f. of the attached cell = 9 V
The current, I flowing in the circuit = e. m. f/(Total resistance)
The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A
The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire
The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V
ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;
![\dfrac{E}{R_{balance}} = \dfrac{V}{R_{cell}}](https://tex.z-dn.net/?f=%5Cdfrac%7BE%7D%7BR_%7Bbalance%7D%7D%20%3D%20%5Cdfrac%7BV%7D%7BR_%7Bcell%7D%7D)
Where:
E = e.m.f. of the balance point cell
= Resistance of 75 cm of potentiometer wire = 0.75×10 = 7.5 Ω
= Resistance of the cell in the circuit = 3 Ω
V = e.m.f. attached cell = 9 V
![\dfrac{E}{7.5} = \dfrac{9}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7BE%7D%7B7.5%7D%20%3D%20%5Cdfrac%7B9%7D%7B3%7D)
E = 7.5*3 = 22.5 V
The e.m.f. = 22.5 V