charge = current * time
q = 15 * 300 [5 min = 300 sec ]
q = 4500 C
Answer:
option C
Explanation:
Given,
Refractive index of medium 1 = n₁
Refractive index of medium 2 = n₂
For total internal reflection to take place light should move from denser medium to the rarer medium.
Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2
n₁ > n₂
The correct answer is option C
Isothermal Work = PVln(v₂/v₁)
PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J
Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,
1.7 kJ = (5487.24)In(v₂/19)
1700 = (5487.24)In(v₂/19)
In(v₂/19) = (1700/5487.24) = 0.3098
In(v₂/19) = 0.3098
(v₂/19) =

v₂ = 19*

v₂ = 25.8999
v₂ ≈ 26 L Option b.
Answer:
0.0133A
Explanation:
Since we have two sections, for the Inductor region there would be a current
. In the case of resistance 2, it will cross a current
Defined this we proceed to obtain our equations,
For
,


For
,


The current in the entire battery is equivalent to,


Our values are,




Replacing in the current for t= 0.4m/s


