To increase it's size
Waxing is the opposite of waning, which is to decrease.
Answer:
Coil 2 have 235 loops
Explanation:
Given
The number of loops in coil 1 is n
₁=
159
The emf induced in coil 1 is ε
₁
=
2.78
V
The emf induced in coil 2 is ε
₂
=
4.11
V
Let
n
₂ is the number of loops in coil 2.
Given, the emf in a single loop in two coils are same. That is,
ϕ
₁/n
₁=
ϕ
₂
n
₂⟹
2.78/159
=
4.11/
n
₂
n₂=
n₂=235
Therefore, the coil 2 has n
₂=
235 loops.
Answer:
Delta_temp = 18[°F]
Explanation:
°F = 9/5*(10)
°F = 18
Note: It is important to clarify that it is only a temperature increase, that it is only a temperature increase. The question is not related to converting from 10°C to fahrenheit degrees
Answer:
Explanation:
Given

Volume Flow rate 
Length 
radius 

According to Hagen-Poiseuille Equation
difference in Pressure is given






Answer:
a I think hope this helps