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: it is C
The thermistor
Explanation:
Thermistors (temperature-sensitive, or thermal, resistors) are used as temperature-measuring devices and in electrical circuits to compensate for temperature variations of other components. They are also used to measure radio-frequency power and radiant power, such as infrared and visible light.
To measure the intensity of the electromagnetic wave after knowing the frequency, we can calculate it by using fast fourier transformations (FFT).
<span>It is measured in calories </span>
Answer:
The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
Explanation:
Thickness of the wall is L= 20cm = 0.2m
Thermal conductivity of the wall is K = 2.79 W/m·K
Temperature at the left side surface is T₁ = 50°C
Temperature of the air is T = 22°C
Convection heat transfer coefficient is h = 15 W/m2·K
Heat conduction process through wall is equal to the heat convection process so

Expression for the heat conduction process is

Expression for the heat convection process is

Substitute the expressions of conduction and convection in equation above


Substitute the values in above equation

Now heat flux through the wall can be calculated as

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
Answer:
E=0
Explanation:
The electric field at the centre of the shell is zero because total enclosed charge in the nucleus is zero