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3 years ago

What does Faraday's law of induction states?

Engineering

1 answer:

Karolina [17]3 years ago

6 0

Explanation:

This relationship, known as Faraday's law of induction (to distinguish it from his laws of electrolysis), states that the magnitude of the emf induced in a circuit is proportional to the rate of change of the magnetic flux that cuts across the circuit.

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Steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is con- densed, and leaves as liquid

AnnZ [28]

Answer:

5.328Ibm/hr

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

for this case we can define the following equation for mass flow using the first law of thermodynamics

$m=\frac{Q}{h1-h2}$

where

Q=capacity of the radiator =5000btu/hr

m = mass flow

then using thermodynamic tables we found entalpy in state 1 and 2

h1(x=0.97, p=16psia)=1123btu/lbm

h2(x=0, p=16psia)=184.5btu/lbm

solving

$Q=\frac{5000}{1123-184.5} =5.328Ibm/hr$

3 0

3 years ago

(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows

maxonik [38]

Answer:

diameter of the sprue at the bottom is 1.603 cm

Explanation:

Given data;

Flow rate, Q = 400 cm³/s

cross section of sprue: Round

Diameter of sprue at the top $d_{top}$ = 3.4 cm

Height of sprue, h = 20 cm = 0.2 m

acceleration due to gravity g = 9.81 m/s²

Calculate the velocity at the sprue base

$V_{base}$ = √2gh

we substitute

$V_{base}$ = √(2 × 9.81 m/s² × 0.2 m )

$V_{base}$ = 1.98091 m/s

$V_{base}$ = 198.091 cm/s

diameter of the sprue at the bottom will be;

Q = AV = (π $d_{bottom}^2$ /4) × $V_{base}$

$d_{bottom}$ = √(4Q/π $V_{base}$ )

we substitute our values into the equation;

$d_{bottom}$ = √(4(400 cm³/s) / (π×198.091 cm/s))

$d_{bottom}$ = 1.603 cm

Therefore, diameter of the sprue at the bottom is 1.603 cm

6 0

2 years ago

The typical area of a commercial airplane's passenger window is 80.0 in^2 . At an altitude of 3.00 × 104 ft above the sea level,

nikdorinn [45]

Answer:

The force over the plane windows are 764 lbf in the EE unit system and 3398 N in the international unit system.

Explanation:

The net force over the window is calculated by multiplying the difference in pressure by the area of the window:

F = Δp*A

The pressure inside the plane is around 1 atm, hence the difference in pressure is:

Δp = 1atm - 0.35 atm = 0.65 atm

Expressing in the EE unit system:

Δp = 0.65 atm * 14.69 lbf/in^2 = 9.55 lbf/in^2

Replacing in the force:

F = 9.55 lbf/in^2 * 80 in^2 = 764 lbf

For the international unit system, we re-calculate the window's area and the difference in pressure:

A = 80 in^2 * (0.0254 m/in)^2 = 0.0516 m^2

Δp = 0.65 atm * 101325 Pa = 65861 Pa = 65861 N/m^2

Replacing in the force:

F = 65861 N/m^2 *0.0516 m^2 = 3398 N

3 0

3 years ago

Water flows through a multisection pipe placed horizontally on the ground. The velocity is 3.0 m/s at the entrance and 2.1 m/s a

Alex_Xolod [135]

Answer:

b. 2.3 kPa.

Explanation:

This situation can be modelled by Bernoulli's Principle, as there are no energy interaction throughout the multisection pipe and current lines exists between both ends. Likewise, this system have no significant change in gravitational potential energy since it is placed horizontally on the ground and is described by the following model:

$P_{1} + \rho \cdot \frac{v_{1}^{2}}{2} = P_{2} + \rho \cdot \frac{v_{2}^{2}}{2}$

Where:

$P_{1}$ , $P_{2}$ - Pressures at the beginning and at the end of the current line, measured in kilopascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Fluid velocity at the beginning and at the end of the current line, measured in meters per second.

Now, the pressure difference between these two points is:

$P_{1} - P_{2} = \rho \cdot \frac{v_{2}^{2}-v_{1}^{2}}{2}$

If $\rho = 1000\,\frac{kg}{m^{3}}$ , $v_{1} = 3\,\frac{m}{s}$ and $v_{2} = 2.1\,\frac{m}{s}$ , then:

$P_{1} - P_{2} = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \frac{\left(2.1\,\frac{m}{s} \right)^{2}-\left(3\,\frac{m}{s} \right)^{2}}{2}$

$P_{1} - P_{2} = -2295\,Pa$

$P_{1} - P_{2} = -2.295\,kPa$ (1 kPa is equivalent to 1000 Pa)

Hence, the right answer is B.

7 0

3 years ago

technician A says that incandescent bulbs resist vibration well. Technician B says that HID headlamps require up to approximatel

bekas [8.4K]

Answer:

Both a and b.

Explanation:

HID headlamps require high voltage ignition to start just like street lamps. It requires almost 25,000 volts to start HID head lamps but require only 80 to 90 volts to keep it operating.

6 0

2 years ago

What does Faraday's law of induction states?​

What does Faraday's law of induction states?