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2 years ago

A) Define the term magnetic flux

Engineering

2 answers:

Radda [10]2 years ago

3 0

Answer:

Magnetic flux is a measurement of the total magnetic field which passes through a given area. It is a useful tool for helping describe the effects of the magnetic force on something occupying a given area.

sp2606 [1]2 years ago

3 0

Magnetic flux is a measurement of the total magnetic field which passes through a given area, If we use the field-line picture of a magnetic field then every field line passing through the given area contributes some magnetic flux. The angle at which the field line intersects the area is also important.

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Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charg

babunello [35]

Answer:

Explanation:

To find the electrical force per unit area on each sheet we start defining our variables,

$\sigma_A = -4.10^{-5}C/m^2$

$\sigma_B= -7.10^{-5}C/m^2$

$\sigma_C = -3.1^{-5}C/m^2$

We find the electric field for each one, this formula is given by,

$E= \frac{\sigma_i}{2\epsilon_0}$

Substituting each value from the three charged sheets, we have

$E_A= \frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_B= \frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})$

$E_C= \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

The electric field is

$E_{NET}= E_A+E_B+E_C$

$E_{NET}=\frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})+\frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})+ \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_{NET} = 0$

Force on each sheet is,

$F=E_{NET}\sigma ds$

$F=0$

The total force is 0

5 0

3 years ago

Using the Breguet range and endurance equation, estimate the amount of kerosene

love history [14]

The amount of kerosene fuel needed for an aircraft weighing 10 metric tons to fly from Boston to Los Angeles, assuming a distance of 5,000 [km] is 10000 kg.

<h3>What is Breguet equation?</h3>

Breguet equation is used to determine the range of airplane fly in some specified set of parameters.

$R=\dfrac{h_f}{g}\dfrac{L}{D}\times \eta \ln \left(\dfrac{W_{initial}}{W_{final}}\right)$

Here, (L/D) is lift to drag ratio,<em> g</em> is gravitation acceleration,<em> h </em>is height and <em>W</em> is weight.

The weight of aircraft is 10 metric tons (this is the dry weight that includes passengers and cargo) The distance of 5,000 [km], with velocity at 300 [m/s] from Boston to Los Angeles has to be cover by the airplane.

The initial weight is 10 metric tons or 10000 kg. Thus, the final weight with total fuel burn is,

$W_{final}=W_{initial}-W_{fuel}\\W_{final}=1-\dfrac{W_{fule}}{W_{initial}}$

The lift-to-drag ratio is 15 and the overall efficiency is 0.3, The standard air density at an altitude of approximately 6,000 [m] is half.

Thus, put the values, in above formula,

$5000=\dfrac{6000}{9.81}(15)\times(0.3) \ln \left(\dfrac{10000}{1-\dfrac{W_{fuel}}{10000}}}\right)\\$

$\ln \left(\dfrac{10000}{1-\dfrac{W_{fuel}}{10000}}}\right)=1.8167\\\ln \left(\dfrac{1}{10000-W_{fuel}}}\right)=1.8167$

Solving this equation, we get,

$W_{fuel}=10000\rm\; kg\\$

Thus, the amount of kerosene fuel needed for an aircraft weighing 10 metric tons to fly from Boston to Los Angeles, assuming a distance of 5,000 [km] is 10000 kg.

Learn more about the Breguet equation here:

brainly.com/question/15218094

#SPJ1

7 0

2 years ago

suppose we number the bytes in a w-bit word from 0 (less significant) to w/8-1 (most significant). write code for the followign

sammy [17]

Solution:

typedef unsigned char *byte_pointer;

static int int_of_bit (byte_pointer x, int loc)

{

return(x[loc] << loc*8);

}

static int replace_byte(unsigned int a, int loc, unsigned int b)

unsigned int a_loc = int_of_bit((byte_pointer) &a , loc);

unsigned int b_loc = (b << loc*8);

a -= a_loc;

a += b_loc;

return a;

}

Explanation:

This takes two ints in hex format, one with 8 bits (0x00000000) and one with 2 bits (0x00) then places the 2 bit hex into the 8 bit at a given location.

EX: replace_byte(0x00000000, 1, 0xFF) return 0x0000FF00

First thing first, it looks like you have some mixup:

" one with 8 bits (0x00000000) and one with 2 bits (0x00)*- what you mean is nibble not bits. Each hex character (0-9, A-F) represent 4 bits (16 possible combination), and is called a "nibble".

0x0000000 is 4 bytes. 0x00 is 2 bytes.

Next, you say "takes two ints in hex format" - your function takes two ints in any format. You're just choosing tp express them in hexidecimal. C++ doesn't care if you specify numbers in hex, decimal, octal, binary, etc.

8 0

3 years ago

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$