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3 years ago

Write a paragraph on computer 473

1 answer:

8 0

Whether you're looking for a high powered computer to play PC games, a small computer to stream movies and stay connected while traveling, or you're looking for an affordable computer to use for word processing and social media, the latest computers and tablets balance portability with the performance and power you need. And the range of printers, computer software and accessories allow you to create the ideal setup to fit your lifestyle.

Hope this helps!

You might be interested in

Explain and show work:

belka [17]

Answer:

Explanation:

The sum of the digits of the number is ...

(4+1+3)+(4+6+5)+(7+8+9) = 8+15+24 = 47

The sum of those digits is 4+7=11, and those digits sum to 1+1 = 2.

That is, the value of the number mod 9 (or 3) is 2.

The ones digit is odd, so the value of the number mod 2 is 1.

This combination of modulo values tells you the mod 6 result is 5.

_____

<em>Additional comment</em>

We can look at the (mod2, mod3) values of the numbers 0 to 5:

0 ⇒ (0, 0)

1 ⇒ (1, 1)

2 ⇒ (0, 2)

3 ⇒ (1, 0)

4 ⇒ (0, 1)

5 ⇒ (1, 2) . . . . the mod {2, 3} results we have for the number of interest.

This process of adding up the digits repeatedly is referred to as "casting out 9s." The result of it is the modulo 9 value of the number (with 0 mapped to 9). Checking the mod 9 result of arithmetic operations is one quick way to spot certain kinds of errors. It can also be used as part of a divisibility test for 3 or 9.

3 0

3 years ago

. What is the combined thrust if both stages are used to launch the rocket?

Rudiy27

Answer:

340 mph

Explanation:

3 0

3 years ago

Read 2 more answers

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to b

coldgirl [10]

Answer:

A) q'_free = 3146.41 W/m²

B) q'_forced = 7521.41 W/m²

Explanation:

We are given;

Free convection coefficient; h_fr = 5 W/m²K

Force convection coefficient; h_forced = 30 W/m²K

Emissivity; ε = 0.95

Temperature of surrounding which is equal to temperature of air; T_s = T_air = 200°C = 473K

Initial temperature; T_i = 25°C = 298K

A) Now, since the convection feature is disabled, the mode of heat transfer associated with this condition is through free convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_free = q'_free convection + q'_radiation

q'_free convection = h_free(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Where, σ is stephan boltzmann constant and has a constant value of 5.67 × 10^(−8) W/m²K⁴

Thus, rewriting;

q'_free = q'_free convection + q'_radiation

We have;

q'_free = [h_free(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_free = [5(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_free = 875 + 2271.41

q'_free = 3146.41 W/m²

B) Now, in this case, since the convection feature is disabled, the mode of heat transfer associated with this condition is through forced convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_forced = q'_forced convection + q'_radiation

Where;

q'_forced convection = h_forced(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Thus, rewriting;

q'_forced = q'_free convection + q'_radiation

We have;

q'_forced = [h_forced(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_forced = [30(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_forced = 5250 + 2271.41

q'_forced = 7521.41 W/m²

7 0

4 years ago

An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$

From the question the, the mass flow rate at the inlet $m_{1}}$ is 2Kg/s $\\$

Since 5% flow is delivered into the feedwater heating, $\\$

$m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s$ $\\$

Also for the outlet 3 the remaining 95% will flow out. Hence

$m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s$ $\\$

Now, from $m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$ we substitute values

$W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}$

$W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)$

$\\$

$W_{out} = 2285.19 kW$ .

Hence the power produced is 2285kW

7 0

3 years ago

For this homework you are writing a main function, with 3 subfunctions. 2. The "main" function is to be named arm_prop a. Inputs

ra1l [238]

Answer:

MATLAB CODE:

clc;

clear all;

x=[1 2 3 4]; %Input vector of variable size

[gm,ra,hm]=arm_prop(x) %Main function calling , output values are hm, ra, hm and imput is x

%% Main function

function [gm,ra,hm]=arm_prop(x)

[gm]=geo_mean(x); %Calling Sub function 1

[ra]=rms_avg(x); %Calling Sub function 2

[hm]=har_mean(x); %Calling Sub function 3

%Sub function 2 =Rms average

function [gm]=geo_mean(x)

n=numel(x); %calculating number of elemnts in vector x

temp=1; %Temporary value for multiplying each and every value of x

for i=1:1:n % For loop run for number of elements in x

temp=temp*x(i); %Multliplyng each and every element of x

end

gm=temp^(1/n); % Calculating geometric mean

end

%Sub function 2 =Rms average

function [ra]=rms_avg(x)

temp=0; %Temporary value for adding square of each and every value of x

N=numel(x); %calculating number of elemnts in vector x

for i=1:1:N % For loop run for number of elements in x

temp=temp+x(i)^2; %Adding square of each and every element of x

end

ra=(temp/N)^(1/2); % Calculating Rms average

end

%Sub function 3 =Harmonic mean

function [hm]=har_mean(x)

temp=0; %Temporary value for adding reciprocal of each and every value of x

N=numel(x); %calculating number of elemnts in vector x

for i=1:1:N % For loop run for number of elements in x

temp=temp+(1/x(i)); %Adding sreciprocal of each and every element of x

end

hm=N/temp; % Halculating Harmonic mean

end

Results:

gm =

2.2134

ra =

2.7386

hm =

1.9200

Here in one main function. 3 sub functions are written. Every sub function returns one output. As all 3 sub functions are inside main function, main function returns 3 output values gm, ra, hm.

Explanation:

3 0

4 years ago