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2 years ago

50 points what shape is mars

Engineering

2 answers:

Artyom0805 [142]2 years ago

5 0

A sphere.............

salantis [7]2 years ago

3 0

Answer:

its a sphere or a circle, whichever you prefer.

Explanation:

The average temperature on Mars is minus 80 degrees Fahrenheit that's way below freezing! But for some reason, a lot of people think its hot because of its appearance.Its surface is rocky, with canyons, volcanoes, dry lake beds and craters all over it. Red dust covers most of its surface. Which gives it a hot desert look, which may be why people think its hot. Mars has clouds and wind just like Earth.

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Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l

sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is $0.44 \frac{W}{m^{2} \times K}$ , the average convection heat transfer coefficient over the entire plate is $0.293 \frac{W}{m^{2} \times K}$ and the total heat flux transfer to the plate is $61.6 KJ$ .

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.

Hot carbon dioxide exhaust gas

physical properties

$r= 1.05 \frac{kg}{m^{3}}$

$c_p = 1.02 \frac{kJ}{Kg \times K}$

$m= 231 \times 10^{7} \frac{N \times s }{m^2}$

υ = $21.8 \times 10^{6} \frac{m^2}{s}$

$k = 32.5 \times 10^{3} \frac{W}{m \times K}$

$\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}$

$Pr = 0.725$

Apart from these other data arr given below,

$v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C$

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

$Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

where h= Average heat transfer coefficient

L= Length of a plate

k= Thermal Conductivity of carbon dioxide

Re = Reynold's Number

Pr = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

is referred as local convection heat transfer coefficient.

To find convection heat transfer coefficient at 1 m from leading edge,

$Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

Here, first we have to find Re and Pr,

$Re = \frac{r \times v \times L}{m}$

$Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}$

$Re = 20.63 \times 10^{-10}$

Pr number is take from physical property data and Pr is 0.725.

Putting value of Re and Pr in main equation,

we get

$Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 0.44 \frac{W}{m^{2} \times K}$

(b) To find average convection heat transfer coefficient,

it can be find out as case (a), only difference is that instead of L=1 m, L=1.5 m would come,

Therefore,

$Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

Finally,

$h = \frac{0.44}{1.5}$

$h = 0.293 \frac{W}{m^{2} \times K}$

$Q = h \times (T_g - T_s)$

$Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140 \\ Q= 61.6 KJ$

8 0

2 years ago

Two previously undeformed rod-shaped specimens of copper are to be plastically deformed by reducing their cross-sectional areas.

mezya [45]

I am not sure I am stuck on this and I have been for 45 min someone please help me and this girl or boy!!

4 0

2 years ago

Anything that is made to meet a need or desire is?

slavikrds [6]

Answer:

I think it is process or technology

7 0

2 years ago

Read 2 more answers

Determine the required dimensions of a column with a square cross section to carry an axial compressive load of 6500 lb if its l

ycow [4]

Answer: 0.95 inches

Explanation:

A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

PE= π^2EI/Le^2

6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches

3 0

2 years ago

Read 2 more answers

In this exercise, you will write a Point structure that represents a space in two-dimensional space. This Point should have both

Afina-wow [57]

Answer:

Check the explanation

Explanation:

Points to consider:

We need to take the input from the user

We need to find the manhatan distance and euclidian using the formula

(x1, y1) and (x2, y2) are the two points

Manhattan:

$|x_1 - x_2| + |y_1 - y_2|$

Euclidian Distance:

$\sqrt{(x1 - yl)^2 + (x2 - y2)^2)}$

Code

#include<stdio.h>

#include<math.h>

struct Point{

int x, y;

};

int manhattan(Point A, Point B){

return abs(A.x - B.x) + abs(A.y- B.y);

}

float euclidean(Point A, Point B){

return sqrt(pow(A.x - B.x, 2) + pow(A.y - B.y, 2));

}

int main(){

struct Point A, B;

printf("Enter x and Y for first point: ");

int x, y;

scanf("%d%d", &x, &y);

A.x = x;

A.y = y;

printf("Enter x and Y for second point: ");

scanf("%d%d", &x, &y);

B.x = x;

B.y = y;

printf("Manhattan Distance: %d\n", manhattan(A, B));

printf("Euclidian Distance: %f\n", euclidean(A, B));

}

Sample output

8 0

3 years ago