How to create ground responce analysis using matlab

Read two numbers from user input. Then, print the sum of those numbers. Hint -- Copy/paste the following code, then just type co

Softa [21]

Answer:

I am Providing Answer in C Language Program.

Explanation:

Please find attachment regarding code of taking two numbers input and adding them.

I would like to recommend you please use software which supports C language.

#include <stdio.h>

int main () {

int a, b, sum;

printf ("\ nEnter two no:");

scanf ("% d% d", & d, & e);

sum1 = d + e;

printf ("Sum:% d", sum1);

return (0);

}

4 0

3 years ago

If a 9V battery produces a current of 3 A through a load, what is the resistance of the load

Elden [556K]

3 ohms hope this helps :D ❤

7 0

3 years ago

Read 2 more answers

A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi

Anon25 [30]

Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

Height = depth = 5 mm

Width = 10 mm

Length L = 45 mm

Load = 290 N

The flexural strength of a specimen is given by

$\sigma = \frac{3 F L}{2 bd^{2} }$

$\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }$

$\sigma =$ 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

4 0

3 years ago

A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055

Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

$K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}$

So the mean conductivity

$K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )$

$T_m=\dfrac{160+273+40+273}{2}$

$T_m=373 K$

$K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )$

$K_m=0.112 \frac{W}{mK}$

So heat conduction through cylinder

$Q=kA\dfrac{\Delta T}{L}$

$Q=0.112\times \pi \times 0.15^2\times 120$

Q=0.95 W/m

4 0

3 years ago

An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a speed of 10 ft/s thr

Sladkaya [172]

Answer:

Part A

The mass plow rate, is approximately 97.0 lbm/s

Part B

The power used to overcome friction, is approximately 1.9 hp

Explanation:

The efficiency of the pump, η = 80%

The power input to the pump, P = 20 hp

The speed of the water through the pipe, v = 10 ft./s

The diameter of the pipe, d = 5.2 inches = 13/30 ft.

The free surface of the pool above the lake, h = 80 ft.

The density of the water, ρ = 62.4 lbm/ft.³

Part A

The mass plow rate, $\dot m$ = Q × ρ

Where;

ρ = 62.4 lbm/ft³

Q = A × v

A = The cross-sectional area of the pipe

∴ Q = π·d²/4 × v = π × ((13/30 ft.)²)/4 × 10 ft.s ≈ 1.4748 ft.³/s

∴ The mass plow rate, $\dot m$ ≈ 1.4748 ft.³/s × 62.4 lbm/ft.³ = 97.02752 lbm/s

The mass plow rate, $\dot m$ ≈ 97.0 lbm/s

Part B

The power to pump the water at the given rate, $P_w$ = $\dot m$ ·g·h

∴ $P_w$ = 97.02752 lbm/s × 32.1740 ft./s² × 80 ft. ≈ 14.1130725 Hp

$P_w$ ≈ 14.1130725 Hp

The power output of the pump, $P_{out}$ = 0.8 × 20 hp = 16 hp

Therefore, the power used to overcome friction, $P_f$ = $P_{out}$ - $P_w$

∴ $P_f$ ≈ 16 hp - 14.1130725 Hp ≈ 1.8869275 hp

The power used to overcome friction, $P_f$ ≈ 1.9 hp

7 0

2 years ago