Please help for our stem project :(

Write a program to control the operation of the RED/GREEN/BLUE LED (LED2) as follows: 1. If no button is pressed, the LED should

aalyn [17]

Answer:

See explaination

Explanation:

int RED=10; int BLUE=11; int GREEN=12; int BUTTON1=8; int BUTTON2=9; void setup() { pinMode(RED, OUTPUT); pinMode(BLUE, OUTPUT); pinMode(GREEN, OUTPUT); pinMode(BUTTON1, INPUT); pinMode(BUTTON2, OUTPUT); } void loop() { int BTN1_STATE=digitalRead(BUTTON1); int BTN2_STATE=digitalRead(BUTTON2); if(BTN1_STATE==HIGH) { digitalWrite(BLUE, HIGH); delay(1000); // Wait for 1 second digitalWrite(BLUE, LOW); } if(BTN2_STATE==HIGH) { digitalWrite(RED, HIGH); delay(4000); // Wait for 4 seconds digitalWrite(RED, LOW); } if(BTN1_STATE==HIGH && BTN2_STATE==HIGH) { digitalWrite(GREEN, HIGH); delay(2000); // Wait for 2 second digitalWrite(GREEN, LOW); } }

4 0

3 years ago

A thick oak wall initially at 25°C is suddenly exposed to gases for which T =800°C and h =20 W/m2.K. Answer the following questi

Schach [20]

Answer:

a) What is the surface temperature, in °C, after 400 s?

T (0,400 sec) = 800°C

b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s

c) What is the temperature, in °C, 1 mm from the surface after 400 s?

T (1 mm, 400 sec) = 798.35°C

Explanation:

oak initial Temperature = 25°C = 298 K

oak exposed to gas of temp = 800°C = 1073 K

h = 20 W/m².K

From the book, Oak properties are e=545kg/m³ k=0.19w/m.k Cp=2385J/kg.k

Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.

From energy balance: $\frac{T - T_{\infty}}{T_i - T_{\infty}} = Exp (\frac{-hA}{evCp})t$

Initial temperature wall = $T_i$

Surface temperature = T

Gas exposed temperature = $T_{\infty}$

6 0

3 years ago

Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value

masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

$q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} } )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}$

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

$q_{n} =uK(\frac{g(\rho_{L}-\rho _{v}) }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4$

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0

3 years ago

Technician A says the final drive assembly always has a gear ratio of 1:1. Technician B says the final drive assembly provides f

Olenka [21]

Answer:

Technician B only is correct

Explanation:

The last stage of gears found between the vehicle transmission system and the wheels is the final drive ratio. The function of the final drive gear assembly is to enable a gear reduction control stage to reduce the rotation per minute and increase the wheel torque, such that the vehicle performance can be adjusted and the final gear ratio can be between 3:1 and 4.5:1 not 1:1

Therefore, technician B only is correct

5 0

3 years ago

The properties of the air in the inlet section with A1 = 0.25ab m2 in a converging-diverging channel are given as U1 = 25a,b m/s

NeX [460]

Answer:

nice cake

Explanation:

3 0

3 years ago