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3 years ago

Driving Distraction Brainstorming Session

1 answer:

5 0

texting, phone calls, putting on makeup, brushing hair, movies playing in car, loud music, children, and that's pretty much all I could think of

please give <u>BRAINLIEST ANSWER └[T‸T]┘</u>

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You just purchased a 400-L rigid tank for a client who works in the gas industry. The tank is delivered pre-filled with 3 kg of

solniwko [45]

Answer:

the pressure reading when connected a pressure gauge is 543.44 kPa

Explanation:

Given data

tank volume (V) = 400 L i.e 0.4 m³

temperature (T) = 25°C i.e. 25°C + 273 = 298 K

air mass (m) = 3 kg

atmospheric pressure = 98 kPa

To find out

pressure reading

Solution

we have find out pressure reading by gauge pressure

i.e. gauge pressure = absolute pressure - atmospheric pressure

first we find absolute pressure (p) by the ideal gas condition

i.e pV = mRT

p = mRT / V

p = ( 3 × 0.287 × 298 ) / 0.4

p = 641.44 kPa

gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 641.44 - 98

gauge pressure = 543.44 kPa

6 0

3 years ago

A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20

viktelen [127]

Answer:

a) 69,630KW

b) 203 KW

Explanation:

The data obtained from Tables A-4, A-5 and A-6 is as follows:

$h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2} = 255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\$

$x_{4} = \frac{s_{4} - s_{f} }{s_{fg} } \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\$

The power produced and consumed by turbine and pump respectively are:

$W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW$

7 0

3 years ago

What is electromagnetic induction?

Over [174]

The process of using magnetic fields to produce voltage.

6 0

3 years ago

Developed an automated program in any language which take 12 dependent variable and corresponding independent variables and show

otez555 [7]

Answer:

Explanation:

var generator = new Random(1);

// Now the nextGaussian() function returns a normal distribution of random numbers with the following parameters: a mean of zero and a standard deviation of one

var draw = function() {

var num = generator.nextGaussian();

var standardDeviation = 60;

var mean = 2003;

// Multiply by the standard deviation and add the mean.

var x = standardDeviation * num + mean;

noStroke();

fill(214, 159, 214, 10);

ellipse(x, 200, 16, 16); };

Hope this will be helpful

3 0

3 years ago

Using the celsius_to_kelvin function as a guide, create a new function, changing the name to kelvin_to_celsius, and modifying th

aleksandr82 [10.1K]

Answer:

# kelvin_to_celsius function is defined

# it has value_kelvin as argument

def kelvin_to_celsius(value_kelvin):

# value_celsius is initialized to 0.0

value_celsius = 0.0

# value_celsius is calculated by

# subtracting 273.15 from value_kelvin

value_celsius = value_kelvin - 273.15

# value_celsius is returned

return value_celsius

# celsius_to_kelvin function is defined

# it has value_celsius as argument

def celsius_to_kelvin(value_celsius):

# value_kelvin is initialized to 0.0

value_kelvin = 0.0

# value_kelvin is calculated by

# adding 273.15 to value_celsius

value_kelvin = value_celsius + 273.15

# value_kelvin is returned

return value_kelvin

value_c = 0.0

value_k = 0.0

value_c = 10.0

# value_c = 10.0 is used to test the function celsius_to_kelvin

# the result is displayed

print(value_c, 'C is', celsius_to_kelvin(value_c), 'K')

value_k = 283.15

# value_k = 283.15 is used to test the function kelvin_to_celsius

# the result is displayed

print(value_k, 'is', kelvin_to_celsius(value_k), 'C')

Explanation:

Image of celsius_to_kelvin function used as guideline is attached

Image of program output is attached.

4 0

3 years ago