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Hunter-Best [27]
3 years ago
11

Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's

numbers on one line, each number followed by one space. (2 pts)
Ex:Enter weight 1:236.0Enter weight 2:89.5Enter weight 3:142.0Enter weight 4:166.3Enter weight 5:93.0You entered: 236.000000 89.500000 142.000000 166.300000 93.000000
Engineering
1 answer:
atroni [7]3 years ago
6 0

Answer:

See below in the explanation section the Matlab script to solve the problem.

Explanation:

prompt='enter the first weight w1:  ';

w1=input(prompt);

wd1=double(w1);

prompt='enter the second weight w2:  ';

w2=input(prompt);

wd2=double(w2);

prompt='enter the third weight w3:  ';

w3=input(prompt);

wd3=double(w3);

prompt='enter the fourth weight w4:  ';

w4=input(prompt);

wd4=double(w4);

prompt='enter the first weight w5:  ';

w5=input(prompt);

wd5=double(w5);

x=[wd1 wd2 wd3 wd4 wd5]

format short

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A signal containing both a 5k Hz and a 10k Hz component is passed through a low-pass filter with a cutoff frequency of 4k Hz. Wh
abruzzese [7]

Answer:

The 5 kHz signal will be less attenuated and the 10 kHz will be completely attenuated.

Low Pass Filter:

The purpose of a filter is to remove the undesired frequency components in a signal. A low-pass filter is one in which we allow low frequencies components to pass through it and block the higher frequency components. The frequency  where it starts to attenuate the signal is called cut-off frequency.

Explanation:

We have a signal that contains 5 kHz and 10 kHz frequency component.

The cut-off frequency is 4 kHz which means the filter will attenuate the frequency components of 4 kHz or greater than that.

So, in theory both 5 kHz and 10 kHz frequency components will be attenuated.

But filters are not ideal and they usually have a little margin also called roll off which means it will allow some of the frequency components greater than cut-off frequency of 4 kHz.

Conclusion:

So to conclude, the 5 kHz signal will be less attenuated and the 10 kHz will be completely attenuated.

3 0
4 years ago
Two kilograms of air in a piston-cylinder assembly undergoes an isothermal
Lubov Fominskaja [6]

Answer:

a) 0.39795 kJ/K

b) 79.589.37 kJ

Explanation:

m = Mass of air = 2 kg

Temperature = 200 K

P₁ = Initial pressure = 300 kPa

P₂ = Final pressure = 600 kPa

R = mass-specific gas constant for air = 287.058 J/kgK

a) For isentropic process

∴ Entropy is generated in the process is 0.39795 kJ/K

b)

∴ Amount of lost work is 79.589.37 kJ

7 0
3 years ago
An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a
DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π ×  0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶  × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

4 0
3 years ago
What are the rigging devices used to move loads such as steel plates and sheet piles without the use of slings, but grip the loa
scoundrel [369]

The rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.

<h3>What are the rigging devices?</h3>

The rigging devices are used to lift the objects and items when the safety is required. This device is used in the industries.

Types of rigging devices

  • Rigging hooks-These rigging device is used when the heavy load need to be lift.
  • Lifting clamps-Lifting clamp are used to lift the device with jaw tension to secure the load. In this, there is no use of slings.
  • Pulley and blocks-In the load is lifts with the help of block and pulley arrangement. This is a widely used rigging device.

Thus, the rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.

Learn more about the rigging devices here;

brainly.com/question/8430576

#SPJ1

4 0
2 years ago
Witch measuring tool would be used to determine the diameter of a crankshaft journal
djverab [1.8K]
Answer =

dial bore gauge

a “dial bore gauge” measures the inside of round holes, such as the bearing journals . can mesure up to 2” and 6” diameter holes .

when ( “ ) is next to a number it means inches fwi - but hope this helped have a good day :)
3 0
3 years ago
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