List in order first three steps to square a board

How far do you jog each morning? You prefer to jog in different locations each day and do not have a pedometer to measure your d

Mashutka [201]

Answer:

The program is given below with appropriate comments for better understanding

Explanation:

#Program

# foot stride = 2.5 feet

# 1 mile = 5280 feet

no_stride_first_min = int(input('Enter the number strides made durng the first minute of jogging: '))

no_stride_last_min = int(input('Enter the number strides made durng the last minute of jogging: '))

avg_stride_one_min = (no_stride_first_min + no_stride_last_min)/2 # calculates the average stride per minute

jogging_duration = float(input('Enter the total time spent jogging in hours and minute: '))

jogging_duration_hours = int(jogging_duration) # gets the hour

jogging_duration_min = jogging_duration - int(jogging_duration) # gets the minute

tot_jogging_duration_min = jogging_duration_hours*60 + jogging_duration_min # calculates total time in minutes

dist_feet = (avg_stride_one_min*2.5)*tot_jogging_duration_min # calculates the total distance in feet

dist_miles = dist_feet/5280 # calculates the total distance in mile

print('Distance traveled in miles = {0:.2f} miles'.format(dist_miles))

7 0

2 years ago

A smoking lounge is to accommodate 19 heavy smokers. The minimum fresh air requirement for smoking lounges is specified to be 30

igor_vitrenko [27]

Answer 1: minimum required flow rate of fresh air is 0.57 m^3/ses

Explanation: since the minimum requirement per person is 30 L/sec

Converting to m^3 it becomes

30/1000 = 0.03 m^3/sec

For 19 heavy smoker we will require

19 * 0.03 = 0.57m^3/sec

Answer 2: diameter of the duct will be 0.3m

Explanation: since flow rate is

Q =0.57m^3/sec

Also

Q = AV (continuity equation)

Where A is the duct area and V is the velocity of air flow in m/sec

0.57m^3/sec = A * 8m/sec

A = 0.57/8 = 0.071m^2

Area of the duct is that of a circle

A = 3.142 *(d^2 ÷4)

d^2 = (0.017 * 4)/3.142 = 0.09

d is square root of 0.09

d = 0.3m

7 0

3 years ago

The penalty for littering 15 lb or less is _____. A. $25 B. $50 C. $100 D. $150

Marina CMI [18]

Answer:

D the closest

Explanation:

i looked it up and 8t says 200

5 0

3 years ago

A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the

Margaret [11]

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

$C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2)$

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

$T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

$C_p = 520 \:\: J/kg.K$

So, the inlet temperature of argon is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$

$T_{in} = \frac{1}{2} \times 119 + 300$

$T_{in} = 360K$

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

$C_p = 5193 \:\: J/kg.K$

So, the inlet temperature of helium is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$

$T_{in} = \frac{1}{2} \times 12 + 300$

$T_{in} = 306K$

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

$C_p = 1039 \:\: J/kg.K$

So, the inlet temperature of nitrogen is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$

$T_{in} = \frac{1}{2} \times 60 + 300$

$T_{in} = 330K$

Note: Answers are rounded to the nearest whole numbers.

5 0

2 years ago

Serves as a protective barrier to prevent contact with engergized ("hot") parts within the unit

erik [133]

Answer:

thanks hot hot

Explanation:

4 0

3 years ago