1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
postnew [5]
3 years ago
14

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

istance. Create an application to determine the distance jogged given the average number of strides ran during the fist minute, average number ran during the last minute, and the total minutes jogging. Design a modularized solution (with methods) to display the distance traveled. Pedometers measure the distance you run. However, you can also do a good estimate of the distance if you know your foot stride, how many strides you complete per minute, and the number of minutes you job. Foot stride is the distance covered by one average step length. Since everyone has a different foot size, strides differ. Manny people average 3 feet per setup when jogging. For this application, assume the foot stride is 2.5 feet. There are 5,280 feet in a mile. To establish how many strides per minutes, allow the user to input the number of strides made during the first minute jogging and the number of strides made string the last minutes of jogging. Use the average of those values to represent the strides per minute. Allow the user to input the total time spent jogging in hours and minutes. Write code that will display to distance traveled in miles.
Engineering
1 answer:
Mashutka [201]3 years ago
7 0

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))

You might be interested in
Due at 11:59pm please help
sergeinik [125]
I believe it’s c table
7 0
2 years ago
. Consider the single-engine light plane described in Prob. 2. If the specific fuel consumption is 0.42 lb of fuel per horsepowe
Trava [24]

Answer:

Hence the Range and Endurance of single engine plane is given by

650.644 miles and 5.3528 hrs at standard sea level.

Explanation:

Given :

A single engine light plane with ,

Specific fuel consumption 0.42lb/hr/hp.

Fuel capacity =44 gal.

Gross weight =3400 lb.

To find :

Range and Endurance of the plane.

Solution:

Consider  all standard measures of standard single engine propeller plane

as

Wing span =35.8 fts.

Wing swing area=174 sq ft

parasite drag coefficient  =Cd.o.=0.025

Oswald's eff. factor= 0.8

ρ=0.002377= corresponds to standard sea level constant.

Now

Formula for Range is given by, Breguent formula.

R=(η/c)  *(Cl/Cd)*ln(W1/W0)

here η is Oswald's constant,

Now calculating lift(Cl) and drag coefficient (Cd)

Cl=W/(1/2*ρ*v^2*S)

W=Gross weight

ρ=0.002377

Assume v=200 ft/sec normally,

S=174 Sq .ft.

CI=3400/(1/2*0.002377*200*200*174)

=6800/16543.9

=0.4110

Now calculating drag constant,

AR=(wing span)^2/wing swing area

=(35.8)^2/174

=7.37

Now

Drag Coefficient

Cd=Cd.o.+ (Cl^2)/(pie*e*AR)

=0.025+(0.4110)^2/(3.142*0.8*7.36)

=0.0342

Given that 44 gal fuel capacity and in Aviation weight of fuel is 5.64 lb/gal

hence weight of fuel=W1=3400- (44*5.64)

=3151.84

Now

for specific fuel consumption=0.42  lb/hp/hr

=0.42  lb*(1/550 ft)*(1/3600)sec

=2.12 *10^-7 lb/ft/sec

Now further calculating range

R=(η/c)  *(Cl/Cd)*ln(W1/W0)

={0.8/(2.12*10^-7)}*(0.4110/0.0342)*ln(3151.84/3400)

=0.024908/0.072504

=0.34354*10^7

=3.4353 *10^6 fts.

1mi =5280 ft

=(3.4353/5280)*10^6

=650.644 miles

Now

For Endurance

E=(η/c)*{(Cl^3/2)/Cd}*(2*ρ*S)^1/2*[1/(W1)^1/2  -1/(W0)^1/2].

=(0.8/2.12*10^-7)*{(0.4110^3/2)/0.0342}*(2*0.002377*174)^1/2*[1/(3151.84)^1/2  -1/(3400)^1/2]

=3.7735*10^6*7.7043*0.8272*0.0006629

=0.01927*10^6

=1.927*10^4 sec

here 1hr =3600 sec

E=(1.927/3600)*10^4

=5.3528 hrs

7 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is
Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α  =  \frac{1}{\pi} (\frac{K}{\sigma Y})^2     ....................1

put here value we get

α  =  \frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2

α  = 15.043 mm

so minimum length of a surface crack is 15.043 mm

7 0
3 years ago
The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pu
marishachu [46]

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

3 0
3 years ago
The underground storage of a gas station has leaked gasoline into the ground. Among the constituents of gasoline are benzene, wi
vovangra [49]

Answer:

a) benzene = 910 days

b) toluene = 1612.67 days

Explanation:

Given:

Kd = 1.8 L/kg (benzene)

Kd = 3.3 L/kg (toluene)

psolid = solids density = 2.6 kg/L

K = 2.9x10⁻⁵m/s

pores = n = 0.37

water table = 0.4 m

ground water = 15 m

u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s

a) For benzene:

R=1+\frac{\rho * K_{d}  }{n}, \rho = 2.6\\ R=1+\frac{2.6*1.8}{0.37} =13.65

The time will take will be:

t=\frac{xR}{a} , x=12,a=0.18\\t=\frac{12*13.65}{0.18} =910days

b) For toluene:

R=1+\frac{2.6*3.3  }{0.37} = 24.19

t=\frac{12*24.19}{0.18} =1612.67days

6 0
2 years ago
Read 2 more answers
Other questions:
  • What is a construction worker with limited skills called?
    12·1 answer
  • You read a research study that concludes that the higher a student's self-esteem, the better he performs in school. This sort of
    5·1 answer
  • A evolução da malha rodoviária do Brasil é um marco notável
    9·1 answer
  • In Visual Basic/Visual Studio, characteristics of controls, such as the Name of the control, or the Text displayed on the contro
    10·1 answer
  • Description: Write a function that takes in a list of numbers and a list of indices. Note that indexList may not only contain va
    8·1 answer
  • A spherical seed of 1 cm diameter is buried at a depth of 1 cm inside soil (thermal conductivity of 1 Wm-1K-1) in a sufficiently
    14·1 answer
  • Consider a section of muscle tissue of a cylindrical shape with a radius of 1.5 cm. During highly rigorous exercise, metabolic p
    8·1 answer
  • The technique of smoothing out joint compound on either side of a joint is known as which of the following
    14·1 answer
  • 4 main causes of erosion
    12·1 answer
  • A building permit allows a builder to?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!