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stepladder [879]
3 years ago
7

python Write a program that takes a date as input and outputs the date's season. The input is a string to represent the month an

d an int to represent the day.
Engineering
1 answer:
kupik [55]3 years ago
4 0

Answer:

month = input("Input the month (e.g. January, February etc.): ")

day = int(input("Input the day: "))

if month in ('January', 'February', 'March'):

season = 'winter'

elif month in ('April', 'May', 'June'):

season = 'spring'

elif month in ('July', 'August', 'September'):

season = 'summer'

else:

season = 'autumn'

if (month == 'March') and (day > 19):

season = 'spring'

elif (month == 'June') and (day > 20):

season = 'summer'

elif (month == 'September') and (day > 21):

season = 'autumn'

elif (month == 'December') and (day > 20):

season = 'winter'

print("Season is",season)

Explanation:

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Technician A says when tightening lug or wheel nuts a torque wrench must be used. Technician B says that an air wrench can be us
ASHA 777 [7]
A, is your answer hope it helps
4 0
3 years ago
An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh
Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

    (ii) maximum bending stress in compression =  0.7413*10^6 Ib-in

B) (i)  The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

    (ii)  Average shear stress at the web = 18.289 * 10^5 psi

    (iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= \frac{M_{mm}Y }{I}  = \frac{6.48 * 10^6 * 2.375}{53.54} =  0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= \frac{M_{mm}Y }{I} = \frac{6.48 *10^6*(8.5-2.375)}{53.54} = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = \frac{wL}{2} = \frac{1000*6*12}{2} = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - \frac{0.5}{2} ) * \frac{(2.375 - (\frac{0.5}{2} ))}{2}

= 5.878 in^3

t = VQ / Ib  = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 )  psi

  = 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib = \frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}

= 1.143 *10^5

4 0
4 years ago
A particular cloud-to-ground lightning strike lasts 500 µµsec and delivers 30 kA across a potential difference of 100 MV. Assu
sasho [114]

Answer:

a) 15 C charge was delivered by the lightening bolt

b) the lightning delivered 3.0 × 10¹² W of power

c)

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is 416666.67 Wh

d)

the residential retail value of the energy delivered by the strike is $ 40.83

e)

a total of 26 lightening strikes would be required to power an average US home for a year.

Explanation:

Given that;

the lighting strike lasted for t ( time ) = 500 μsecs = 500×10⁻⁶ s

Current I = 30 kA = 30×10³ A

voltage V = 100 mV = 100×10⁶ v

a)

we know that; I = Q/t

so Q = I × t

we substitute

Q =  30×10³ × 500×10⁻⁶

Q = 15 C

Therefore 15 C charge was delivered by the lightening bolt

b)

Power P = V × I

we substitute

Power P = 100×10⁶ × 30×10³

P = 3.0 × 10¹² W

Therefore, the lightning delivered 3.0 × 10¹² W of power

c)

we know that; Power = Energy / Time

Energy = Power × Time

we substitute

E = 3.0 × 10¹²  × 500×10⁻⁶

E = 1500 × 10⁶ J

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is;

⇒ 1500×10⁶ / 3600 Wh

= 416666.67 Wh

d)

given that;  1 KWh → $ 0.098

energy delivered by the strike = 416666.67 Wh = 416.66667 KWh

so the residential retail value of the energy delivered by the strike will be;

416.66667 KWh × $ 0.098

= $ 40.83

∴ the residential retail value of the energy delivered by the strike is $ 40.83

e)

Given that; average monthly residential energy consumption is 900 kWh.

for a year; energy consumption = 12 × 900 kWh = 10,800 KWh = 10800000 Wh

Now

1 lightening strike ⇒ 416666.67 Wh

x lightening strike ⇒ 10800000 Wh

x = 10800000 / 416666.67

x = 25.9199 ≈ 26

Therefore; a total of 26 lightening strikes would be required to power an average US home for a year.  

7 0
3 years ago
Water is being heated in a vertical piston-cylinder device. The piston has a mass of51 kg and a cross-sectional area of 100 cm2.
djverab [1.8K]

Answer:

The temperature at which the water starts boiling is 111.35 °C.

Explanation:

First, we have to determine the total pressure experimented by water inside the piston-cylinder device. The total pressure is the sum of atmospheric and gauge pressures, the gauge pressure is the weight of piston divided by its cross-sectional area, that is:

P_{g} = \frac{m_{p}\cdot g}{A}

Where:

P_{g} - Gauge pressure, measured in pascals.

m_{p} - Mass of the piston, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

A - Cross-sectional area, measured in square meters.

P_{g} = \frac{(51\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(100\,cm^{2})\cdot \left(\frac{1\,m^{2}}{10000\,m^{2}} \right)}

P_{g} = 50015.7\,Pa

P_{g} = 50.016\,kPa

The total pressure of water is:

P_{t} = 100\,kPa + 50.016\,kPa

P_{t} = 150.016\,kPa

Temperature at which water starts boiling is the saturation temperature associated with total pressure, which can be obtained from property tables:

T_{sat} = 111.35\,^{\circ}C

The temperature at which the water starts boiling is 111.35 °C.

6 0
3 years ago
Define the stress and strength? A material has yield strength 100 kpsi. A cantilever beam has length 10 in and a load of 100 Lbf
Firlakuza [10]

Answer:

Stress is a force that acts on a unit area of a material. The strength of a material is how much stress it can bear without permanently deforming or breaking.

Is the beam design acceptable for a SF of 2? YES

Explanation:

Your factor of safety is 2, this means your stress allowed is:

  • σall = YS/FS = 100kpsi/2 = 50kpsi

Where:

  • σall => Stress allowed
  • YS => Yield Strength
  • FS => Factor of safety

Now we are going to calculate the shear stress and bending stresses of the proposed scenario. If the calculated stresses are less than the allowed stress, that means the design is adequate for a factor of safety of 2.

First off we calculate the reaction force on your beam. And for this you do sum of forces in the Y direction and equal to 0 because your system is in equilibrium:

  1. ΣFy = 0
  2. -100 + Ry = 0     thus,
  3. Ry = 100 lbf

Knowing this reaction force you can already calculate the shear stress on the cantilever beam:

  1. τ = F/A
  2. τ = 100lbf/(2in*5in)
  3. τ = 10 psi

Now, you do a sum of moments at the fixed end of your cantilever beam, so you can cancel off any bending moment associated with the reaction forces on the fixed end, and again equal to 0 because your system is in equilibrium.

  1. ΣM = 0
  2. -100lbf*10in + M = 0
  3. M = 1000 lbf-in

Knowing the maximum bending moment you can now calculate your bending stress as follows:

  • σ = M*c/Ix

Where:

  • σ => Bending Stress
  • M => Bending Moment
  • c => Distance from the centroid of your beam geometry to the outermost fiber.
  • Ix => Second moment area of inertia

Out of the 3 values needed, we already know M. But we still need to figure out c and Ix. Getting c is very straight forward, since you have a rectangle with base (b) 2 and height (h) 5, you know the centroid is right at the center of the rectangle, meaning that the distance from the centroid to the outermost fibre would be 5in/2=2.5in

To calculate the moment of Inertia, you need to use the formula for the second moment of Inertia of a rectangle and knowing that you will use Ix since you are bending over the x axis:

  • Ix = (b*h^3)/12 = (2in*5in^3)/12 = 20.83 in4

Now you can use this numbers in your bending stress formula:

  1. σ = M*c/Ix
  2. σ = 1000 lbf-in * 2.5in / 20.83 in4
  3. σ = 120 psi

The shear stress is 10psi and the bending stress is 120psi, this means you are way below the stress allowed which is 50,000 psi, thus the beam design is acceptable. You could actually use a different geometry to optimize your design.

4 0
3 years ago
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