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Kamila [148]
3 years ago
5

The Air Force One (Boeing 747-200) has a long-range mission takeoff gross load 833,000 pounds (Ibm). What is Air Force One's tak

eoff mass in: a. gram (9) b. kilogram (kg) c. tonne (ton) d. Mton
Engineering
1 answer:
Ahat [919]3 years ago
4 0

Answer:

The mass in:

1) Grams =377765.5\times 10^{3}grams

2) Kilograms =377765.5kilograms

3)Tonnes =377.7655tonnes

4) Megatonnes =0.378megatonnes

Explanation: The given mass of the aircraft in pounds is 833,000 pounds.

Part 1)

Since we know that 1 pound equals 453.5 grams thus by ratio we have

833,000 pounds =833000lb\times 453.5\frac{g}{lb}=377765.5\times 10^{3}grams

Part 2)

Since we know that 1000 grams equals 1 kilogram

thus the above mass in kilograms equals\frac{377765.5\times 10^{3}}{1000}=377765.5kg

Part 3)

Since there are 1000 kilograms in 1 tonne

thus the given mass is converted into tonnes as

mass_{tonnes}=\frac{377765.5kg}{1000}=377.765tonnes

Part 4)

Since 1 Mega tonne(Mton) equals 1000 tonnes thus the given mass is converted into mega tonnes as

mass_{M\cdot tonn}=\frac{377.7655tonnes}{1000}=0.378Megatonnes

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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
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Rashid [163]

Answer:

Exit temperature = 32 °C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

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6 0
3 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

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3 years ago
Suppose there are n chairs in a row. We want to compute the number of ways to put 2 students into seats so that they are not nex
icang [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) f_{(n)} = f_{(n-1)} + n-2

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8 0
3 years ago
The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb>ft. Determine t
irina1246 [14]

Answer:

M = 281.25 lb*ft

Explanation:

Given

W<em>man</em> = 150 lb

Weight per linear foot of the boat: q = 3 lb/ft

L = 15.00 m

M<em>max</em> = ?

Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):

∑ Fy = 0  (+↑)     ⇒    q'*L - W - q*L = 0

⇒       q' = (W + q*L) / L

⇒       q' = (150 lb + 3 lb/ft*15 ft) / 15 ft

⇒       q' = 13 lb/ft   (+↑)

The free body diagram of the boat is shown in the pic.

Then, we apply the following equation

q(x) = (13 - 3) = 10   (+↑)

V(x) = ∫q(x) dx = ∫10 dx = 10x   (0 ≤ x ≤ 7.5)

M(x) = ∫10x dx = 5x²  (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft

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M(7.5) = 5(7.5)² = 281.25 lb*ft

8 0
3 years ago
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