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3 years ago

At a 4 percent annual growth rate in GDP per capita, it will take

Engineering

1 answer:

allochka39001 [22]3 years ago

8 0

Answer:

At a 4 percent annual growth rate in GDP per capita, it will take 18 years for GDP per capita to double.

Explanation:

GDP stands for the gross domestic products. It is the basis to monitor the value of good and service in the market.
According to rule of 70:

Number of years to double = $\frac{70}{annual growth rate}$

= $\frac{70}{4}$

≈ 18 years

In this way in total of 18 years the GDP will be double at the constant annual growth rate of 4%.

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Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass

snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0

3 years ago

Use the results of Prob. 5–82 for plane strain near the tip with u 5 0 and n 5 13. If the yieldstrength of the plate is Sy, what

rosijanka [135]

Answer:

Kindly follow the steps as shown below.

Explanation:

8 0

3 years ago

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olchik [2.2K]

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7 0

3 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10

Fiesta28 [93]

Answer:

Recall the formula for the maximum stress, σₐ = 2σ₀ *√ (α/ρₓ)

where

σ₀ = tensile stress = 140 MPa = 1.40x 10⁸Pa

α = crack length = 3.8 × 10–2 mm = 3.8 x 10⁻⁵m

ρₓ = radius of curvature = 1.9 × 10⁻⁴mm = 1.9 × 10⁻⁷m

Substituting these values into the formula, we can calculate the max stress as

====== 2 x 1.40x 10⁸ x √(3.8 x 10⁻⁵/1.9 × 10⁻⁷)

σₐ = 24.4MPa

6 0

3 years ago

An electric dipole is made of two charges of equal magnitudes and opposite signs. The positive charge, q=1.0 μC, is located at t

vfiekz [6]

Answer:

work done by electric field is 0.06 J

Explanation:

Given data:

Two point charge is $+ 1\mu C and -1 \mu C$

0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)

-1 charge positioned is (0 cm , -1 cm, 0.00 cm)

E = 3.0\times 10^6 N/C

From above information, the distance between given two charges d = 2 cm

then d = 0.02m

work needed is W = q E d

$W = 1.0 \times 10^{-6} \times 3.0 \times 10^6 \times 0.02$

W = 0.06 J

Therefore work done by electric field is 0.06 J

8 0

3 years ago