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Crank
3 years ago
12

Advocates of small government complain that two proposed government programs will steal the market from small businesses: One pr

ogram will foster a major increase in the government production of bread; the second will increase the government production of manicures. If the demand for bread is highly inelastic, and the demand for manicures is highly elastic, which program should worry small-government proponents the most
Engineering
1 answer:
VashaNatasha [74]3 years ago
8 0

Answer:

Explanation:

If the demand elasticity of bread is highly inelastic, it means the effect of change in price on demand of bread will be insignificant. Hence if a program fosters major increase in government production of bread, the prices of bread may get down. As law of demand states with lower price demand will be higher of government made bread. But as the demand is inelastic, there will be no significant effect on demand of bread of small business producing bread.

On other hand, If the demand elasticity of manicure is highly elastic, it means the effect of change in price on demand of manicure will be significant. Hence if a program that will increase in government production of manicures, the prices may get low. With lower price demand will be higher of government made manicure. And as the demand is highly elastic, there will be severe significant effect on demand of manicure of small business producing manicure.

Thus the second program worries small government proponents the most.

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Gre4nikov [31]

Answer:

It’s so we can then understand other cultures and other ways that people do things or we can also be prepared for what a country that we have not been to yet is going to relatively be like. This can also help with vacation planning.

Explanation:

it can also help us understand different people from different cultures.

3 0
1 year ago
How would you use an inclined plane to increase speed?​
Elenna [48]

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i would throw it in a 90 degree angle to gain speed

Explanation:

8 0
2 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
15
AveGali [126]

Answer:

b

Explanation:

4 0
2 years ago
A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the
raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

E = G \times H

Where G represents complex rated power and H is the inertia constant of turbo-generator.

E = 100 \times 8 \\\\E = 800 \: MJ

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$ P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}  $

Where M is given by

$ M = \frac{E}{180 \times f} $

$ M = \frac{800}{180 \times 50} $

M = 0.0889 \: MJ \cdot s/ elec \: \: deg

So, the rotor acceleration is

$ P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$  30 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$   \frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}  $

$   \frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2 $

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$ \Delta  \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2 $

Where t is given by

1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2  \\\\t = 0.2 \: sec

So,

$ \Delta  \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2 $

$ \Delta  \delta = 6.75 \: elec \: deg

The change in torque in rpm/s is given by

$ \Delta  \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ  }   $

$ \Delta  \delta =28.12 \: \: rpm/s $

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$ Rotor \: speed = \frac{120 \cdot f}{P}  + (\Delta  \delta)\cdot t  $

Where P is the number of poles of the turbo-generator.

$ Rotor \: speed = \frac{120 \cdot 50}{4}  + (28.12)\cdot 0.2  $

$ Rotor \: speed = 1500  + 5.62  $

$ Rotor \: speed = 1505.62 \:\: rpm

4 0
3 years ago
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