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san4es73 [151]
2 years ago
14

Consider two adjacent states, S1 and S2, that wish to control particulate emissions from power plants and cement plants; New Jer

sey and Pennsylvania, for example. Each state wants to remove all particulates larger than 100 nanometres from the airspace below 1000 feet. Suppose this requires 50 kilotons of abatement in each state. The aggregate cost of abatement when the marginal abatement costs of all industries within the state are equalized is 3A21 for state S1 and it is A2 for state S2, where Ai is measured in kilotons.(a) If the states do not cooperate, what is the cost of abatement in state S1?(b) Would a permit market with 50 kilotons of emissions permits distributed to the firms by means of an ascending clock auction equate the marginal abatement costs of all power and cement firms within state S1? You may assume that no firms are bankrupted.(c) If the states do not cooperate, what is the cost of abatement in state S2?(d) If the states do cooperate by means of a 100 kiloton cap and a permit market that covers both states with permits allocated by an ascending clock auction, what is the total cost of abatement? You may assume that no firms are bankrupted. You may also assume that a joint auction will not change the aggregate cost functions 3A21 for state S1 and A2 for state S2.(e) Why does the sum of the costs of abatement from question 1a and 1c not equal the cost of abatement in question 1d?

Physics
1 answer:
natka813 [3]2 years ago
7 0

Answer:

a. 7500

b. Yes

c. 2500

d. 7500

Explanation:

Please see attachment

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Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th
DENIUS [597]

Answer:

a) vboat = 5.95 m/s  b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As  we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w =  \frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

3 0
2 years ago
Certain meteorites have been examined and found to carry samples of which molecules?
ki77a [65]

Answer:

Sugars...

Explanation:

Several meteorites have been found to carry molecules of sugars that are essential for life. These sugars include Ribose, Arabinose and Xylose. These are found in meteorites that are rich in carbon. These significant discoveries can pave way in finding the origin of life on Earth.

6 0
2 years ago
Diesel engines give more miles per gallon than gasoline engines, but some of this is due to the higher energy content of diesel
Lina20 [59]

Answer:

0.106

Explanation:

For 1 liter of diesel the car can get 19 km, if it takes 0.2 MJ for each km then it would take the total energy of 19*0.2 = 3.8 MJ to move an aerodynamic car 19 km. Since 1 liter of of diesel also contains 36 MJ in internal energy, then the efficiency of the diesel engine is the ratio of its output energy over its input energy:

\frac{3.8}{36} = 0.106

5 0
2 years ago
A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin
KengaRu [80]

a) 219.8 rad/s

b) 20.0 rad/s^2

c) 2.9 m/s^2

d) 7005 m/s^2

e) Towards the axis of rotation

f) 0 m/s^2

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

\omega=\frac{\theta}{t}

where

\theta is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

\omega_{max}=35 rev/s

The angle covered during 1 revolution is

\theta=2\pi rad

Therefore, the maximum angular velocity is:

\omega_{max}=35 \cdot 2\pi = 219.8 rad/s

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

\alpha = \frac{\Delta \omega}{t}

where

\Delta \omega is the change in angular velocity

t is the time elapsed

Here we have:

\omega_0 = 0 is the initial angular velocity

\omega_{max}=219.8 rad/s is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

\alpha = \frac{219.8-0}{11}=20.0 rad/s^2

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the radius of the circle

Here we have

\alpha =20.0 rad/s^2

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

a_t=(20.0)(0.145)=2.9 m/s^2

d)

The magnitude of the radial (centripetal) acceleration is given by

a_c = \omega^2 r

where

\omega is the angular velocity

r is the radius of the circle

Here we have:

\omega_{max}=219.8 rad/s is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

a_c=(219.8)^2(0.145)=7005 m/s^2

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

\alpha=0, since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

a_t=(0)(0.145)=0 m/s^2

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

v=\omega r

where

v is the linear speed

\omega is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

\omega=219.8 rad/s

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

v=(219.8)(0.145)=31.9 m/s

7 0
2 years ago
A transmission channel is made up of three sections. The first section introduces a loss of 16dB, the second an amplification (o
AlekseyPX

Answer:

P_{out} = 0.100 W = 100 mW

Explanation:

The attached image shows the system expressed in the question.

We can define an expression for the system.

The equivalent equation for the system would be

G_{total} = G_{1} + G_{2} + G_{3}\\G_{total} = -16dB+20dB-10 dB = -6 dB

so, the input signal could be expressed in dB terms

P_{in} [dB] = 10 log_{10}(P_{in}) \\P_{in} [dB] = 10 log_{10}(0.4)\\P_{in} [dB] = -3.97 dB (1)

so the output signal could be expressed as.

P_{out} = P_{in} + G_{1} + G_{2} + G_{3}\\P_{out} = -3.97 dB - 6dB = -9.97 dB

The gain should be expressed in dB terms and power in dBm terms so

P_{out} = -9.97 + 30 = 20.03 dBm

using the (1) equation to find it in terms of Watts

P_{out} = 0.100 W = 100 mW

3 0
2 years ago
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