A heavy brick weighs more than a fluffy cushion, but the cushion takes up more space. Which object has more matter?

Assume that a firm has the following marginal benefit of pollution (denoted E for emissions, measured in tons): MB=150-5 E e. No

jeka57 [31]

Answer:

Explanation:

E)cost of pollution is reduction in benefit of the firms.

MB=150-5E,. MB=90-3E

E=30-MB/5. E=30-MB/3

Joint MB,

E=60-8MB/15

MB=112.5-1.875E

Total pollution reduction=24

Total pollution=60-24=36

MB=112.5-1.875*36=112.5-67.5=45

Firm 1

MB=150-5E.

45=150-5E.

E=-105/-5=21

Reduction=30-21=9

Firm 2,

MB=90-3E

45=90-3E

E=-45/-3=15

Reduction =30-15=15

So firm 2 is reducing 15 units of pollution and firm 1 reducing 9 units of pollution.

As each firm have to reduce 12 units of pollution but firm 2 reducing 3 units more while firm 1 reducing 3 less units.

So, firm 2 is selling 3 units of pollution emission permit to firm 1.

F)firm 1 reduce 9 units and firm 2 reduce 12 units of pollution after trade.

Total cost of pollution reduction will total Benefit reduction by pollution reduction.

MB=112.5 -1.875E

TB=112.5E-0.9375E^2

TB at E=60

TB=112.5*60-0.9375*60*60=6750-3375=3375

TB at E=36

TB=112.5*36-0.9375*36*36=4050-1215=2836

Total cost of pollution reduction=3375-2836=540

G)price of permit= cost of extra pollution reduction by firm 2 or total cost from 9 to 12 by firm 2

MB=90-3E

TB=90E- 1.5E^2

TB at E=18

TB=90*18 -1.5*18*18=1620-486=1134

TB at E=15

TB=90*15 -1.5*15*15=1350-337.5=1012.5

Permit price=1134-1012.5=121.5

Total cost to firm 2 =1134

Net total cost to firm 2=1134-121.5=1012.5

Total cost to firm 1=150E-2.5E^2=150*9-2.5*9*9=1350-202.5=1147.5

Net total cost=1147.5+121.5=1269

H) the total cost is lower in cap & trade policy is because the firm who has higher cost of pollution reduction is paying the other firm who has lower cost of pollution reduction to reduce more pollution ,so that his part of pollution reduction can be completed.

And the amount he is paying is equal to the amount that is cost of other firm of reducing additional pollution units.

So the cost the firm is lower as he is paying lower amount than if he reduce pollution by itself.

5 0

3 years ago

in heating a kettle of water on an electric stove, 3.34×10^3 J of thermal energy was provided by the element of the stove. yet,

insens350 [35]

Answer:

The percentage efficiency of the electrical element is approximately 82.186%

Explanation:

The given parameters are;

The thermal energy provided by the stove element, $H_{supplied}$ = 3.34 × 10³ J

The amount thermal energy gained by the kettle, $H_{absorbed}$ = 5.95 × 10² J

The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

$\eta \% = \dfrac{H_{supplied} - H_{absorbed} }{H_{supplied}} \times 100$

Therefore, we get;

$\eta \% = \dfrac{3.34 \times 10^3 - 5.95 \times 10^2}{3.34 \times 10^3} \times 100 = \dfrac{549}{668} \times 100 \approx 82.186 \%$

The percentage efficiency of the electrical element, η% ≈ 82.186%.

4 0

3 years ago

Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have

Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0

3 years ago

What is the relationship between the ridge<br> and the age of sediment?<br> Plzzz helppp:(

leva [86]

Answer:

The sediments accumulating on and around mid-ocean ridges are mostly formed from the calcareous and siliceous tests of pelagic organisms. This research is concerned with understanding how the rate of sediment supply varies from place to place due to varied productivity of pelagic organisms, how the sediments accumulate on the complex topography of a mid-ocean ridge, and with using the sediments to study mid-ocean ridge processes such as faulting and volcanism.

Sediment transport and accumulation

When pelagic materials reach the seafloor, they are redistributed by bottom currents and by sedimentary flows. This work studied the form of the accumulation using sediment profiler records collected with a Deep Tow system from the Scripps Institution of Oceanography deployed over the Mid-Atlantic Ridge in the early 1970s. The records showed that both sets of transport processes are important. The shapes of deposits were studied to see to what extent they conform to the diffusion transport model - many deposits have parabolic surfaces, which are the steady state forms expected from the diffusion transport model under boundary conditions of constant input or output flux to basins.

8 0

3 years ago

A balloon has a volume of 1.75L at a temperature of 298K. What will be the volume of the balloon if you take it out into the win

lozanna [386]

Answer:

1.52 L

Explanation:

P1V1T2=V2P2T1

V2=V1T2/T1

Fill in with given values then solve

We can check this by knowing that V and T at constant P have a proportional relationship. Hence, this is correct.

- Hope that helped! Please let me know if you need further explanation.

7 0

2 years ago