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dexar [7]
3 years ago
12

It has been estimated that 139.2x10^6 m^2 of rainforest is destroyed each day. assume that the initial area of tropical rainfore

st is 20x10^12 m^2.
a) What is the exponential rate of rainforest destruction in units of 1/days?

b) If there were 24.5x10^12 m^2 of tropical rainforest on Earth in 1975, how miuch tropical rainforect would be left on Earth in 2015 if the exponential rate of destruction determined in part (a) stayed constant over this time interval?

c) If tropical rainforests remove 0.83 kg (C)/m^2-year from the atmosphere, how much less carbon [kg(C)] would be removed in 2025 compared to removed in 1975?
Engineering
1 answer:
Dmitry [639]3 years ago
6 0

Answer:

A. 6.96 x 10^-6 /day

B. 22.466 x 10^12 m^2

C. 9.1125 x 10^14 kg of CO2

Explanation:

A. Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Initial area of the rainforest = 20 x 10^12 m^2

Therefore to calculate exponential rate in 1/day,

Rate of rainforest destruction/ initial area of rainforest

= 139.2 x 10^6/20 x 10^12

= 6.96 x 10^-6 /day

B. Rainforest left in 2015 using the rate in A.

2015 - 1975 = 40 years

(40 * 365 )days + 10 days (leap years)

= 14610 days

Area of rainforest in 1975 = 24.5 x 10^12m^2

Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Area of rainforest in 2015 = 14610 * 139.2 x 10^6

= 2.034 x 10^12 m^2

Area left = area of rainforest in 1975 - area of rainforest destroyed in 40years

= 24.5 x 10^12 - 2.034 x 10^12

= 22.466 x 10^12 m^2

C. How much CO2 will be removed in 2025

Recall: Photosynthesis is the process of plants taking in CO2 and water to give glucose and O2.

So CO2 removed is the same as rainforest removed so we use the rate of rainforest removed in a day

Area of rainforest in 1975 = 24.5 x 10^12 m^2

Area of rainforest removed in 2025 = 18262 days * 139.2 x 10^6

= 2.54 x 10^12 m^2

Area of rainforest removed between 1975 - 2025 = 24.5 x 10^12 - 2.54 x 10^12

= 21.958 x 10^12 mC2 of rainforest removed

CO2 = 0.83kg/m^2.year

CO2 removed between 1975 - 2025 = 0.83 * 21.958 x 10^12 * 50 years

= 9.1125 x 10^14 kg of CO2 was removed between 1975 - 2025

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Answer: 3 different types of people using Highway Transportation system are :

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Explanation:

1. Pedestrians - Most of pedestrians use to walk on the footpath along side road that keeps them on a safe side. But there are places without footpath along side road, in that case they have to walk on the road itself. Here, we need to take care for them. We need to wait in case they are crossing road and also check for them while taking a turn.

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4 0
3 years ago
The theoretical maximum specific gravity of a mix at 5.0% binder content is 2.495. Using a binder specific gravity of 1.0, find
PSYCHO15rus [73]

Answer:

The theoretical maximum specific gravity at 6.5% binder content is 2.44.

Explanation:

Given the specific gravity at 5.0 %  binder content 2.495

Therefore

95 % mix + 5 % binder  gives S.G. = 2.495

Where the  binder is S.G. = 1, Therefore

Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495

(95 +5)/(Vx +5) = 2.495

2.495 × (Vx + 5) = 100

Vx =35.08 to 95

Or density of mix = Mx/Vx = 95/35.08 = 2.7081

Therefore when we have 6.5 % binder content, we get

Per 100 mass unit

93.5 Mass unit of Mx has a volume of

Mass/Density = 93.5/2.7081 = 34.526 volume units

Therefore we have

At 6.5 % binder content.

(100 mass unit)/(34.526 + 6.5) = 2.44

The theoretical maximum specific gravity at 6.5% binder content = 2.44.

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3 years ago
What is the purpose of gears?
Brrunno [24]

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6 0
2 years ago
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati
snow_tiger [21]

Answer:

D=0.41m

Explanation:

From the question we are told that:

Discharge rate V_r=0.35 m3/s

Distance d=4km

Elevation of the pumping station h_p= 140 m

Elevation of the Exit point h_e= 150 m

Generally the Steady Flow Energy Equation SFEE is mathematically given by

h_p=h_e+h

With

P_1-P_2

And

V_1=V-2

Therefore

h=140-150

h=10

Generally h is give as

h=\frac{0.5LV^2}{2gD}

h=\frac{8Q^2fL}{\pi^2 gD^5}

Therefore

10=\frac{8Q^2fL}{\pi^2 gD^5}

D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}

D=0.41m

8 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please cal
Sphinxa [80]

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

\frac{d_{2} }{d_{1} } =\frac{1}{2} (\sqrt{1+8F^{2} } -1)

10*2=\sqrt{1+8F^{2} } -1

Solving for F:

F = 7.4162

v=F\sqrt{gd_{1} }

Here, g = gravity = 32.2 ft/s²

v=7.4162*\sqrt{32.2*1} =42.0833ft/s

b) The flow rate:

q=v*L*d_{1} =42.0833*80*1=3366.664ft^{3} /s

c) The Froude number:

v_{2} =\frac{q}{L*d_{2} } =\frac{3366.664}{80*10} =4.2083ft/s

F=\frac{v_{2}}{\sqrt{gd_{2} } } =\frac{4.2083}{\sqrt{32.2*10} } =0.2345

d) The flow energy dissipated:

E=\frac{(d_{2}-d_{1})^{3} }{4d_{1}d_{2}} =\frac{(10-1)^{3} }{4*1*10} =18.225ft

e) The critical depth:

d_{c} =(\frac{(\frac{q}{L})^{2}  }{g} )^{1/3} =(\frac{(\frac{3366.664}{80})^{2}  }{32.2} )^{1/3} =3.8030ft

4 0
3 years ago
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