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natima [27]
2 years ago
13

Which statements describe the relationship between the area and the biodiversity of an ecosystem? Check all that apply. Large ec

osystems always have higher biodiversity than smaller ecosystems. A large area of a forest will likely have higher biodiversity than a smaller area of the same forest. A half acre of rainforest would likely have greater biodiversity than a full acre of desert. Small ecosystems always have low biodiversity. The same area of two different types of ecosystems will have the same biodiversity.
Physics
2 answers:
Anastaziya [24]2 years ago
9 0

Answer:

The correct answers are B. A large area of a forest will likely have higher biodiversity than a smaller area of the same forest.

C. A half acre of rainforest would likely have greater biodiversity than a full acre of desert.

Explanation:

Took the Test Fam

Guest
1 year ago
bet
Guest
1 year ago
thx
snow_tiger [21]2 years ago
3 0

Answer:

A large area of a forest will likely have higher biodiversity than a smaller area of the same forest.

A half acre of rainforest would likely have greater biodiversity than a full acre of desert.

Explanation:

biodiversity of an ecosystem can be simply explain as the variability that exist in the ecosystem. These are difference that exist between living organism in terms of their habitat, their species and so on. However there exist different relationship between the area of ecosystem and the biodiversity of an ecosystem such that An ecosystem with large area of a forest will likely have higher biodiversity than a smaller area of the same forest.

Another one is that A half acre of rainforest would likely have greater biodiversity than a full acre of desert.

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A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh
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Answer:

heat used to rise temperature pan =  30.1%

heat used to rise temperature water =  69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature =  80.0ºC

to find out

heat used to rise temperature of  pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature    ...........1

specific heat of water is 4186 J/kgºC

so

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature    ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

so

total heat = 62.8 + 27 = 89.8 kJ

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heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

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2 years ago
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Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
kobusy [5.1K]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere  ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon  }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Q₁ = 572.8 * 10^{-9} C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2}  = \frac{r}{R} *Q_{1}

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Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  =0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

8 0
2 years ago
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