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dimaraw [331]
3 years ago
6

The total number of species in an area is termed as​

Engineering
2 answers:
lukranit [14]3 years ago
8 0

Answer:

biodiversity, also called biological diversity, the variety of life found in a place on Earth or, often, the total variety of life on Earth. A common measure of this variety, called species richness, is the count of species in an area.

Explanation:

frez [133]3 years ago
5 0

Answer:

biodiversity, also called biological diversity, the variety of life found in a place on Earth or, often, the total variety of life on Earth. A common measure of this variety, called species richness, is the count of species in an area.

#IyotTarzan

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Partnership budiness​
Ne4ueva [31]

Answer:

what does this mean

Explanation:

8 0
3 years ago
an object of mass 2kg is released from a top of inclined plane 30° and height 6m. The coefficient of kinetic friction of the sur
mel-nik [20]

Explanation:

1) Work done = force x distance x cos(θ)

= 0.15 x 6 x cos(30)

= 0.779

2) Ek = ½mv²

v = acceleration due to gravity so 9.81

Ek = ½(2)(9.81)²

Ek = 96.2361

3) v = (√(2em)) / m

= (√(2(96.2361)(2)) / 2

= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.

5 0
2 years ago
¿Qué es la masonería?​
timofeeve [1]

Answer: freemasonry is Being a Mason is about a father helping his son make better decisions; a business leader striving to bring morality to the workplace; a thoughtful man learning to work through tough issues in his life.

Explanation:

3 0
3 years ago
Ohm's law states that the current (I) in amps equals the voltage (E) in volts decided by the resistance (R) in ohm's. If you con
Hitman42 [59]

Answer:

1.14 * 10^-3  amperes

Explanation:

According to ohms law

V = IR   ---- ( 1 )

V = voltage  = 2.4 * 10^3 v

I = current  = ?

R = resistance =  2.1 * 10^6 Ω

from equation 1

I = V / R

 = ( 2.4 * 10^3 ) / (2.1 * 10^6 )

 = 1.14 * 10^-3  amperes

7 0
3 years ago
Consider a power plant with a heat rate of 10,800 kJ/kWh burning bituminous coal with 75 percent carbon and a heating value (ene
grandymaker [24]

Answer:

1. \eta_{p} = 33.34%

2. m_{coal} = 0.396 kg/kWh

3. carbon emission = 0.297 kg/kWh[/tex]

4. carbon-dioxide emission = 1.089 kg/kWh

5. Min flow of cooling water =  = 1457.14 kg/kWh

Given:

Heat Rate = 10,800 kJ/kWh

Heating value = 27,300 kJ/kg

Solution:

In order to calculate the plant efficiency, [rex]\eta_{p}[/tex]

Heat rate in coal fired steam power plant = \frac{3600}{\eta_{p}}

Therefore,

\eta_{p} = \frac{3600}{Heat rate}

\eta_{p} = \frac{3600}{10800}\times 100 = 33.34%

Now,

To calculate mass of coal per kWh, m_{coal}:

m_{coal} = \frac{Required heat rate}{heating value per kg}

m_{coal} = \frac{10800}{27300} = 0.396 kg/kWh

Now,

Rate of emission carbon and carbon-dioxide from the plant:

In accordance to the question, 75% of the bituminous coal being burned is carbon, thus carbon emission is given by:

carbon emission = 0.75\times m_{coal}

carbon emission = 0.75\times 0.396 = 0.297 kg/kWh = 29.7%

Now, for carbon-dioxide emission:

Since, molecular weight of carbon-dioxide = 44 kg

Thus carbon-dioxide emission:

0.297\times \frac{44}{12} = 1.089 kg/kWh

The minimum flow of cooling water per kWh if the allowance in temperature is 10^{\circ}:

\frac{0.85\times \frac{2}{3}\times 10800}{4.2} = 1457.14 kg/kWh

8 0
3 years ago
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