All categories

3 years ago

Altitudinal zonation is best described as _______

Physics

1 answer:

IgorC [24]3 years ago

6 0

Answer:

Explanation:

Attitudinal zonation is best described as the natural layering of ecosystems.

Factors that determine Altitudinal zonation

1. Temperature

2. humidity

3. soil composition

4. solar radiation

Attitudinal zonation supports different vegetation and animal species.

Altitudinal zonation occurs in marine environment, shorelines and in wetlands

It is observed that the changes in vegetation and animal species of a particular altitudinal zone increase latitudes in the concept of of life zones.

Nowadays, altitudinal zonation represents a core concept in mountain research.

You might be interested in

Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of

alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is 29.89 mi/h which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, v, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

$\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx 87.084 \ ft./s$

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈ 87.084 ft./s - 0 = 87.084 ft./s

≈ 87.084 ft./s

$v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h$

The speed of the person at the end of the impact, v₂ ≈ 59.38 mi/h

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

$v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}$

$v = \dfrac{40,000 \times 30 + 150 \times 0}{40,000 + 150} \approx 29.89$

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, the findings does not agree with the expectation

Learn more here:

brainly.com/question/18326789

3 0

3 years ago

This is my question. I need the answer stat.

Kipish [7]

Answer:

Explanatioyour answers look right, but if there has , has to be another answer its a , but your answers are right

5 0

3 years ago

Read 2 more answers

Consider the numbers 23.68 and 4.12. The sum of these numbers has ____ significant figures, and the product of these numbers has

damaskus [11]

Answer:multiplying will give us 7 significant figures and addition will give us 3 significant figures

Explanation:

After multiplying the two numbers they resulting value will give a value in its 4 decimal places because both given values are in 2 decimal places. The 4 dp is gotten by the addition of the decimal places of both given numbers (2+2) and

The result of its addition will give us a value in its 1dp and 3 significant figures since the addition of 23.68 and 4.12 will give us 27.8

7 0

3 years ago

What provides the centripetal force needed to keep Earth in orbit?

kkurt [141]

The centripetal force needed to keep earth in orbit is gravity.

3 0

3 years ago

Read 2 more answers

For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e

Sati [7]

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

5 0

4 years ago