All categories

2 years ago

Which biome contains mostly coniferous trees and receives 35 to 100 cm of rain per year?

Physics

1 answer:

a_sh-v [17]2 years ago

4 0

The question is incomplete as it does not have the options which are:

deciduous forest

taiga (boreal forest)

temperate rainforest

tropical rainforest

Answer:

Taiga (boreal forest)

Explanation:

A Biome refers to the habitat which is occupied by flora and fauna living in similar conditions. These biomes are distinguished based on many features like precipitation, temperature and many other physical factors.

In the given question, the biome which receives an annual rainfall of 35 to 100 cm annually and is mostly covered by the coniferous trees is known as "Taiga biome" which is also known as Boreal forest.

The Taiga biome is one of the largest terrestrial biomes which is present in Eurasia and North America. The biome is characterised by the conifers trees and therefore is also known as the Coniferous trees.

Thus, Taiga (boreal forest) is the correct answer.

You might be interested in

Pamela drove her car 999999 kilometers and used 999 liters of fuel. she wants to know how many kilometers (k)(k)left parenthesis

Vanyuwa [196]

When the relationship between two variables are said to be proportional, it means that one variable is a constant multiple of the other variable. They are related by a constant of proportionality, usually denoted as k.

In this problem, the dependent variable is the distance in kilometers. Your mileage is limited with the amount of fuel you have. Thus, the independent variable is the liters of fuel. When these two are proportional, it could be expressed as

distance = k * liters of fuel, such that
distance/liters of fuel = k

By variation,

distance,1/liters of fuel,1 = distance,2/liters of fuel,2, where 1 denotes situation 1 and 2 denotes situation 2. Therefore,

999999 km /999 liters = x km /121212 liters, where x is the unknown distance. We can now therefore find the value of x.

x = (999999*121212)/999
x = 121333212 kilometers

3 0

1 year ago

Two parallel co-axial disks are floating in deep space (far from sun and planets). Each disk is 1 meter in diameter and the disk

HACTEHA [7]

Answer:

T₂ = 5646 K

Explanation:

Let's start by finding the power received by the first disc, for this we use Stefan's law

P = σ. A e T⁴

Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of the disk, T the absolute temperature and e the emissivity that for a black body is 1

The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is

I = P / A

Writing this expression for both discs

I₁ A₁ = I₂ A₂

I₂ = I₁ A₁ / A₂

The area of a sphere is

A = 4π r²

I₂ = I₁ (r₁ / r₂)²

r₂ = r₁ ± 5

I₁ = I₂ ( (r₁ ± 5)/r₁)²

Let's write the Stefan equation

P / A = σ e T⁴

I = σ e T⁴

This is the intensity that affects the disk, substitute in the intensity equation

σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²

The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white

e₁ T₁⁴ = T₂⁴ (r1 + 5)²/r₁²

T₁ = T₂ {(e₂/e₁)}^{1/4} √(1 ± 1/ r₁)

If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term

(1 ±x) n = 1 ± n x

Where x = 5 / r₁ << 1

We replace

T₁ = T₂ {(e₂/e₁)}^{1/4} (1 ± ½ 5/r1)

T₁ = T₂ {(e₂)}^{1/4} (1 ± 5/2 1/r1)

If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach

T₁ = T₂ {(e₂)}^{1/4} ¼

T₁ = T₂ 0.9 $0.9^{1/4}$

5500 = T₂ 0.974

T₂ = 5646 K

3 0

1 year ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

Dvinal [7]

Nobody will do that for 5 points loll

8 0

2 years ago

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -

Svetllana [295]

Answer:

The amount of charge the space shuttle collects is -1.224nC

Explanation:

The magnitude of Electric potential is given as;

V = kq/r

where;

V is the electric potential in volts

k is coulomb's constant

r is the radius of the sphere or distance moved by the charge

given; V = -1.1 V, k = 8.99 x 10⁹ Nm²/C², r = 10m

Substituting this values in the above equation, we estimate the amount of charge space shuttle collects.

q = (V*r)/k

q = (-1.1 *10)/(8.99 x 10⁹ )

q = -1.224 X 10⁻⁹ C

q = -1.224nC

Therefore, the amount of charge the space shuttle collects is -1.224nC

8 0

2 years ago

A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.

Aleonysh [2.5K]

Answer: $53.31\°$ East of North

Explanation:

We have the following data:

Speed of the wind from East to West: $6.68 m/s$

Speed of the bee relative to the air: $8.33 m/s$

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the Speed of the wind from East to West (in the horintal part) and the speed due North relative to the ground (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

$sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}$

$sin \theta=\frac{6.68 m/s}{8.33 m/s}$

Clearing $\theta$ :

$\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})$

$\theta=53.31\°$

6 0

2 years ago