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3 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]3 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.

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If a rock has a weight of 30 N on Earth, would its weight be more or less if it was on Jupiter (gravity on Jupiter = 25 m/s2)?

goldenfox [79]

I’m sorry this is not going as planned

8 0

3 years ago

Two spherical objects have masses of 3.1 x 10^5 kg and 6.5 x 10^3 kg. The gravitational attraction between them is 65 N. How far

nata0808 [166]

Answer:

4.55 x 10⁹m

Explanation:

Given parameters:

Mass of object 1 = 3.1 x 10⁵kg

Mass of object 2 = 6.5 x 10³kg

Gravitational force = 65N

Unknown:

Distance between them = ?

Solution:

To solve this problem, we use the expression below from the universal gravitational law;

Fg = $\frac{G mass 1 x mass 2}{distance ^{2} }$

G = 6.67 x 10⁻¹¹

65 = $\frac{6.67 x 10^{11} x 3.1 x 10^{5} x 6.5 x 10^{3} }{distance^{2} }$

Distance = 4.55 x 10⁹m

3 0

3 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

3 years ago

An object is 30 cm in front of a converging lens with a focal length of 10 cm. Use ray tracing to determine the location of the

poizon [28]

Answer:

Inverted

Real

Explanation:

u = Object distance = 30 cm

v = Image distance

f = Focal length = 10 cm

Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{30}\\\Rightarrow \frac{1}{v}=\frac{1}{15}\\\Rightarrow v=15\ cm$