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lora16 [44]
2 years ago
10

The rainforest is hot, humid, and has frequent rainfall. In what way do the atmosphere and the hydrosphere contribute to this ty

pe of climate?
Physics
1 answer:
il63 [147K]2 years ago
7 0

Answer:

The atmosphere which contains nitrogen, oxygen, carbon dioxide, and other gases essential for life, contributes to the rainforest climate. The rainforest is hot, humid and helps balance the amount of carbon dioxide in the atmosphere by absorbing it and releasing more oxygen which is important for climate change.

The hydrosphere, which is the combination of various types of water sources and types found in air, in the form of water vapor; water found below the earth's surface and above, also contributes to the rainforest climate. The frequent rainfall dictates the vegetation growth in that area which also creates weather patterns due to the contribution of the atmosphere and the hydrosphere. The hydrosphere is important in the formation of rain. A  rainforest of frequent rainfall indicates a large amount of the combination of various types of water sources and types found in a particular area, which release the water back onto the rainforest.

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What does the speed of sound depend on?
melamori03 [73]
Assuming it's an ideal material, the answer is A.

The speed of sound depends on the medium that the sound travels through.
5 0
2 years ago
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The law of universal gravitation states that a gravitational force exists between all objects in the universe. This force is dir
Ber [7]

Answer: D. it has been demonstrated to be without exception under certain stated conditions.

Explanation:

A <u>Law</u> is an affirmation (something established) based on repeated long-term observation of a phenomenon that has been studied and verified.  

That is: A law is present in all known theories and therefore is considered universal. In addition, a law can not be refuted, nor changed, because its precepts have been proven through various studies.  

<u>Unlike theory</u>, which is the set of rules and principles that describe and explain a particular phenomenon and <u>is subject to changes as new evidence emerges that gives meaning to it.  </u>

Then, based on what is explained above, the law of universal gravitation is a statement that exists because it was rigorously tested and verified, therefore it can not be refuted.

3 0
3 years ago
1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve
hjlf

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

d_x = 12 km east

d_y = 16 km south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

S=\frac{1}{2}gt^2

From which we find the total time of the fall, t:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s

And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is

d=vt=(16 m/s)(2 s)=32 m

So, the closest answer is B.

5 0
3 years ago
A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g
Hitman42 [59]

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

8 0
3 years ago
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If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the
kodGreya [7K]

Answer:

a)

v = 4.048 *10^6 m/s

b)  

Angular frequency =  1.92 * 10^7

Explanation:

As we know

v =  \frac{qBr}{m}

q is the charge on the electron = 3.2 * 10^{-19} C

B is the magnetic field in Tesla = 0.4 T

r is the radius of the circle = 0.21 m

mass of the electrons = 6.64 * 10^{-27} Kg

a)

Substituting the given values in above equation, we get -

v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6m/s

b)  

Angular frequency =

\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7

8 0
2 years ago
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