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Andrei [34K]
3 years ago
6

Earth's atmosphere contains only small amounts of carbon dioxide because

Physics
1 answer:
Mamont248 [21]3 years ago
4 0
Of biogeochemical cycles

Because certain biogeochemical cycles take place to lessen these gases from forming since they are not beneficial for life if too much.
Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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Bob runs 1800 seconds at an average speed of 1.5 m/sec. How far did he go? 25 Points!!!!
german

Distance= Time×Speed

= 1800×1.5

= 2700 m

I am not sure it's right. the question itself is confusing.

4 0
3 years ago
How can we realize that light travel in straight line ?​
Norma-Jean [14]

Answer:

It can be seen from the operation of pin-hole camera, formation of shadows and eclipse.

Explanation:

The phenomenon of light traveling in a straight line is known as rectilinear propagation of light.

One this evidence can be seen from the operation of pin-hole camera, which depends on rectilinear propagation of light

Also two natural effects that result from the rectilinear propagation of light are the formation of Shadows and Eclipse.  

3 0
3 years ago
Although the skier has a jacket on, she is still cold. How can her circulatory system help keep her warm?
Pepsi [2]

by creating body heat so she can stay warm

6 0
3 years ago
Read 2 more answers
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

4 0
3 years ago
A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air
fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

s=u_y t+\frac{1}{2}at^2

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

u_y=0 is the initial vertical velocity

t is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

v_x = 10 m/s

Therefore, the horizontal distance covered after a time t is

d=v_x t

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

d=(10)(2.02)=20 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
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