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3 years ago

What is the primary function of earth's atmosphere (1 point)?

1 answer:

5 0

<span>Without the atmosphere that protects the Earth, humans, animals and all plant life would not be able to survive. The atmosphere works to hold oxygen in the area around the earth. Humans and most animals are dependent on oxygen. Plants are also dependent on oxygen because they take the leftover carbon dioxide from humans when they inhale the oxygen.</span>

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A rock is rolling down a hill. At position 1, its velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, its veloci

alukav5142 [94]

As this happens over twelve seconds, you would take the total difference in velocities and divide it by twelve to find the change per second

44.0 m/s - 2.0 m/s = 42.0 m/s

42.0 m/s / 12 s = 3.5 m/s2

the acceleration of the rock would be 3.5 m/s2

4 0

3 years ago

Read 2 more answers

How many neutrons are in an atom of Carbon-13 or (613C)?

vivado [14]

Answer: 6 OR 7

Explanation:

3 0

3 years ago

What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?

myrzilka [38]

Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

Ef= final energy at the top of the ramp= KEf+PEf

Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0 -------------(1)

KEf = mgh = 200×9.8×h

Where h= Sin 22 = h/d= h/4.1

0.375×4.1=h

or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 m $Vf^{2}$ = 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 m $Vi^{2}$ =0.5×200× $Vi^{2}$ =100 $Vi^{2}$ j

Put these values in eq 1, we get;

0-100 $Vi^{2}$ +3018.4-0=0

-100 $Vi^{2}$ =-3018.4

==> $Vi^{2}= \frac{3018.4}{100}$ = 30.184

==> Vi = $\sqrt{30.184} = 5.5 m.sec$

7 0

4 years ago

1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (H

JulijaS [17]

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

$dV = \overrightarrow{E} . d\overrightarrow{r}\\\\$