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3 years ago

Which biome's yearly rainfall mainly evaporates?

Physics

2 answers:

tino4ka555 [31]3 years ago

7 0

India's monsoon area i beleive

Tasya [4]3 years ago

3 0

The desert biome. How this helps

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Newton began his academic career in 1667. For how long was he a working scientist? Was he a very productive scientist?

rjkz [21]

Newton's 3 laws are...

inertia: things tend to continue to do what they are doing.

Change: to make something change you need a force to change it. the force needed = the mass times its acceleration

<span> Resistance: When you push on something, it pushes back.

From yahoo answers
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2 years ago

Once again, move the balloon to the right and let it go. Note how fast the balloon moves. Next, brush the balloon against the en

lilavasa [31]

Answer:

Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons

Explanation:

From Plato. Hope this helps!

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2 years ago

Read 2 more answers

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$