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3 years ago

9

How does the sun affect the water cycle?

2 answers:

Temka [501]3 years ago

7 0

The answer is B!

Reasoning:
- A is just wrong.
- B happens after the water has evaporated and cooled. The sun cannot cool water vapor. In fact, the sun can’t cool anything.
- C is also just plainly wrong because the sun cannot cool.
- Therefore, D is the answer.

If you could mark me as brainiest, that’d help me level up :D

Savatey [412]3 years ago

4 0

i think the answer is D.

heating surface water causes it to evaporate and enter the air as a gas.

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What is the speed of a horse in meters per second that runs a distance of 1.2 miles in 2.4 minutes

Ilia_Sergeevich [38]

Time t=2.4 minutes=2.4×60=144 seconds
distance s=1.2 miles=1.2×1609=1930.8 meters
speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]

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3 years ago

Why are stars considered to be the building blocks of the universe?

eduard

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3 years ago

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Sliding friction is affected by the weight of the object.<br><br> True or false

Dmitry [639]

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3 years ago

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Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to

katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

$OB=0.40 km$

And the Max distance towards bookstore is,

$OA=3.65 km$

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

$AB=\sqrt{OB^{2}+OA^{2}}$

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

$AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km$

Therefore the distance between there destination is 3.67 km.

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3 years ago

A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light i

slega [8]

Answer:

The appropriate solution is:

(a) $\frac{1}{4}(I_o)$

(b) $\frac{1}{4} (u_o)$

(c) $\frac{1}{2}B_o$

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

⇒ $\frac{I}{I_o} =\frac{r_o_2}{r_2}$

On applying cross-multiplication, we get

⇒ $I=I_o\times \frac{1_2}{2^2}$

⇒ $=I_o\times \frac{1}{4}$

⇒ $=\frac{1}{4} (I_o)$

(b)

As we know,

⇒ $\frac{u}{u_o} =\frac{I}{I_o}$

By putting the values, we get

⇒ $u=\frac{1}{4}(u_o)$

(c)

⇒ $\frac{B^2}{B_o^2} =\frac{u}{u_o}$

$=\frac{I}{I_o}$

⇒ $B=B_o\times \sqrt{\frac{1}{4} }$

⇒ $=\frac{1}{2}(B_o)$

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3 years ago