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3 years ago

Acid rain breaks down rocks by reacting with their minerals. This process is an example of?

Physics

2 answers:

lawyer [7]3 years ago

4 0

Chemical erosion
;)
Good luck!
M.

Kryger [21]3 years ago

3 0

This process is an example of chemical weathering

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When energy from the sun hits the air above

valentina_108 [34]

Answer: C I think.

Explanation:

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3 years ago

What is the speed of a wave that has a frequency of 173 Hz and a wavelength of 2.59 meters? Express your answer to the nearest w

devlian [24]

Answer:

448 m/s is the correct answer.

Explanation:

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3 years ago

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Which term is applied to an object through which light passes?

Elena L [17]

Answer:

Explanation:

transparent_objects that allows light to pass through and can you see through them

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3 years ago

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Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and

dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)

8 0

3 years ago

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A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco

Sunny_sXe [5.5K]

Answer:

(a)

$x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km$

(b) $r=41.32km$

$\alpha =24.23^o$

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies $45^o$ south of east. The y-component will be negative and the x-component will be positive.

$x_1=25cos45^o=17.68km$

$y_1=-25sin45^o=-17.68km$

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

$x_2=40cos60^o=20km$

$y_2=40sin60^o=34.64km$

Therefore, total displacements:

$x=x_1+x_2=(17.68+20)km=37.68km$

$y=y_1+y_2=(-17.68+34.64)km=16.96km$

Magnitude of displacements,

$r=\sqrt{x^2+y^2}=41.32km$

Direction,

$\alpha =tan^{-1}(\frac{y}{x})=24.23^o$

6 0

3 years ago