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2 years ago

10

what evidence would you expect to find on the moon if it had been subjected to plate tectonics? (select all that apply.)

1 answer:

irina1246 [14]2 years ago

5 0

Volcanoes and mountains are the evidences that show plate tectonics.

<h3 /><h3>What evidence would you expect to find on the Moon if the Moon had been subjected to plate tectonics? </h3>

The presence of volcanos and mountains are the evidences that provides information that the moon has plate tectonic. If there is no volcanoes and mountains on the moon then we can say that there is no plate tectonic on the moon.

So we can conclude that volcanoes and mountains are the evidences that show plate tectonics.

Learn more about tectonics here: brainly.com/question/2325633

#SPJ11

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<u>Answer:</u> The amount of heat needed is 29000 Cal.

<u>Explanation:</u>

The process involved in this problem are:

$(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

<u>For process 1:</u>

$q_1=mC_{p,s}\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of ice = 200 g

$T_2$ = final temperature = $0^oC$

$T_1$ = initial temperature = $-30^oC$

Putting all the values in above equation, we get:

$q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal$

<u>For process 2:</u>

$q_2=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

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$L_f$ = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

$q_2=200g\times 80Cal/g=16000Cal$

<u>For process 3:</u>

$q_3=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water = 200 g

$T_2$ = final temperature = $50^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal$

Calculating the total heat absorbed, we get:

$Q=q_1+q_2+q_3$

$Q=3000+16000+10000=29000Cal$

Hence, the amount of heat needed is 29000 Cal.

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