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3 years ago

A student drew the following model:

Physics

2 answers:

Gelneren [198K]3 years ago

8 0

Answer: C Tectonic Plates

Explanation:

Morgarella [4.7K]3 years ago

5 0

Answer:

C. Tectonic plates.

Explanation:

The series given in the question is mantle...convection...volcano...cooling...crust..motion and then tectonic plate is next.

here the process started in mantle where hot lava was moving and due to convection, this molten lava came on the the surface in the form of volcano and cooled down and rocky crust was formed and the moving crust is known as tectonic plates.

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An astronaut on an alien planet drops a rock into a crater which is 100 meters deep. The rock hits the bottom of the crater 4 se

Semmy [17]

Answer:

The gravity on this planet is stronger than that of earth.

Explanation:

First we need to find the acceleration due to gravity value of this planet to compare its gravity force with that of the earth. Hence, we will use second equation of motion:

h = Vi t + (0.5)gt²

where,

h = height or depth of crater = 100 m

Vi = Initial Velocity of rock = 0 m/s

t = time = 4 s

g = acceleration due to gravity on this planet = ?

Therefore,

100 m = (0 m/s)(4 s) + (0.5)(g)(4 s)²

g = (200 m)/(16 s²)

g = 12.5 m/s²

on earth:

ge = 9.8 m/s²

Since,

ge < g

Therefore,

<u>The gravity on this planet is stronger than that of earth.</u>

6 0

3 years ago

She left the cubes in the water for three hours which of the following describes a heat flow that took place during those three

quester [9]

Answer:

veffvevfevve

Explanation:

3 0

2 years ago

Read 2 more answers

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

3 years ago

If you rub an inflated balloon against your hair and place it against a door, by what mechanism does it stick? explain.

Snowcat [4.5K]

When you rub inflated balloon with your hair or your kitten's fur, charge is induced on all over the balloon's surface. This is called "charging by friction" because you developed charges by rubbing to bodies with each other. It will also stick on your wall you can check it out. This is because of "unlike charges attract each other". Rubbed balloon and wall possessed unlike charges which made them stick together.

8 0

3 years ago

A person climbs out of swimming pool and stands in the open air.Explain why evaporation of water from the surface of the persons

Lorico [155]

Answer:

The person feels cool at first because the swimming pool water is usually cool and he/she has that water on his body. But when it evaporates, the cool air directly touches his body and that's why he/she feels cold.

Thank You! Please mark me Brainliest!

8 0

2 years ago