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Papessa [141]
3 years ago
13

Why isn’t earth a closed system? A. It reflects and radiates energy B. Matter continually cycles between the spheres C. Large am

ounts of matter come to earth from space D. Earthquakes, volcanoes, and Hot Springs release energy from Earth’s interior
Physics
2 answers:
SOVA2 [1]3 years ago
8 0

Answer:

C. Large amounts of matter come to earth from space

Explanation:

A closed system is a system where only energy is transferred to the surrounding. No matter will be transferred into the system or out of the system.

By that definition earth is not closed system. since large amount of matter from distant supernovas, asteroids etc. has been deposited on earth from the very beginning.

We mine Gold, uranium for our use which was deposited on earth by supernovas So all deposits of iron, aluminum etc. come from stars massive than our sun.

The amount of all these mineral deposits are pretty high, needless to mention the addition of mass by meteorites and the loss of mass by sending space crafts to space

So we can say earth is not a closed system

Note:

But for our current calculations we treat earth as a closed system since no mineral deposit is made on earth at the moment by supernova burst (life on earth will come to an end if that happens now). The loss of mass by sending objects into space and the addition of mass by meteorites are very negligible compared to the mass of earth. So for simplicity purpose we consider earth as a closed system

Irina18 [472]3 years ago
5 0

Earth is actually considered a closed system so this question is confusing.

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Answer:

from food eaten (i.e contain carnlbohyadrates)

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With the simplified model of the eye, what corrective lens (specified by focal length as measured in air) would be needed to ena
Ymorist [56]

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Please see the attached picture for the complete answer.

Explanation:

6 0
4 years ago
A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
3 years ago
Help me calculate the kinetic energy (just the middle column) ASAP! SHOW WORK! ON PAPER
lukranit [14]

Answer:

Explanation:

I can tell you what the answers for the middle column are, but if you don't know how to solve total energy problems, they won't make any sense to you at all.

First row, KE = 0

Second row, KE = 220500 J

Third row, KE = 183750 J

Fourth row, KE = 205800 J

That's also not paying any attention to significant digits because your velocity only had 1 and that's not enough to do the problem justice. I left all the digits in the answer. Round how your teacher tells you to.

3 0
3 years ago
A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
GREYUIT [131]

Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

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So, the electron's speed is

v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s

6 0
3 years ago
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