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3 years ago

Choose all the answers that apply.

Physics

2 answers:

Lynna [10]3 years ago

7 0

Answer:

Sometimes come very close to the Earth

Explanation:

It is the only answer, I got it wrong on my test and this is the correct answer. :)

Salsk061 [2.6K]3 years ago

6 0

"Asteroid" is the name we give to the huge number of small bodies
that orbit the sun, here in the inner solar system.

Their orbits are scattered all over the place. Most of them spend
most of the time between the orbits of Mars and Jupiter, but there are
many asteroids that sometimes come very close to Earth.

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Suppose you observe that at night, the air just above the ground feels cooler than the air above it. then in the middle of a sun

yanalaym [24]

The answer is convection

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4 years ago

A 60cm long string of an ordinary guitar is tuned to produce the note a4 (frequency 440hz) when vibrating in its fundamental mod

pishuonlain [190]

Answer:

0.78 m

Explanation:

The relationship between wavelength and frequency of a wave is given by

$v=f \lambda$

where

v is the speed of the wave

f is the frequency

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For the sound wave in this problem, we have

$f=440 Hz$ is the frequency

v = 344 m/s is the speed of sound in air

Substituting into the equation and re-arranging it, we find the wavelength:

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3 years ago

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maxonik [38]

The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other; and negatively charged objects and neutral objects attract each other.

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3 years ago

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In terms of matter and resources, Earth is essentially a(n) ________ system ; in terms of energy, Earth is a(n) ________ system.

Savatey [412]

Answer:

Explanation:

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3 years ago

The gnaphosid spider Drassodes cupreus has evolved a pair of lensless eyes for detecting polarized light. Each eye is sensitive

BartSMP [9]

Answer:

Part A

The intensity is $I = 618 W/m^2$

Part B

The intensity is $I_1= 81.884 W/m^2$

Explanation:

From the question we are told that

The intensity of the light detected by first eye is $I = 700 W/m^2$

Now at initial state according the question the light ray is perpendicular to the eye so it means that it is at 90° the eye

Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light )would detect when the head is rotated by 20° its previous orientation

This is mathematically evaluated as

$I = I_i cos^2 ( 20^o)$

$I = 700\ cos^2 (20)$

$I = 618 W/m^2$

Now the second question is to obtain the intensity the first eye (the first eye in this case is the one that is not focused on the light )would detect when the head is rotated by 20° its previous orientation

Now in this case the angle between the eye and the light is 90-20 = 70°

$I_1 = 700 \ cos^2 (70)$

$I_1= 81.884 W/m^2$

5 0

3 years ago