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ivanzaharov [21]

3 years ago

8

When the northern hemisphere is tilted toward the sun that part of the earth is closer to the sun and thus hotter? True or false

1 answer:

Ratling [72]3 years ago

8 0

<u>False</u>

<u>Explanation:</u>

Most of us have a belief that Earth is very close to the sun in summer season and so it is hotter. Also, since Earth is far away from the sun in winter season, it is colder. Though this idea looks true, it is incorrect.

The orbit of the Earth isn't a perfect circle. It is quite tilted. During some months of the year, Earth is very close to the sun than at other times. We have winter when the Earth is very close to the sun and summer when it is far away in the Northern Hemisphere. So when we compare the distance of the Sun from the Earth, this change in Earth's distance throughout the year does not affect our weather much.

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3 years ago

Charging a ballon and rubbing it on wool is an example of ___?

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3 years ago

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tia_tia [17]

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3 0

2 years ago

Describe the movement of particles at the melting point of a substance with temperature and energy.

Gnesinka [82]

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7 0

2 years ago

If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move

alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: $m=22kg$

coefficient of friction: $\mu =0.27$

and we also know the acceleration of gravity is $g=9.81m/s^2$

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

Normal force $N$ (acting up from the object)

weight: $w=mg$ (acting down from)

so the sum of forces in the vertical axis "y" are:

$F_{y}=N-w\\F_{y}=N-mg$

from Newton's second Law we know that $F=ma$ , so:

$ma_{y}=N-mg$

and since the object is not accelerating in the vertical direction (the movement is only horizontal) $a_{y}=0$ , and:

$0=N-mg\\N=mg$

-----------

now let's analyze the horizontal forces

frictional force: $f= \mu N$ and since $N=mg$ --> $f=\mu mg$

force to move the object: $F$

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

$F=ma_{x}=F-f\\ma_{x}=F-\mu mg$

and we are told that the crate moves at a steady speed, thus there is no acceleration: $a_{x}=0$

and we get:

$0=F-\mu mg\\F=\mu mg$

substituting known values:

$F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N$

3 0

3 years ago