All categories

3 years ago

How many stars can be seen without a telescope on earth

1 answer:

5 0

There are only about 5,000 stars visible to the naked, average, human eye, . And, because the Earth itself gets in the way, you can only see about a half of those from where you stand.

You might be interested in

PLS HELP

Ahat [919]

"This was the idea that non-living objects can give rise to living organisms."

7 0

3 years ago

wite a paragraph explaining the difference between things that have matter and things that don't have matter.

mihalych1998 [28]

Answer: Matter is how much space or opacity an object takes up. In short anything that take up space. Things like balls, trees, and even people are all made of matter. Mass is how much matter a object has. Air also has mass also, even though we can't see it. Things like cars, buildings, even planets have mass.

Explanation: Paragraphs are sometimes 4-6 sentences.

6 0

3 years ago

The Celsius temperature scale is based on which of the following

Lapatulllka [165]

Freezing.................

6 0

3 years ago

A charged particle is observed traveling in a circular path in a uniform magnetic field. If the particle had been traveling four

8090 [49]

Answer:

The new radius of the trajectory of the particle is four times the previous radius

Explanation:

In order to know what is the radius of the trajectory of the charged particle, if its speed is four times as fast, you take into account the following formula, which describes the radius of a charged particle in a magnetic field:

$r=\frac{mv}{qB}$ (1)

If the speed of the particle is for time as fast, that is, v' = 4v, you obtain, in the equation (1):

$r'=\frac{mv'}{qB}=\frac{m(4v)}{qB}=4\frac{mv}{qB}=4r$

The new radius of the trajectory of the particle is four times the previous radius

8 0

2 years ago

A swimming pool has the shape of a box with a base that measures 30 m by 12 m and a uniform depth of 2.2 m. How much work is req

Alex777 [14]

Answer:

853776 J

Explanation:

The work-energy needs to pump water out of the pool is the product of the weight of water and distance h

E = Wh = mgh

Since water mass is a body of water we can treat it as the product of density 1000kg/m3 and volume, which is the product of base area and uniform height h

$m = \rho V = \rho \int\limits^{2.2}_0 {A} \, dh$

Therefore:

$E = mgh = g\rho A\int\limits^{2.2}_0 {h} \, dh\\E = 9.8*1000*30*12[h^2/2]^{2.2}_0 = 1764000(2.2^2 - 0^2) = 853776 J$

5 0

2 years ago