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hram777 [196]
3 years ago
15

The Milky Way measures _________ light years in diameter.

Physics
1 answer:
Snowcat [4.5K]3 years ago
7 0

Answer:

The Milky Way is about 1,000,000,000,000,000,000 km (about 100,000 light years or about 30 kpc) across. The Sun does not lie near the center of our Galaxy. It lies about 8 kpc from the center on what is known as the Orion Arm of the Milky Way.

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IBM has a computer it calls the Blue Gene/L that can do 136.8 teracalculations per second. How many calculations can it do in a
Norma-Jean [14]
There are 1,000,000 micro seconds in one second so multiple 136.8 by 1000000 and you'll get 136,800,000 Tera calculations per second.
4 0
3 years ago
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The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
A teacher wants to demonstrate that the radioactive source emits alpha beta and gamma radiation. Describe a method the teacher c
Mila [183]

By using an electric field, it is feasible to differentiate between these different forms of radiation.

<h3>What is a radioactive source?</h3>

A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.

The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.

Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.

To learn more about the radioactive source refer;

brainly.com/question/12741761

#SPJ1

8 0
2 years ago
Nuclear fusion has not yet been harnessed as an energy source because
maksim [4K]

Before getting an answer for it first we have to understand nuclear fusion.

Nuclear fusion is a thermo-nuclear reaction in which two  light unstable nuclei will form a heavy stable nuclei. In this process there will be some mass defect which will be converted into energy as per Einstein's mass energy equivalence theorem.

The theorem is stated as E = mc^{2} where c is the velocity of light and m is the mass converted into energy.

One take an example of fusion in sun where 4 hydrogen atoms combine to form a helium nucleus which are explained below-

                                                      H_{1}^{1}+H_{1} ^{1} =H_{1} ^{2} +e_{+1} ^{0} +energy

                                                      H_{1} ^{2} +H_{1} ^{1} =He_{2} ^{3} +energy

                                                      He_{2} ^{3}+H_{1} ^{1} =He_{2} ^{4} +e_{+1} ^{0} +energy

                                                       -----------------------------------------------------------------------

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Here e_{+1} ^{0} is the positron.

In this process very high temperature is needed which is approximately equal to the temperature of the sun i.e 10^{6} K

Such temperature is very difficult to initiate the reaction on the earth surface. It should be carried out in an sustainable way also .Otherwise It will cause nuclear hazards.

6 0
3 years ago
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It would be A. Because think of the explanations Jasons friend could say to them that would be a negative 'statement'.
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