Answer:
A m 23 3.3 32 m 0.37
Explanation:
So ur still thinking of me Just like I know you should. I cant not give you everything You know I wish I could.
If placed in water Saturn would float. This is because it is mostly made up of gas, which is less dense than water.
We need to find the volume of a spherical shell with a radius of
6.37 million meters and a thickness of 0.95 mile.
The technically correct way to do this is to find the volume of the
outside of the shell, then find the volume of the inside of the shell,
and subtract the inside volume from the outside volume. That's
the REAL way to do it.
But look. This 'shell' (the 0.95 mile of water) is only about 1530 meters thick,
on a sphere with a radius of 6.37 million meters. The depth of the water is like
0.024 percent of the radius ! There's not a whole lot of difference between the
sphere outside the water and the sphere inside it.
So I want to do this problem the easier way ... Let's say that the volume
of the water is going to be
(the surface area that it covers on the Earth)
times
(the thickness of the coating of water) .
The area of a sphere is 4 pi Radius² .
That's
(4 pi) x (6.37 x 10⁶ m)²
= (4 pi) x (40.58 x 10¹² m²)
We're only interested in 70% of the total surface area.
= (0.7) x (4 pi) x (40.58 x 10¹²) m²
= 3.57 x 10¹⁴ square meters of Earth's surface.
The volume of the water covering that area is
(the area) times (average depth of 0.95 mile) .
We have to change that 0.95 mile to meters.
The question reminds us that 1 mile = 1609 meters .
So the volume of the water is
(the area) times (0.95 x 1609 meters).
But we're not there yet. The question isn't asking for the volume.
It's asking for the mass of the water.
We're ready to get the volume in cubic meters.
We're supposed to know that each cubic meter is 1,000 liters,
and the mass of 1 liter of water is 1 kilogram.
So each cubic meter of volume is 1,000 kilograms of mass.
Now we're ready to dump all the numbers into the machine and
turn the crank. The mass of all this water will be
(the surface area) x (0.95 x 1609 meters) x (1,000 kg/m³)
= (3.57 x 10¹⁴ m²) x (1528.6 m) x (1,000 kg/m³)
= 5.457 x 10²⁰ kilograms .
This is my answer, and I'm stickin to it.
But ... just like all the other problems you get in high school, the
answer doesn't matter. The teacher doesn't need the answer,
and YOU don't need the answer. The reason you got this problem
for an assignment is to give you practice in HOW TO FIND the
answer ... how to plan what you're going to do with the problem,
and then how to carry it out.
I don't know how much effort you put into this problem, but somewhere
along the way, you chickened out and posted it on Brainly. So far, the
result of that decision was: The person who got all the practice was ME.
I got the good stuff, and all YOU got was the answer.
I hope my work is clear enough that you can go through it, and pick up
some of the good stuff for yourself.
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
Answer:
<h2>Electricity has many uses in our day to day life. It is used for lighting rooms, working fans and domestic appliances like using electric stoves, A/C and more. All these provide comfort to people. In factories, large machines are worked with the help of electricity.</h2>
