Explanation:
The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.
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.
You would multiply the speed by the time. So the answer would be 840 miles.
Answer:
200 N
Explanation:
Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.
So, E = σ/ε = FL/eA
Now, since at break extension = e.
So making e subject of the formula, we have
e = FL/EA = FL/Eπr² where r = radius of metal wire
Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r
So, e' = F'(2L)/Eπ(2r)²
e' = 2F'L/4Eπr²
e' = F'L/2Eπr²
Since at breakage, both extensions are the same, e = e'
So, FL/Eπr² = F'L/2Eπr²
F = F'/2
F' = 2F
Since F = 100 N,
F' = 2 × 100 N = 200 N
So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.
Answer:
B. holding a coffee mug
Explanation:
Something must move a distance for work to be done.
