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mina [271]
3 years ago
14

Assuming that 70 percent of the Earth’s surface

Physics
1 answer:
Aneli [31]3 years ago
3 0
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.
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What causes a satellite to move tangentially to a circular path?
yawa3891 [41]

Explanation : We know that two forces acting on the satellite.

The speed of the satellite and the force of the gravity on the satellite then the result , the satellite moves around the earth in a circular motion.

A satellite to move tangentially to a circular path, while revolving around the circular path because the centripetal force is acting on it.

So, we can say gravity is responsible for the motion of the satellite in a circular path.

5 0
3 years ago
Read 2 more answers
If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val
baherus [9]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}

\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \  body \ (v) = 20 \ m/s}

\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \  body \ (u) = 0}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \  body \ ( F)}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

☯ <u>Using 1st equation of motion </u>

\\

\dashrightarrow\:\: \sf{v = u + at}

\\

\dashrightarrow\:\: \sf{20 = 0 + a(4)}

\\

\dashrightarrow\:\: \sf{20 = 4a}

\\

\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}

\\

\dashrightarrow\:\: \sf{a = 5}

\\

☯ <u>Now, Finding the force exerted </u>

\\

\dashrightarrow\:\: \sf{F = ma}

\\

\dashrightarrow\:\: \sf{F = 40 \times 5}

\\

\dashrightarrow\:\: \sf{F = 200 \ N}

\\

☯ <u>Hence</u>, \\

\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}

8 0
3 years ago
The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma
Vesna [10]

Answer:

Number of turns per unit length will be n=2.786\times 10^4turns/m

Explanation:

We have given that strength of the magnetic field produced by the solenoid B = 7 T

Current in the solenoid i = 200 A

Let the number of turns per unit length is n

Magnetic field due to solenoid is given as B=\mu ni here \mu is permeability of free space n is number of turns per unit length and i is current

So 7=4\pi \times 10^{-7}\times  n\times 200

n=2.786\times 10^4turns/m

7 0
3 years ago
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1. An object in free fall will have an initial velocity equal to zero when: a. It is thrown vertically down
Ne4ueva [31]

Answer:

b. It is dropped

Explanation:

If the initial velocity is zero, the object move from rest. That happens if the object is dropped

6 0
3 years ago
A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

7 0
3 years ago
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