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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.
You might be interested in
class 7 Physics answera plot measuring 2 km in length and 1 by 2 km in breadth has area of how many hectare ​
motikmotik

Answer:

Area of plot in hectare = 100 hectare

Explanation:

Given;

Length of plot = 2 km

Width of plot = 1/2 km = 0.5 km

Find:

Area of plot in hectare

Computation:

Area of plot = Length x width

Area of plot = 2 x 0.5

Area of plot = 1 square kilometer

1 square kilometer = 100 hectare

Area of plot in hectare = 1 x 100

Area of plot in hectare = 100 hectare

7 0
3 years ago
how much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m
Vesna [10]

2.89watts.

<h3>What is meant by sound intensity?</h3>
  • The average rate at which sound energy moves across a unit area normal to a given direction is used to determine a sound's intensity. This rate is generally stated in ergs per second per square centimeter.
  • Decibels are the units used to measure sound intensity, often known as sound power or sound pressure. The decibel (dB) unit is named after Alexander Graham Bell, who also created the audiometer and the telephone. An audiometer is a tool to gauge a person's hearing capacity for various noises.
  • Our ability to measure the flow of sound energy as a time-averaged vector quantity makes sound intensity measuring an effective method. We can identify sound sources and tell direct sound from reverberant sound in a room using the characteristics of sound intensity.

How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:

Formula: I\frac{P}{4\pi r^{2} }

I=1.6x10-3 w/m2

r=12m

I\frac{P}{4\pi r^{2} }

P=I4\pi r^{2}

   =(1.6x10-3\frac{W}{m^{2} } ) (4\pi (12m)^{2} )

   =2.89watts.

To learn more about sound intensity, refer to:

brainly.com/question/17062836

#SPJ9

8 0
2 years ago
The picture below shows an experiment to find out the highest frequencies which students can hear. When the teacher turns the di
Ede4ka [16]
Wheres the picture? Theres no picture below
5 0
3 years ago
A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the
Gnoma [55]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating  system = 6.74 J

We know that energy is given by

(a) Ke=\frac{1}{2}mv_{max}^2

6.74=\frac{1}{2}\times 0.267\times v_{max}^2

v_{max}=7.1m/sec

(b) Now \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec

We know that \omega =\sqrt{\frac{k}{m}}

35.68=\sqrt{\frac{k}{0.267}}

k=339.9N/m

(c) We know that energy is given by

E=\frac{1}{2}KA^2

6.74=\frac{1}{2}\times 339.9\times A^2

A=19.91cm

4 0
3 years ago
Someone please helppp :|||
juin [17]

Answer:

b

Explanation:

4 0
3 years ago
Read 2 more answers
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