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Sergeu [11.5K]
3 years ago
7

Charles law increases keep pressure constant then you observe blank

Physics
1 answer:
OLga [1]3 years ago
7 0

If you decrease the pressure of a fixed amount of gas, its volume will increase.

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Why don’t we feel the gravitational force of a large object such as a skyscraper semi-truck?
Kobotan [32]

Answer:

Se the explanation below

Explanation:

We do not feel these forces of these bodies, because they are very small compared to the force of Earth's attraction. Although its mass is greater than that of a human being, its mass is not compared to the Earth's mass. In order to understand this problem we will use numerical data and the universal gravitation formula, to give validity to the explanation.

<u>Force exerted by the Earth on a human being</u>

<u />

F=G*\frac{m_{1}*m_{2}}{r^2}

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 5.97*10^24[kg]

r = distance from the center of the earth to the surface or earth radius = 6371 *10^3 [m]

<u />

Now replacing we have

F = 6.673*10^{-11} *\frac{80*5.97*10^{24}}{(6371*10^{3})^{2}  } \\F = 785[N]

<u>Force exerted by a building on a human being</u>

<u />

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 300000 [ton] = 300 *10^6[kg]

r = distance from the building to the person = 2[m]

F = 6.673*10^{-11}*\frac{80*300*10^6}{2^{2} }  \\F= 0.4 [N]

As we can see the force exerted by the Earth is 2000 times greater than that exerted by a building with the proposed data.

8 0
3 years ago
Do geochemists need to have a knowledge of physical science? Explain your<br> answer. please help:))
saul85 [17]
No they don’t lol I hate school
6 0
3 years ago
Assuming that 70 percent of the Earth’s surface
Aneli [31]
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.
3 0
3 years ago
Kellie found a sample of a pure element. She carefully studied the sample and made a list of its properties.
aivan3 [116]
1. (C.) -Si- is silicone which at room temp is a solid and it's a hard crystalline Brittle rock basically...

2. And it will Be D. NOT C... i can't stand user's who throw out answers for free points.... 
3 0
3 years ago
A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm
I am Lyosha [343]

Answer:

Rate of change of magnetic field is 3.466\times 10^3T/sec        

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop r=\frac{0.13}{2}=0.065m

Length of the circular loop L=2\pi r=2\times 3.14\times 0.065=0.4082m

Wire is made up of diameter of 2.6 mm

So radius r=\frac{2.6}{2}=1.3mm=0.0013m

Cross sectional area of wire A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2

Resistivity of wire \rho =2.18\times 10^{-8}m

Resistance of wire R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm

Current is given i = 11 A

So emf  e=11\times 1.67\times 10^{-3}=0.0183volt

Emf induced in the coil is e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}

0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=3.466\times 10^3=T/sec

8 0
3 years ago
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