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bixtya [17]
3 years ago
8

The illustration above shows the item needed for an investigation. which item is the independent variable? which items are the c

onstants? what might a dependent variable be?

Physics
1 answer:
kirill115 [55]3 years ago
4 0
The independent variable are the pots that you use (different thing same method)

A possible dependent variable is the tomatoes, the amount of time you cook them (assuming thats what you're supposed to do) the temperature, etc, what you need to keep the same.

I'm unsure of the constants though, sorry. Hope this helped a bit!
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A 5 kg toy is tied to a rope where the tension measures 150 N. What is the weight of the object?
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Formula from physics to get the answer.
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3 years ago
Gases tend to deviate from the ideal gas law at
IgorLugansk [536]
<h2>Answer: high pressures</h2>

The Ideal Gas equation is:

P.V=n.R.T  

Where:

P is the pressure of the gas

n the number of moles of gas

R=0.0821\frac{L.atm}{mol.K} is the gas constant

T is the absolute temperature of the gas in Kelvin.  

According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them.

Now, real gases can behave approximately to an ideal gas, under the conditions described above.

However, when <u>temperature is low</u> these gases deviate from the  ideal gas behavior, because the molecules move slowly, allowing the repulsion or attraction forces to take effect.

The same happens at <u>high pressures</u>, because the volume of molecules is no longer negligible.

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4 years ago
After a scientist wrote up her findings in a paper, she submitted it to a journal for peer review. After the review, it was acce
sineoko [7]
It gets published 
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8 0
4 years ago
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6 0
3 years ago
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One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R
Drupady [299]

(a) 120.8 m/s^2

The gravitational acceleration at a generic distance r from the centre of the planet is

g=\frac{GM'}{r^2}

where

G is the gravitational constant

M' is the mass enclosed by the spherical surface of radius r

r is the distance from the centre

For this part of the problem,

r=R=1.17\cdot 10^6 m

so the mass enclosed is just the mass of the core:

M'=M=2.48\cdot 10^{24}kg

So the gravitational acceleration is

g=\frac{(6.67\cdot 10^{-11})(2.48\cdot 10^{24}kg)}{(1.17\cdot 10^6 m)^2}=120.8 m/s^2

(b) 67.1 m/s^2

In this part of the problem,

r=3R=3(1.17\cdot 10^6 m)=3.51\cdot 10^6 m

and the mass enclosed here is the sum of the mass of the core and the mass of the shell, so

M'=M+4M=5M=5(2.48\cdot 10^{24}kg)=1.24\cdot 10^{25}kg

so the gravitational acceleration is

g=\frac{(6.67\cdot 10^{-11})(1.24\cdot 10^{25}kg)}{(3.51\cdot 10^6 m)^2}=67.1 m/s^2

8 0
4 years ago
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