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trapecia [35]
3 years ago
12

What matches ????????????????

Physics
1 answer:
dangina [55]3 years ago
8 0

Answer:

1st: Radiation

2nd: Conduction

3rd: Convection

Explanation:

I actually learned this before in school. Yay

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If you have 100 W expended over 20 s how much energy did it take?
netineya [11]

Power = (work or energy) / (time)

100 W  =  (energy) / (20 sec)

Energy = 2,000 watt-sec

<em>Energy = 2,000 J</em>

8 0
3 years ago
Your environment can play a role in whether or not a specific gene is expressed
Hatshy [7]
I don't think so. How are you supposed to write the answer?
4 0
3 years ago
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The mixture you separated was a mixture of iron filings, sand, and salt. Based on your understanding of matter, is this mixture
V125BC [204]

Answer: Heterogeneous mixture - the parts are not uniformly mixed.

A mixture contains components having distinct chemical properties. There are two types of mixtures: homogeneous and heterogeneous. In a homogeneous mixture there is uniform distribution of components. we cannot distinguish one portion of the mixture from another. for example salt mixed in water. In heterogeneous mixture, the components are not uniformly mixed. hence, we are able to distinguish different parts of a mixture, like the mixture of iron, sand and salt given in this question.

6 0
3 years ago
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What is epsilon zero and why does it come into use, particularly in the case of Gauss's Law?
Paha777 [63]
<span>Epsilon zero is permittivity of free space means how much air or vacuum permits electric field to travel from one charge to other.It is constant in the coulomb law. This allow Gauss's a lot easier to solve rather than using K</span>
5 0
3 years ago
wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
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