2 years ago

HELPPPPPP ! Pleaseeee anyone ?

2 answers:

7 0

1- periodic table 2- elements 3- atomic number 4- properties 5- periods 6- groups 7- group number 8- valence electrons, give me brainliest pleaseee !!

kenny6666 [7]2 years ago

4 0

Answer:

1 Periodic table

2 Elements

3 atomic number

4 properties

5 Periods

6 groups

7 group number

8 Valence electrons

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2 years ago

Define mass .Also write the table showing the relation of 'kg' with it's sub- multiples and multiples.

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2 years ago

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Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he

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Answer:

E = 2k $\frac{\lambda}{ r}$

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

λ = q_ {int} /l

q_ {int} = l λ

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E A = l λ /ε₀

is area of cylinder is

A = 2π r l

we substitute

E = $\frac{ l \ \lambda}{ \epsilon_o \ 2\pi \ r \ l }$

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3 years ago

If a 50 W light bulb and a 90 W light bulb operate from 120 V, which bulb has a greater current in it?

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The 50 W bulb would have a greater current than the 90 W bulb

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2 years ago

a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc

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Answer:

P = 0.25 W

Explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

$P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W$

So, 0.25 W power is delivered to the bulb.

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3 years ago