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4 years ago

Why polarization does not occur in dry cell?

Physics

1 answer:

Arisa [49]4 years ago

7 0

Answer:

Manganese oxide prevents polarisation in dry cells. - Polarization is a defect that occurs in simple electric cells due to the accumulation of hydrogen gas around the positive electrode. ... - MnO2 reacts with H2 and forms water as byproduct, so depolarization doesn't occur.

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What phenomenon causes the fuzzy image? What kind of lens is used to correct this, and how is it corrected?

Amiraneli [1.4K]

Sample Response: This phenomenon is called chromatic aberration. This happens when light of different wavelengths focuses at different points. A converging lens is used to help light of different wavelengths focuses at a common point.

8 0

3 years ago

Read 2 more answers

an object is 40M below the surface of sea water of density 1020 kg/m3 calculate the pressure difference

olga nikolaevna [1]

Answer:

P = 400248 [Pa]

Explanation:

We can calculate the hydrostatic pressure exerted by a column of liquid by means of the following equation.

$P=Ro*g*h$

where:

Ro = density = 1020 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = depth = 40 [m]

Therefore:

$P=1020*9.81*40\\P=400248 [Pa]$

8 0

3 years ago

A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

zzz [600]

Answer:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Explanation:

We know that the electric force equation is:

$F=k\frac{q_{1}*q_{2}}{r^{2}}$

k is the electric constant $9*10^{9} Nm^{2}/C^{2}$
r is the distance between the particles
q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

$F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}$

$F_{31}=k\frac{q*2q}{r_{31}^{2}}$

$F_{31}=k\frac{2q^{2}}{r_{31}^{2}}$

r(31) is the distance between 3 and 1

2. Now, let's find the electric force between the third particle and the second particle.

$F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}$

$F_{32}=k\frac{q*(-q)}{r_{32}^{2}}$

$F_{32}=-k\frac{q^{2}}{r_{32}^{2}}$

r(32) is the distance between 3 and 2.

Now, $r_{31}+r_{32}=2a$ or $r_{32}=2a-r_{31}$

The net force must be zero so:

$F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}$