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2 years ago

Plz help asap.

Physics

1 answer:

stich3 [128]2 years ago

5 0

Answer:

In a primary cell, two electrodes (one of copper and other of zinc) of metal atoms are used. These electrodes are dipped in an electrolyte solution that causes the metals to produce their respective positive and negative ions.

In this way, the flow of charges takes place and supply the electricity to the source.

Unlike a primary cell, a dry cell contains paste of an electrolyte instead of the solution. The contents of electrolyte paste react with each other through a chemical process and convert the chemical energy into electrical energy.

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What is the acceleration of a rocket if 200 Newtons are applied to its 5.5 kg

snow_tiger [21]

Given
Force=200N
Mass=5.5kg

F=ma
a=F/m=200/5.5=38.18metre per second

7 0

2 years ago

A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so

joja [24]

ANSWER:100dB

from the sound intensity level,the sound intensity is calculated as:

$\beta$ ₁= $\beta$ =(10dB)㏒₁₀(l₁/l₀)

inserting numbers:

120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²

Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which

can be used to solve for l₁ and get

l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²

since the sound intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in

ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:

l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²

since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:

$\beta$ ₂= $\beta$ =(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)

$\beta$ ₂=(10dB)㏒₁₀(2.04×10¹⁰)

$\beta$ =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB

3 0

3 years ago

Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o

VikaD [51]

Answer:

For the point inside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

For the point outside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

x = -x0:

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}$

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}$

We have to integrate it over the ring, which is an angular integration.

$E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}$

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

x = +x0:

We will only change the boundaries of the last integration.

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$