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3 years ago

In a chemical reaction, which is the MOSTLY LIKELY reactant for these products? A:3O2 B:CO2 C: 2KCl D:CaCO3

Physics

2 answers:

sammy [17]3 years ago

7 0

The answer is A : 302 I really hope this helps!

fomenos3 years ago

5 0

The chemical reaction mostly likely reactant of those products is A: 302

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A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

And so, for r<a,

$V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

- Potential at the surface of the outer cylinder:

$V(b)=0$

Therefore, the potential difference is simply equal to

$V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

The electric field is just the derivative of the electric potential:

$E=-\frac{dV}{dr}$

so we can find it by integrating the expression for the electric potential. We find:

$E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

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7 0

3 years ago

A 4113 N piano is to be pushed up a(n) 3.99 mfrictionless plank that makes an angle of 25.5â—¦with the horizontal.Calculate the

Serhud [2]

Answer:

W = 14.8 kJ

Explanation:

W = F S cos ∅

W = 4113 x 3.99 x cos 25.5

W = 16410.87 x 0.9025 = 14810.8 J or 14.8 kJ

3 0

3 years ago

What hazard is associated with ionizing radiation?

Talja [164]

Acute health effects such as skin burns or acute radiation syndrome can occur when doses of radiation exceed certain levels.

8 0

3 years ago

Read 2 more answers

Radio waves, visible light, and x-rays are exempted of electromagnetic waves that always differ from each other in?

leva [86]

I think wavelength is the correct answer. Correct me if Im wrong. Hope this helps!

6 0

4 years ago

Classify each of the following statements as a characteristic (a) of electric forces only, (b) of magnetic forces only, (c) of b

ioda

A- electric forces only - WHEN The force exerted on a stationary charged object is nonzero.

B- magnetic forces ONLY - The magnitude of the force depends on the charged object's direction of motion.

C- both electric and magnetic forces - The force is proportional to the magnitude of the charge of the object on which the force is exerted.

D- neither electric nor magnetic forces- The force exerted on a moving charged object is zero.

(vi)- The force exerted on a charged object is proportional to its speed- magnetic forces only

To know more about magnetic forces visit : brainly.com/question/10353944

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4 0

1 year ago