1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nataly862011 [7]
2 years ago
8

A very long conducting cylinder (length L) of radius R(R<R and (b)r

Physics
1 answer:
mina [271]2 years ago
4 0

Answer:

Explanation:

We have to find electric potential V at a distance r.

a) For r>R,

The electric field in the cylinder is given by

E.A equating it to the other electric field given by

б.A/ε₀

Here the area of cylinder is given by= 2*3.14*r*L

While for the outside, the area= 2*3.14*R*L

Equating both, we get

E= бR/rε₀

Now,

The potential difference is given as:

ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.

Where ΔV= V₀-V

So solving we get

V₀=V-бR/ε₀ln (r/R)

b) For r<R i.e. inside the cylinder

There will be no electric field produced as E=0

So ultimately Vin= V

c) V=0 at r= infinity.

You might be interested in
For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t
Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at (\frac{35}{2},0)

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

y=2+\frac{1}{x}

y=2+x^{-1}

y'=-x^{-2}

y'=-\frac{1}{x^{2}}

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

y'=-\frac{1}{x^{2}}

y'=-\frac{1}{(1)^{2}}

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

y-y_{1}=m(x-x_{1})

y-3=-1(x-1})

y-3=-1x+1

y=-x+1+3

y=-x+4

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

-x+4=0

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

y'=-\frac{1}{x^{2}}

y'=-\frac{1}{(2)^{2}}

m=y'=-\frac{1}{4}

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

y-y_{1}=m(x-x_{1})

y-2.5=-\frac{1}{4}(x-2})

y-2.5=-\frac{1}{4}x+\frac{1}{2}

y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}

y=-\frac{1}{4}x+3

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

-\frac{1}{4}x+3=0

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

y'=-\frac{1}{x^{2}}

y'=-\frac{1}{(2.5)^{2}}

m=y'=-\frac{4}{25}

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

y-y_{1}=m(x-x_{1})

y-2.4=-\frac{4}{25}(x-2.5})

y-2.4=-\frac{4}{25}x+\frac{2}{5}

y=-\frac{4}{25}x+\frac{2}{5}+2.4

y=-\frac{4}{25}x+\frac{14}{5}

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

-\frac{4}{25}x+\frac{14}{5}=0

and solve for x

x=\frac{35}{20}

so, when fired from (2.5, 2.4) the rocket will hit the target at (\frac{35}{2},0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

y'=-\frac{1}{x^{2}}

y'=-\frac{1}{(4)^{2}}

m=y'=-\frac{1}{16}

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

y-y_{1}=m(x-x_{1})

y-2.25=-\frac{1}{16}(x-4})

y-2.25=-\frac{1}{16}x+\frac{1}{4}

y=-\frac{1}{16}x+\frac{1}{4}+2.25

y=-\frac{1}{16}x+\frac{5}{2}

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

-\frac{1}{16}x+\frac{5}{2}=0

and solve for x

x=40

so, when fired from (4,2.25) the rocket will hit the target at (40,0)

I uploaded a graph that represents each case.

8 0
3 years ago
Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45
victus00 [196]
<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>
4 0
3 years ago
If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :
gizmo_the_mogwai [7]

Let's see

Momentum be P

\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c

\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c

\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c

\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}

\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c

On comaparing

  • c=1

So

  • b-3c=1
  • b-3=1
  • b=1+3
  • b=4

and

  • -a-b=-1
  • -a-4=-1
  • -a=-1+4=3
  • a=-3

So the unit is

  • DV⁴/F³
5 0
2 years ago
Read 2 more answers
if bananas are curved because they grow towards the sun ☀️so say if a was born towards the sun would i have been a curved baby o
Ivenika [448]

ಠ_ಠ Hey, hang on.. you might've made a discovery. Nobody has tested it so how do we know? ಠ_ಠ

3 0
3 years ago
Read 2 more answers
During which segments is kinetic energy decreasing?<br> A)1,2,3<br> B) 4,5<br> C)1,3,5<br> D)2,4
Nutka1998 [239]

Answer:

c

Explanation:

kinetic energy is energy an object has due to its movement. for instance, if someone was riding down a hill, when the motion of the bike begins to decrease so does the kinetic energy

3 0
3 years ago
Read 2 more answers
Other questions:
  • A 2.0-kg mass (mA) and a 4.0-kg mass (mB) are connected to a lightweight cord that passes over a frictionless pulley. The pulley
    10·1 answer
  • Most often frequency is measured by the number of waves that pass a point in one blank
    11·1 answer
  • Flowchart symbols
    11·1 answer
  • What will the motion of the cart be like when there is no force at all ? ( There is no friction in this model. )
    10·1 answer
  • The subject of the second part of messiah is:
    10·1 answer
  • How many regions does the deltoid muscle have
    5·2 answers
  • (II) A sled is initially given a shove up a frictionless 23.0° incline. It reaches a maximum vertical height 1.22 m higher than
    5·1 answer
  • What is true about empirical and molecular formulas?
    7·1 answer
  • What is most likely the author’s motive for writing this article?
    9·2 answers
  • A golfer hits a ball and gives it an initial velocity of 40 m/s , at an angle of 30 degrees above the horizontal.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!