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2 years ago

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Trace the path of a ray emitted from the tip of the object through the focal point of the mirror and then the reflected ray that

results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.

1 answer:

Valentin [98]2 years ago

5 0

Answer:

1 / f = 1 / p + 1 / q

Explanation:

Mirror problems can be solved analytically with the constructor equation

1 / f = 1 / p + 1 / q

An approximate solution can also be given using graphical methods, where two of three rays must be drawn,

1) A ray that passes through the focal length and bounces horizontally

2) A ray that passes through the center of curvature (c = 2f) of the mirror and does not deviate

3) A beam that comes horizontal and bounces through the focal

The attached shows the ray that passes through the focal length and bounces horizontally.

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Define neuron and nerve<br>cell

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2 years ago

Read 2 more answers

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]

a)

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:

a)

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,

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

b)

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

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

c)

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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2 years ago

In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t

Vitek1552 [10]

Answer:

So the ratio will be $\frac{T_L}{T_H}=-0.171$

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So $Q=450j$

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency $\eta =\frac{W}{Q}=\frac{290}{450}=0.6444$

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine $=\frac{0.6444}{0.55}=1.171$

Efficiency of Carnot engine is $\eta =1-\frac{T_L}{T_H}$

$1.171 =1-\frac{T_L}{T_H}$

$\frac{T_L}{T_H}=-0.171$

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3 years ago

Read 2 more answers

Need help on physics/momentum

koban [17]

Answer:

vf=11.67m/s

Explanation:

vf:

Δp=m(vf-vi)

80=12(vf-5)

80=12vf-60

140=12vf

vf=11.67m/s

pf:

p=m*v

p=12*11.67

p=140.04 N*s

Hope this helps

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2 years ago