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3 years ago

A boulder on top of a cliff has a potential energy of 12400J. How high above is the boulder?

Physics

1 answer:

Bezzdna [24]3 years ago

6 0

Do you also know what the mass of the boulder is?

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2 years ago

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The electric potential difference across the membrane of a body cell is 0.070 V (higher on the outside than on the inside). The

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Answer:

The electric field is $8.75 \times 10^{6}~v~m^{-1}$ and the ditection is from outer to inner side of the membrane.

Explanation:

We know the electric field ( $\vec{E}$ ) is given by $\vec{E} = - \nabla V$ , 'V' being the potential.

In 1-D, it can be written as

$E=\dfrac{V}{d}$

where 'd' is the separation of space in between the potential difference is created.

Given, $V = 0.070~V~$ and the thickness of the cell membrane is $d = 8.0 \times 10^{-9}~m$ .

Therefore the created electric field through the cell membrane is

$E = \dfrac{0.07~V}{8 \times 10^{-9}~m} = 8.75 \times 10^{6}~m~s^{-1}$

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2 years ago

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Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light

professor190 [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$\theta_{max} =18.38^o$

New $n_{cladding} =1.491$

Explanation:

From the question we are told that

The refractive index of the core is $n_{core} = 1.497$

The refractive index of the cladding is $n_{cladding} = 1.421$

Generally according to Snell's law

$n_{core} * sin(90- \theta) = n_{cladding} * sin (90)$

Where $\theta_{max}$ is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection

$\theta_{max} = 90 - sin^{-1} [\frac{n_{cladding}}{n_{core}} ]$

$\theta_{max} = 90 - sin^{-1} [\frac{1.421}{1.497}} ]$

$\theta_{max} =18.38^o$

Given from the question the the largest angle is 5°

Generally the refraction index of the cladding is mathematically represented as

$n_{cladding} = n_{core} * sin (90 - 5)$

$n_{cladding} =1.491$

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2 years ago