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2 years ago

6

An experimenter finds that no photoelectrons are emitted from a particular metal unless the wavelength of light is less than 295

nm. her experiment will require photoelectrons of maximum kinetic energy 2.4 ev. what frequency light should be used to illuminate the metal?

1 answer:

wariber [46]2 years ago

7 0

If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.

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A circular hole in an aluminum plate is 2.739 cm in diameter at 0.000°C. What is the change in its diameter when the temperature

prisoha [69]

Answer:

$L = 2.746 cm$

Explanation:

As we know that thermal expansion coefficient of aluminium is given as

$\alpha = 24 \times 10^{-6} per ^oC$

now we also know that after thermal expansion the final length is given as

$L = L_o(1 + \alpha \Delta T)$

here we know that

$L_o = 2.739 cm$

$\alpha = 24 \times 10^{-6}$

$\Delta T = 108 - 0= 108^oC$

now we will have

$L = 2.739(1 + 24 \times 10^{-6} (108))$

$L = 2.746 cm$

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3 years ago

a wildlife biologist examines for a genetic trait he suspects may be linked to sensitivity to industrial toxins in the environme

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The probability he finds the trait in none of the dogs

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2 years ago

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in

FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, $V=108\ cm^3$

We know that the volume of the cylinder is given by :

$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

$\dfrac{h}{r}=\dfrac{108}{\pi 1.83}$

$\dfrac{h}{r}=59 \pi$

Hence, this is the required solution.

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2 years ago

Heat is transferred from the sun to the earth via electromagnetic waves (see Chapter 24). Because of this transfer, the entropy

Sever21 [200]

Answer:

1. Decreases,

2. Increases,

3. Increases

Explanation:

The heat which is a product of sun's energy, is transferred from the sun to the earth through radiation, conduction or convention. This heat passes through the earth atmosphere, then warms it , before becoming heat energy.

Therefore, Heat is transferred from the sun to the earth via electromagnetic waves . Because of this transfer, the entropy of the sun DECREASES, the entropy of the earth INCREASES and the entropy of the sun-earth system INCREASES.

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3 years ago

What is the formula for silver nitrate

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Answer:

Explanation:

Agno3

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3 years ago