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2 years ago

What type of mirror can produce both converging and diverging rays?

Physics

2 answers:

Ivahew [28]2 years ago

4 0

Answer:

A convex mirror is a diverging mirror (f is negative) and forms only one type of image. It is a case 3 image—one that is upright and smaller than the object, just as for diverging lenses.

Explanation:

hope this helps have a good night :)

belka [17]2 years ago

4 0

Convex mirror
i hope this helps :) i learned about this last year

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What are successfulness of the Competition policy in South Africa?

Mars2501 [29]

Answer:

The product choices along with its competitive prices were provided to the consumers. Practices such as horizontal collusion and resale price maintenance was declared unlawful in 1984. Prevention of monopoly growth was the aim of the competition policy act.

Explanation:

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3 years ago

An astronaut is walking in space. Which of these would have the greatest speed as observed by the astronaut?

Aleks04 [339]

The answer is C) an electromagnetic wave

An electromagnetic wave, which includes electromagnetic radiation such as visible light, moves the fastest of all of the options listed by a significant margin, especially through space. In fact, light travelling through space is technically the theoretical limit of how fast something can travel.

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3 years ago

Read 2 more answers

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

2 years ago

Read 2 more answers

How to answer this question?

laila [671]

Answer:

measure the vector diagram first

5 0

2 years ago

The process that enhances some properties of an object at the expense of other properties is called:

lidiya [134]

Answer:

The answer is: letter a, pop-out effect.

Explanation:

The "pop-out effect" is a phenomenon which allows the person's precognitive processes to detect a<em> visual stimulus that is potentially the most meaningful one</em> in a person's spatial field of attention. The pop-up effect occurs when a person distinguishes one object from the rest.

For example, when a child chooses among pictures in different colors, it is common for the child to point at colored pictures rather than grayscale pictures. This is an example of a pop-out effect. <u>The properties of the colored pictures is more preferred by the child thus, causing him not to choose or mind the grayscale images.</u>

Thus, this explains the answer.

6 0

3 years ago