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3 years ago

A 2 cm tall object is placed 30cm in front of a concave mirror with focal length of 25 cm. What is the height of the image?

Physics

1 answer:

viktelen [127]3 years ago

8 0

Answer:

$- 10$ cm

Explanation:

$h_{o}$ = height of the object = 2 cm

$h_{i}$ = height of the image

$d_{o}$ = distance of the object from the mirror = 30 cm

$f$ = focal length of the mirror = 25 cm

$d_{i}$ = image distance

using the mirror equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$

$\frac{1}{30} + \frac{1}{d_{i}} = \frac{1}{25}$

$d_{i} = 150$ cm

using the equation

$\frac{h_{i}}{h_{o}} = \frac{- d_{i}}{d_{o}}$

$\frac{h_{i}}{2} = \frac{- 150}{30}$

$h_{i} = - 10$ cm

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Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, $\rho = 10.49 g/cm^{3}$

Density of Pd, $\rho = 12.02 g/cm^{3}$

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Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, $\rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}$

where

$V_{c} = a^{3}$ = Volume of crystal lattice

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n = 4 = no. of atoms in FCC

Therefore,

$\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}$

Therefore, the length of the unit cell is given as:

$a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}$ (1)

Average atomic weight is given as:

$A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}$

where

$C_{Ag}$ = 79 %

$A_{Ag}$ = 107

$C_{Pd}$ = 21%

$A_{Pd}$ = 106

Therefore,

$A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78$

In the similar way, average density is given as:

$\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}$

$\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}$

Therefore, edge length is given by eqn (1) as:

$a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm$

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Hydrogen bombs and stars both produce energy with nuclear ?

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Answer:

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3 years ago

During a race, a sprinter accelerated 1.8 m/s 2 in 2.5 seconds.How many meters per second did the sprint increase with this amou

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During a race, a sprinter accelerated 1.8 m/s 2 in 2.5 seconds. The sprint increase with this amount of acceleration by 4.5 m/s.

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Acceleration is the time rate of change of velocity.

Acceleration a = velocity v / time t

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The sprint increase with this amount of acceleration by 4.5 m/s.

Learn more about acceleration.

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2 years ago

At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the l

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Answer:

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3 years ago

An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th

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Answer:

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1/4v^2 = μgx

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Now substitute x

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3 years ago