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4 years ago

O que é um espelho ?

Physics

2 answers:

AveGali [126]4 years ago

6 0

Um espelho é uma superfície reflexiva que reflete uma imagem clara.

yKpoI14uk [10]4 years ago

3 0

<span>medio espejo-</span> una superficie reflectante, ahora por lo general de vidrio cubierto con una amalgama de metal, que refleja una imagen clara

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Which of the following is the brightness of a star as we see if from Earth?

My name is Ann [436]

I think that it is apparent magnitude

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3 years ago

If the person starts over and moves his hand more quickly the waves will have a

Mama L [17]

Answer:

larger push force

Explanation:

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4 years ago

the fundamental frequency for the 3rd chord on a five-string guitar is 240 Hz. what frequency would produce the 10th harmonic?

Mamont248 [21]

The frequency of the 10th harmonic is 800 Hz

Explanation:

The frequency of the nth-harmonic for the standing waves in a string is given by the equation

$f_n = n f_1$

where

$f_1$ is the fundamental frequency of the string

In this problem, we are given the frequency of the 3rd harmonic:

$f_3 = 240 Hz$

Which can be rewritten in terms of the fundamental frequency

$f_3 = 3 f_1$

So we find $f_1$ :

$f_1 = \frac{f_3}{3}=\frac{240}{3}=80 Hz$

Now that we have the fundamental frequency, we can find the frequency of the 10th harmonic, with n = 10 :

$f_{10} = 10f_1 = (10)(80)=800 Hz$

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4 years ago

A hoop, a solid cylinder, a spherical shell, and a solid sphere are placed at rest at the top of an inclined plane. All the obje

iogann1982 [59]

Answer:

Solid sphere will reach first

Explanation:

When an object is released from the top of inclined plane

Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane

initial gravitational potential energy = final total kinetic energy

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now we have

$I = mk^2$

here k = radius of gyration of object

also for pure rolling we have

$v = R\omega$

so now we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})$

$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$

$v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}$

so we will say that more the value of radius of gyration then less velocity of the object at the bottom

So it has less acceleration while moving on inclined plane for object which has more value of k

So it will take more time for the object to reach the bottom which will have more radius of gyration

Now we know that for hoop

$mk^2 = mR^2$

k = R

For spherical shell

$mk^2 = \frac{2}{3}mR^2$

$k = \sqrt{\frac{2}{3}} R$

For solid sphere

$mk^2 = \frac{2}{5}mR^2$

$k = \sqrt{\frac{2}{5}} R$

So maximum value of radius of gyration is for hoop and minimum value is for solid sphere

so solid sphere will reach the bottom at first

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3 years ago

lawyer [7]

Answer:

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3 years ago