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3 years ago

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When an object with a height of 0.10 meter is placed at a distance of 0.20 meter from a convex spherical mirror, the image will

appear to be 0.06 meter behind the mirror. What’s the height of the image?

1 answer:

goblinko [34]3 years ago

5 0

Answer:

0.03 m

Explanation:

Mirrors work on the principle of reflection.

Reflection occurs when a ray of light hits a surface and bounces back into the origina medium at a different angle.

We can solve this problem by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

y = 0.10 m is the height of the object

p = 0.20 m is the distance of the object from the mirror

q = -0.06 m is the distance of the image from the mirror (negative because it appears behind the mirror, so it is a virtual image)

Solving for y', we find the size of the image:

$y'=-\frac{q}{p}y=-\frac{-0.06}{0.20}(0.10)=0.03 m$

And the positive sign indicates that the image is upright.

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A proton moves from a location where V = 87 V to a spot where V = -40 V. (a) What is the change in the proton's kinetic energy?

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Answer: a) 127 eV; b) there is no change of kinetic energy.

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If we replace the proton by a electron we have a very different situaction, the electrons are located in a lower potental then they can not move to higher potential if any external force does work on the system.

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a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

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Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

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In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$

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The total acceleration is the resultant of the centripetal acceleration ( $a_c$ ) and the tangential acceleration ( $a_t$ ):

$a=\sqrt{a_c^2+a_t^2}$

We know that:

a = 4.4g

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So, we can find the tangential acceleration:

$a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2$

The angular acceleration is related to the tangential acceleration by

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where r = 10.9 m is the length of the centrifuge. Substituting,

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