Greg is giving a slide show presentation to a large audience. How might a laser best help him with the presentation?

Physics

2 answers:

gizmo_the_mogwai [7]3 years ago

3 0

Explanation :

The full form of laser is Light Amplification by Stimulated Emission of Radiation. It has particles of very high energy. It points in one direction.

Greg is giving a slide show presentation to a large audience. It is very important to make the presentation informative. By using a laser, we can focus on the important part of the presentation. So that we can highlight the important points.

Bogdan [553]3 years ago

3 0

Answer: When giving a slide show presentation we usually use a projector and a projection screen to show it.

Sometimes, Greg will need to "point" to some graph, image, equation, or something like that in order to drag the audience's attention in such a part.

If Greg went in front of the screen to signalize the part he wants, he will project a shadow on the screen, and this may be troublesome for some of the audience.

A laser comes handy here because Greg does not even need to move to point the laser in the part of the screen he wants, and it does not project a shadow or anything in the screen (The laser emits a "column" of coherent photons, that don't interact with the light of the projector), so it's very useful to signalize points as he speaks, which also allows him to not lose the flow of the presentation.

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laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m$

Distance from second drop is 0.83999 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m$