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jekas [21]
3 years ago
15

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

le light with no greater than 1.0 mW total power. They're relatively safe because the eye's blink reflex limits exposure time to 250 ms.
Requried:
a. Find the intensity of a 1-mW class 2 laser with beam diameter 2.0 mm .
b. Find the total energy delivered before the blink reflex shuts the eye.
c. Find the peak electric field in the laser beam.
Physics
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

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fomenos

Answer:

The horizontal distance is 0.64 m.

Explanation:

Initial velocity, u =2.5m/s

The maximum horizontal distance is

R = \frac{u^2}{2g}\\\\R = \frac{2.5\times 2.5}{9.8}\\\\R = 0.64 m

3 0
3 years ago
A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The
VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

\dfrac{1}{2}mu^2 = m g d sin \theta

u^2 = 2 g d sin30^0

u= \sqrt{2\times 9.8 \times 10 sin30^0}

u = 9.89 m/s

Using second law of motion  

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,  

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

3 0
3 years ago
LVULAN
3241004551 [841]

For this case we have that by definition, the kinetic energy is given by the following formula:

k= \frac {1} {2} * m * v ^ 2

Where:

m: It is the mass

v: It is the velocity

According to the data we have to:

m = 100 \ kg\\v = 9 \frac {m} {s}

Substituting the values we have:

k = \frac {1} {2} * (100) * (9) ^ 2\\k = \frac {1} {2} * (100) * 81\\k = 50 * 81\\k = 4050

finally, the kinetic energy is 4050 \ J

Answer:

Option A

7 0
3 years ago
Which of the following is numerically the same as the specific gravity? Mass Weight Density Volume
max2010maxim [7]
The answer is density

4 0
3 years ago
Read 2 more answers
Please help!
Trava [24]

Answer:

3 N to the right

Explanation:

There are two forces acting on the car:

- A force of 10 N towards the right

- A force of 7 N towards the left

Therefore, the net force is given by the difference between the two, since they are in opposite directions:

F=10 N-7 N=3 N

And the direction is to the right, since the force to the right has greater magnitude than the force to the left.

7 0
3 years ago
Read 2 more answers
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