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3 years ago

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Hey, guys need your help. I do school online and I have this project due. The problem is, I didn't know I needed most of these s

upplies till now. Can somebody please test this for me and give me a good explanation of what happened so I can understand what's going on? Thanks in advance! :)

1 answer:

Alexxandr [17]3 years ago

6 0

You need to post a picture so someone can help

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

Alexus [3.1K]

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>

6 0

3 years ago

Please<br>help<br>me<br>with this problem

uranmaximum [27]

Answer:

speeding up

Explanation:

because its speeding up, theres going to be more newtons in the back

i really hope this is right, tell me if so

3 0

2 years ago

If you connected 10 of these 12V (10 W) lamps in parallel, to the 12V source, how much current would the source have to supply

Flauer [41]

Answer:

$T=8.33A$

Explanation:

From the question we are told that:

Number of battery $n=10$

Voltage source $E=12V$

Lamp Power $P=10W$

Generally the equation for Resistance is mathematically given by

$R=\frac{V^2}{P}$

$R=\frac{12^2}{10}$

$R=14.4ohms$

Therefore

$R_{eq}=\frac{14.4}{10}$

$R_{eq}=1.44$

Generally the equation for Current is mathematically given by

$T=\ffrac{V}{Req}$

$T=\frac{12}{1.44}$

$T=8.33A$

6 0

2 years ago

Which element could be described by the list of properties above?

Murljashka [212]

Answer:

list the properties

Explanation:

7 0

3 years ago

Read 2 more answers

An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t

viktelen [127]

Answer:

a) F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ ), b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

F = $k \frac{q_2q_2}{r^2}$

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron r₁ = x

right side -electron r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

F_net = k q Q ( $\frac{1}{r_1^2}+ \frac{1}{r_2^2}$ )

F _net = kqQ ( $\frac{1}{x^2} + \frac{1}{(d-x)^2}$ )

let's substitute the values

F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( $\frac{1}{x^2} + \frac{1}{(0.30-x)^2}$ )

F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given

x (m) F (N)

0.05 27.0 10-16

0.10 8.10 10-16

0.15 5.76 10-16

0.20 8.10 10-16

0.25 27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0

3 years ago