Direct current made it possible to distribute electric power over greater areas.

Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m

Leno4ka [110]

Answer: $1.593*10^{17} photons$ released if the laser is used 0.056 s during the surgery

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

$E =\frac{hc}{\lambda}$

Where $\lambda$ is the wavelength, $h$ is the Planck's constant and $h$ is the speed of light

$h = 6.626*10^{-34} \frac{m^{2} kg }{s}$ -> Planck's constant

$c = 3*10^{8} \frac{m}{s}$ -> Speed of light

So, replacing in the equation:

$E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}$

Then, the energy of each released photon by the laser is:

$E = 3.867*10^{-19} \frac{J}{photons}$

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

$\frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}$

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

$2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}$

And by doing a simple rule of three, if $2.844*10^{18} photons$ are released every second, then in 0.056 s:

$0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons$ are released during the surgery

5 0

3 years ago

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how

liraira [26]

Answer:

A

$x = 0.198456 \ m$

B

$h = 1.3061 \ m$

C

$v = 5.06 \ m/s$

D

$d = 4.0273 \ m$

Explanation:

Considering the first question

From the question we are told that

The spring constant is $k = 32.50 N/m$

The potential energy is $PE = 0.640 \ J$

Generally the potential energy stored in spring is mathematically represented as $PE = \frac{1}{2} * k * x^2$

=> $0.640= \frac{1}{2} * 32.50 * x^2$

=> $x = \sqrt{0.03938}$

=> $x = 0.198456 \ m$

Considering the second question

From the question we are told that

The mass of the dart is m = 0.050 kg

Generally from the law of energy conservation

$PE = mgh$

=> $0.640 = 0.050 * 9.8 * h$

=> $h = 1.3061 \ m$

Considering the third question

The height at which the dart was fired horizontally is $H = 3.90\ m$

Generally from the law of energy conservation

$PE = KE$

Here KE is kinetic energy of the dart which is mathematical represented as

$KE = \frac{1}{2} * mv^2$

=> $0.640 = \frac{1}{2} * 0.050 * v^2$

=> $v^2 = 25.6$

=> $v = 5.06 \ m/s$

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

$t = \frac{ 2 * H }{g}$

=> $t = \frac{ 2 * 3.90 }{9.8 }$

=> $t = 0.7959 \ s$

Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

$d = v * t$

=> $d = 5.06 * 0.7959$

=> $d = 4.0273 \ m$

5 0

2 years ago

Find the current i in:

lawyer [7]

Answer:

I = 24 A

Explanation:

This is Parallel Circuit and it is the first principle of parallel circuit that voltage will be equal in all components in the circuit

It includes 10 resistors Therefore the voltage across,

R1 = R2 = R3 = R4 = R5 = R6 = R7 = R8 = R9 = R10 = voltage in battery

We will apply Ohm's Law to each resistor to find its current because we know the voltage across each resistor is 12 V and the resistance of each resistor is 5Ω

I (R1) = E (R1) / R1

I (R1) = 12v / 5Ω

I (R1) = 2.4 A

The value resistance E of all resistors are same therefore by applying the formula above the value of current in all resistors will be 2.4 A

The Total current in the circuit will be

I (total) = I (1) + I (2) + I (3) + I (4) + I (5) + I (6) + I (7) + I (8) + I (9) + I (10)

I (total) = 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4

I (total) = 24 A

5 0

3 years ago

What happens when infrared hits a shiny surface

saw5 [17]

Will reflect ........

3 0

3 years ago

How does quantum mechanics relate to the universe?

Rom4ik [11]

It is the description of all the microscopic objects in everything in the univverse

4 0

3 years ago