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hichkok12 [17]
3 years ago
10

A laser used for many applications of hard surface dental work emits 2780-nm wavelength pulses of variable energy (0-300 mJ) abo

ut 20 times per second.part a. Determine the number of photons in one 80-mJ pulse.part b. Determine the average power of photons in one 80-mJ pulse during 1 s.
Physics
1 answer:
Bess [88]3 years ago
6 0

Answer:

a

    n =  1.119 *10^{18} \ photons

b

  P  =  1.6 \ W

Explanation:

From the question we are told that

    The wavelength is  \lambda  =  2780 nm =  2780 *10^{-9} \ m

     The  energy  is  E =  80 mJ  =  80 *10^{-3} \ J

This energy is mathematically represented as

     E   = \frac{n  *  h *  c }{\lambda }

Where  c is the speed of light with a value  c =  3.0 *10^{8} \ m/s

             h is the Planck's  constant with the value  h  =  6.626 *10^{-34} \ J \cdot s

             n is the number of pulses

So

      n =  \frac{E * \lambda }{h * c }

substituting values

       n =  \frac{80 *10^{-3} *  2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }

       n =  1.119 *10^{18} \ photons

Given that the pulses where emitted 20 times in one second then the period of the pulse is

       T  =  \frac{1}{20}

      T = 0.05 \ s

Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as

       P  =  \frac{E}{T}

substituting values

       P  =  \frac{ 80 *10^{-3}}{0.05}

        P  =  1.6 \ W

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navik [9.2K]

Answer:

V = IR

R = V/I

I = V/R

Explanation:

7 0
2 years ago
Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it
d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s,  1)    v = \sqrt{0.55^2 + 19.6 y},  2)  v = 2.05 m / s,

3)  d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

         Q = v A

where Q is the flow rate, A is area and v is the velocity

         

the area of ​​a circle is

        A = π r²

radius and diameter are related

        r = d / 2

substituting

       A = π d²/4

       Q = π/4   v d²

let's reduce the magnitudes

       v = 0.55 m / s = 55 cm / s

let's calculate

       Q = π/4   55   1.96²

       Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

        v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

        v = \sqrt{0.55^2 + 2  \ 9.8\  y}

        v = \sqrt{0.55^2 + 19.6 y}

2) ask to calculate the velocity for y = 0.2 m

        v = \sqrt{0.55^2 + 19.6 \ 0.2}

        v = 2.05 m / s

3) We write the continuous equation for this point 2

        Q = v₂ A₂

        A₂ = Q / v₂

let us reduce to the same units of the SI system

        Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

        A₂ = 165.95 10⁻⁶ / 2.05

        A₂ = 80,759 10⁻⁶ m²

area is

        A₂ = π/4   d₂²

        d₂ = \sqrt{4  A_2 / \pi }

        d₂ = \sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }

        d₂ = 10.14 10⁻³ m

        d₂ = 1.014 cm

4 0
3 years ago
Amy has a mass of 30 kg, and she is riding a skateboard traveling 5 m/s. What is her momentum
soldi70 [24.7K]
We Know, P = m*v
Here, m = 30 Kg
v = 5 m/s

Substitute it into the expression, 
P = 30*5 Kgm/s
P = 150 Kgm/s

So, your final answer is 150 Kg.m/s

Hope this helps!
3 0
3 years ago
Read 2 more answers
A 2290 kg car traveling to the west at 22.3 m/s slows down uniformly. How long would it take the car to come to a stop if the fo
Elina [12.6K]

Answer:

5.72 s

Explanation:

From Newton's law, F = ma

The East is +ve direction, Hence,

F = +8930 N

m = 2290 kg

a = ?

8930 = 2290 × a

a = 8930/2290 = 3.90 m/s²

So, we will find the time it takes the car to stop using the equations of motion

a = 3.90 m/s²

u = initial velocity of the car = - 22.3 m/s (the velocity is to the west)

v = final velocity of the car = 0 m/s (since the car comes to rest)

t = time taken for the car to come to rest = ?

v = u + at

0 = - 22.3 + (3.90)(t)

3.9t = 22.3

t = 5.72 s

5 0
3 years ago
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Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
kumpel [21]

Answer:

7800kg/m³

Explanation:

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

Given the density of iron = 7.8 g/cm3.

The SI units must be in kg/m³

7.8g = 7.8/1000 kg

7.8g = 0.0078kg

1cm³ = 0.000001m³

7.8g/cm³

= 0.0078/0.000001 kg/m³

= 7800kg/m³

Hence the density in SI unit is 7800kg/m³

4 0
3 years ago
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