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2 years ago

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A 1050 W carbon-dioxide laser emits light with a wavelength of 10μm into a 3.0-mm-diameter laser beam. What force does the laser

beam exert on a completely absorbing target?

1 answer:

avanturin [10]2 years ago

3 0

The force exerted by the laser beam on a completely absorbing target is $3.5 \times 10^{-6} \ N$ .

The given parameters;

<em>power of the laser light, P = 1050 W</em>
<em>wavelength of the emitted light, λ = 10 μm </em>

The speed of the emitted laser light is given as;

v = 3 x 10⁸ m/s

The force exerted by the laser beam on a completely absorbing target is calculated as follows;

P = Fv

$F = \frac{P}{v} \\\\F = \frac{1050}{3\times 10^8} \\\\F = 3.5 \times 10^{-6} \ N$

Thus, the force exerted by the laser beam on a completely absorbing target is $3.5 \times 10^{-6} \ N$ .

Learn more here:brainly.com/question/17328266

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A. When it is in a magnetic field, it becomes a temporary magnet.

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An iron bolt is attracted to a magnet because when in a magnetic field, the iron becomes a temporary magnet.

This is because the iron aligns their electrons in the magnetic fields.

This causes that attraction between the magnet and the iron.
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Which characteristic must a food have to receive a “natural” label from the FDA ?

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3 years ago

0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distanc

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3 years ago

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

B)

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

C)

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

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3 years ago

Two identical particles of charge 6 μμC and mass 3 μμg are initially at rest and held 3 cm apart. How fast will the particles mo

krek1111 [17]

Answer:

Explanation:

The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .

Their potential energy at distance d

= kq₁q₂ / d

= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J

= 16.2 J

Their total kinetic energy will be equal to this potential energy.

2 x 1/2 x mv² = 16.2

= 3 x 10⁻⁶ v² = 16.2

v = 5.4 x 10⁶

v = 2.32 x 10³ m/s

When masses are different , total P.E, will be divided between them as follows

K E of 3 μ = (16.2 / 30+3) x 30

= 14.73 J

1/2 X 3 X 10⁻⁶ v₁² = 14.73

v₁ = 3.13 x 10³

K E of 30 μ = (16.2 / 30+3) x 3

= 1.47 J

1/2 x 30 x 10⁻⁶ x v₂² = 1.47

v₂ = .313 x 10³ m/s

3 0

3 years ago