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2 years ago

Two flat glass surfaces are separated by a 150-nm gap of air.

Physics

1 answer:

elena-s [515]2 years ago

7 0

Answer:

a. 600 nm

b. 300 nm

Explanation:

According to the question it is given that there is two flat glass surface that is separated by 150nm air gap

Based on the above information

The computation is shown below

a. The bright thickness is

$t = \frac{\lambda}{2} (m - \frac{1}{2} )\\\\\lambda = \frac{2t}{(m - \frac{1}{2} )} \\\\\frac{2(150)}{(1 - \frac{1}{2} )} \\\\$

= 600 nm

b. The dark thickness is

$t = \frac{m\lambda }{2} \\\\\lambda = \frac{2t}{m} \\\\\frac{2(150)}{1}$

= 300 nm

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A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does th

NeX [460]

(a) The diameter of the largest orbit just before the protons exit the cyclotron is 39 cm.

(b) The number of orbits completed by the proton during this 1.0 ms is 14000 revolutions.

The energy that an object has as a result of motion is known as kinetic energy. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.

K.E = 1/2mv²

mv² = 2 K.E

v = sqrt( 2*KE / M)

(a) The KE of the medical isotopes = 6.5 MeV

v = sqrt( 2* 6.5* 1.6* 10^-13 / 1.67* 10^-27 )

v = 3.53 × 10⁷ m/s

Now the centripetal force:

mv² / R = qvB

r = q*v / mv²

r = ( 1.67* 10^-27 * 3.53 × 10⁷ ) / ( 1.9* 1.6* 10^-19 )

r = 0.19415 m

diameter d = 2r,

d= 2(0.19415 m)

d= 0.39 m ≅ 39 cm

(b) The time period to complete a revolution around the spiral trajectory is:

T = 2πr / v

T = 2*3.14* 0.1941 / 3.53*10^7

T = 0.7 × 10⁻⁷ s

Thus, the number of orbits that the proton does to complete the revolution in 1 ms is:

n = t / T

n = 10^-3 / (0.7*10^-7)

n = 14285.71 ≅ 14000 revolutions

The complete question is :

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is 1.2 T.

a. What is the diameter of the largest orbit, just before the protons exit the cyclotron?

b. A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1.0 ms?

To know more about cyclotron refer to: brainly.com/question/14985809

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2 years ago

A 15.0 mW laser puts out a narrow beam 2.00 mm indiameter.

Murljashka [212]

Answer:

1341.03 V/m

Explanation:

The power output per unit area is the intensity and also the is the magnitude of the Poynting vector.

$S = \frac{P}{A}$ = cε₀ $E^{2} _{rms}$

⇒ $\frac{P}{A}$ = cε₀ $E^{2}_{rms}$

Where;

P is the power output

A is the area of the beam

c is speed of light

ε₀ is permittivity of free space 8.85 × 10⁻¹² F/m

$E_{rms}$ is the average (rms) value of electric field

Making electricfield $E_{rms}$ the subject of the equation

$E^{2}_{rms}$ = P / Acε₀

$E_{rms}$ = √(P / Acε₀)

But area A = πr²

$E_{rms}$ = √(P / πr²cε₀)

Given:

Output power, P = 15 mW = 0. 015 W

Diameter, d = 2 mm = 0.002 m

⇒ Radius, $r = \frac{d}{2} = \frac{0.002}{2} = 0.001 m$

Solving for average (rms) value of electric field;

$E_{rms} = \sqrt{\frac{0.015 W}{\pi * (0.001 m)^2 * (3 * 10^8 m/s) * (8.85 * 10^-12) C^2/Nm^2} }$

$E_{rms}$ = 1341.03 V/m

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3 years ago

you ask people to rate there depression on a scale of 1 to 10 their respnses are a (n)___in your study. a.variable b.discrete v

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The answer is A) variable. The depression level is a value that varies with the respondents and may inform you about the relationship between depression and other variables, such as alcohol use or sleep habits. It is also likely a discrete variable, since the answer is probably an integer (1, 2, ... 10), rather than a continuous spectrum (virtually any number between 1 and 10).

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3 years ago

A small bead of mass m is constrained to slide without friction inside a circular vertical hoop of radius r which rotates about

Stolb23 [73]

Answer:

Explanation:

The bead is moving on a vertical circular path so it must have a centripetal force towards the centre.

This force is equal to m v² / r

v is velocity of bead and r is radius of the circular path.

The vertical hoop is also rotating about a vertical axis passing through the centre at frequency f so the bead will experience a cetrifugal force due to rotation of the hoop. Its value is

m ω² r . Only at the point o degree and 180 degree , these forces are opposite to each other so at these points , the bead will be in equilibrium .

mv² / r = m ω² r

v² = ω² r²

v = ω r

= 2π f r

= 2 x 3.14 x 2 x 0.22

v = 2.76 m /s

For the bead to rise upto point θ = 90 degree , height achieved is radius R

required velocity = √ 2gR

= √ 2x 9.8x.22

= 2.076 m/s

This velocity is less than the velocity calculated earlier so the bead will be able to ride the required height.

v = 2.76 m/s

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3 years ago

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shepuryov [24]

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3 years ago