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Lady bird [3.3K]
3 years ago
6

- A thin film of oil * (n = 1.45) on a puddle of water, producing different colors. What is the minimum thickness of a place whe

re the oil creates constructive interference for light with a wavelength equal to 545 nm?

Physics
1 answer:
Darya [45]3 years ago
4 0

Answer:

There will be a phase change at the first interface and no phase change at the second interface:

If the film is 1/4 wavelength thick this restriction will hold

The wavelength of the light in oil is 545 nm / 1.45 = 376 nm

376 nm / 4 = 94 nm

"D" is correct

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Suppose I have a vector that is 7 units long and that makes an angle of +30 degrees from the positive x-axis. I want to add to t
Vinil7 [7]

Answer:

sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

Explanation:

We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis

So x component of the vector =7cos30^{\circ}=7\times 0.866=6.06

y component of the vector =7sin30^{\circ}=7\times 0.5=3.5

So vector will be 6.06i+3.5j

Now other vector of length of 7 units and makes an angle of 120° with positive x-axis

So x component of vector  =7cos120^{\circ}=7\times -0.5=-3.5i

y component of the vector =7sin120^{\circ}=7\times 0.866=6.06j

Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

5 0
4 years ago
show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through t
timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

              (b) r_{p}= 9.95 x 10⁻¹ m

              (c) ratio = 1800

Explanation: (a) <u>Potential</u> <u>difference</u> is defined as the energy a charged particle has between two points in a circuit. It is calculated as

\Delta V=\frac{pe}{q}

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = \frac{1}{2}m.v^{2}

so

V.q=\frac{1}{2} m.v^{2}

V=\frac{m.v^{2}}{2q}

Calculating:

V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}

V = -109.44 x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is -109.44 x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

F_{mag} = F_{c}

qvB=\frac{mv^{2}}{r}

r=\frac{mv}{qB}

The radius of the curvature, for a proton, will be:

r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

(c) Radius of curvature, if it was a electron:

r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}

r = 54.33 x 10⁻⁵m

ratio = \frac{9.95.10^{-1}}{54.33.10^{-5}}

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

5 0
4 years ago
I’ll name u BRAINLIEST if u get this right
nekit [7.7K]
V = m1 u1 - m2 u2 / m1
v = 0.01 * 500 - 2 * 1.4 / 0.01
v = 220 m/s
7 0
3 years ago
A carnival ride has a 2.0m radius and rotates once each .90s. Find the centripetal acceleration
Paraphin [41]
We know that centripetal acceleration is nothing but the ratio of the square of the tangential velocity to that of the radius vector.
a=v*v/r=ωωrr/r
=ωωr
=2πf2πfr
=2π2πr/TT
=97m/secsec
3 0
3 years ago
The greater the mass of an object.
yKpoI14uk [10]

Answer:

Mass is the measure of an object's matter (what it's made up of). The greater an object's mass, the greater its gravitational force. The earth has a strong attracting force for objects with smaller mass (including the moon), and the sun has an attracting force on the earth and other planets in our solar system.

7 0
3 years ago
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