The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away from the slit, when light of wavelength 530 nm is used. find the slit width.

Answer:

The force exerted by the child is 460.8 N

Explanation:

We have given that pressure [tex]P=1.2\times 10^5pa[/tex]

Radius of the circular piston r = 0.035 m

So area [tex]A=\pi r^2=3.14\times 0.035^2=0.003846m^2[/tex]

We have to find the force exerted by the child

We know that force is given by

[tex]Force=pressure\times area[/tex]

So force will be [tex]Force=1.2\times 10^5\times 0.00384=460.8N[/tex]

So the force exerted by the child is 460.8 N

Pressure

The definition of pressure is given as the force per unit area

The unit of pressure is Newton

Explanation:

Given data

Pressure = 1.2 × 105 pa

diameter = 0.035 m

We know that the expression for pressure is given as

Pressure = Force/Area

Making Force the subject of the formula

Force = Pressure * Area

But we do not know the area

Now let us find the area

Area = πd^2/4

Area= (3.142*(0.035)^2)/4

Area = 3.142*0.001225/4

Area= 0.00384895/4

Area= 0.000962 m^2

Force = 1.2 × 10^5*0.000962

Force = 115.44 Newton

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Answer:

1.6 x 10^-4  Ω

Explanation:

The photoelectric threshold wavelength for this copper surface is 264 nm

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

[tex]\large {\boxed {E = h \times f}}[/tex]

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]

[tex]\large {\boxed {E = qV + \Phi}}[/tex]

E = Energi of A Photon ( Joule )

m = Mass of an Electron ( kg )

v = Electron Release Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Charge of an Electron ( Coulomb )

V = Stopping Potential ( Volt )

Let us now tackle the problem !

Given:

λ = 254 nm = 2,54 × 10⁻⁷ m

V = 0.181 Volt

c = 3 × 10⁸ m/s

h = 6.63 × 10⁻³⁴ Js

q = 1.6 × 10⁻¹⁹ C

Unknown:

λ₀ = ?

Solution:

[tex]E = qV + \Phi[/tex]

[tex]h f = qV + h f_o[/tex]

[tex]h \frac{c}{\lambda} = qV + h \frac{c}{\lambda_o}[/tex]

[tex]6.63 \times 10^{-34} \times \frac{3 \times 10^8}{2.54 \times 10^{-7}} = 1.6 \times 10^{-19}(0.181) + 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{\lambda_o}[/tex]

[tex]7.83 \times 10^{-19} = 2.896 \times 10^{-20} + \frac{1.989 \times 10^{-25}}{\lambda_o}[/tex]

[tex]\frac{1.989 \times 10^{-25}}{\lambda_o} = 7.83 \times 10^{-19} - 2.896 \times 10^{-20}[/tex]

[tex]\lambda_o \approx 2.64 \times 10^{-7} ~ m[/tex]

[tex]\large {\boxed{\lambda_o \approx 264 ~ nm} }[/tex]

Grade: College

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light

In a parallel circuit, the voltage remains constant throughout. An analogy using a bike path explanation further simplifies this concept.

In the world of physics, it is true to say: 'Voltage remains the same throughout a parallel circuit'. To further illustrate this concept, consider a bike path as an analogy for a circuit.

If a parallel circuit was a bike path, then each 'fork' in the path represents a separate path for the electricity. Think of the voltage as a measure of energy.

Even though cyclists (or charges) can take different paths (resistors), they all start at the same height (self-same voltage).

Therefore, they all have the same amount of potential energy (voltage) at any point.

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The magnetic flux through each loop of a coil can be calculated using the given expression. The magnetic moment of the coil can be found by multiplying the current and the area of the loop. Faraday's law states that a changing magnetic flux induces an electromotive force in a circuit.

The magnetic flux through each loop of a 75-loop coil can be calculated using the expression: (8.8t−0.51t³)×10−²t⋅m², where t represents time in seconds. The magnetic flux is a measure of the magnetic field passing through a surface. It can also be calculated using the product of the magnetic field strength and the area of the loop.

The magnetic moment of the coil can be found using the formula μ = IA, where I is the current and A is the area of the loop. Substituting the given values, we find that the magnetic moment equals 2.5 × 10⁻⁶ A·m².

Faraday's law states that a changing magnetic flux through a loop induces an electromotive force (emf) in the circuit. Therefore, if the current through the coil varies with time, the magnetic flux will change accordingly and an emf will be induced in the circuit.

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The energy of a photon with a frequency of 6.0 x 1020 Hz is 3.976 x 10-13 joules.

To find the energy of a photon, we can use the equation E = hf, where

E represents the energy, h is Planck's constant (6.626 x 10-34 J·s), and f is the frequency of the photon. Substituting the given frequency (f = 6.0 x 1020 Hz) into the equation and using Planck's constant, we calculate the energy (E) as follows:

E = (6.626 x 10-34 J·s) (6.0 x 1020 Hz) = 3.976 x 10-13 J. Although this amount of energy might seem very small, it's important to remember that visible light and other forms of electromagnetic radiation consist of vast numbers of photons, which collectively can have a significant amount of energy.

Answer:

The thermal kinetic energy of the molecules is raised

Explanation:

The magnetic force on a charged particle moving through a magnetic field perpendicular to its velocity can cause the particle to move in a circular path. The radius of this path can be calculated using the formula r = mv/qB, resulting in an approximate radius of 42 millimeters for the given parameters.

In physics, one of the forces that can cause a particle to undergo uniform circular motion is the magnetic force. This happens when a charged particle moves in a direction perpendicular to a magnetic field. The magnetic force, being perpendicular to the velocity of the particle, acts as a centripetal force, constantly changing the direction of the particle's velocity, thus causing the particle to move in a circular path. The radius of this circular path can be obtained using the equation for the magnetic force acting on a moving charged particle, F = qvB, where q is the charge of the particle, v is the speed of the particle, and B is the magnetic field strength.

So the magnetic force equals the centripetal force when qvB = mv²/r, which allows us to solve for r, the radius of the path: r = mv/qB.

Substituting the given values: r = (2.05*10^-12 kg * 67.3*10^3 m/s) / (3.32*10^-6 C * 7.85 T), we find the radius r is about 0.042 meters or 42 millimeters. Note we have included unit conversions for mass m from picograms (pg) to kilograms (kg), speed v from kilometers per second (km/s) to meters per second (m/s), and charge q from microcoulombs (μC) to coulombs (C).

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The radius of the particle's circular path is 42 millimeters.

Given:

A uniform magnetic field with magnitude B = 7.85 mT

A negatively charged particle with charge q = -3.32 μC = -3.32 × 10⁻⁶ C and mass m = 2.05 pg = 2.05 × 10⁻¹⁵ kg

The particle moves with a speed of v = 67.3 km/s = 67.3 × 10³ m/s perpendicular to the magnetic field

The magnetic force on a charged particle moving in a magnetic field is given by the equation:

F = qvB sinθ

where θ is the angle between the velocity of the particle and the magnetic field. In this case, θ = 90°, so sinθ = 1. Therefore, the magnetic force on the particle is:

F = qvB = (-3.32 × 10⁻⁶ C)(67.3 × 10³ m/s)(7.85 × 10⁻³ T) = 1.75 × 10⁻² N

The magnetic force is also the centripetal force that keeps the particle moving in a circular path. Therefore, we can equate the magnetic force to the centripetal force to find the radius of the particle's circular path:

F = mv²/r

Solving for r, we get:

r = mv²/F = (2.05 × 10⁻¹⁵ kg)(67.3 × 10³ m/s)²/(1.75 × 10⁻² N) = 0.042 m = 42 mm

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Answer:

True.

Explanation:

A transformer is used either to increase or decrease the voltage.

It has two separate sets of coils of wire. The coils connected to the input voltage are called  primary and the output coils are called  secondary.

If number of windings (or coils) of wire in primary are less than the windings in secondary, it is a step up transformer because output voltage will be more than the input voltage. In this case the output voltage given out by the secondary will be more than the primary.

If the primary coils are more than the secondary, it is called a step down transformer. In this case, secondary voltage (output voltage) will be less than the primary.

Answer: The energy released in the given nuclear reaction is 4.98 MeV.

Explanation:

The equation for the alpha decay of Th-232 nucleus follows:

[tex]_{90}^{232}\textrm{Th}\rightarrow _{88}^{228}\textrm{Ra}+_{2}^{4}\textrm{He}[/tex]

We are given:

Mass of [tex]_{90}^{232}\textrm{Th}[/tex] = 232.038054 u

Mass of [tex]_{88}^{228}\textrm{Ra}[/tex] = 228.0301069 u

Mass of [tex]_{2}^{4}\textrm{He}[/tex] = 4.002602 u

To calculate the mass defect, we use the equation:

[tex]\Delta m=\text{Mass of reactants}-\text{Mass of products}[/tex]

Putting values in above equation, we get:

[tex]\Delta m=(m_{Th})-(m_{Ra}+m_{He})\\\\\Delta m=(232.038054)-(228.0301069+4.002602)=0.0053451u[/tex]

To calculate the energy released, we use the equation:

[tex]E=\Delta mc^2\\E=(0.0053451u)\times c^2[/tex]

[tex]E=(0.0053451u)\times (931.5MeV)[/tex] (Conversion factor: [tex]1u=931.5MeV/c^2[/tex] )

[tex]E=4.98MeV[/tex]

Hence, the energy released in the given nuclear reaction is 4.98 MeV.

Answer:

Newton's Third Law

Explanation:

Did the assignment

Answer: The correct answer is "their skin senses it as warmth".

Explanation:

Electromagnetic wave: the direction of electric field and magnetic field are perpendicular to the direction to the propagation of the wave.  It travels with the speed of the light.

For example, X-rays, gamma rays, infrared rays and visible region.

Infrared radiation: The range of the wavelength is from 700 nm to 1 mm. It is longer than the visible light. It is a part of the electromagnetic spectrum.

The infrared radiations cannot be seen by naked eyes of human. It can be felt by the human being. They can sense it as heat.

It is given in the problem that most objects emit infrared energy. Human can feel this infrared radiation. Their skin senses it as warmth.

We first observe that the particle moves in the direction of the vector

[tex](3\,\vec\imath + 4\,\vec\jmath + 3\,\vec k) - (\vec\imath - \vec k) = 2\,\vec\imath + 4\,\vec\jmath + 4\,\vec k[/tex]

so the initial velocity vector [tex]\vec v_0[/tex] is parallel to this vector. Given its initial speed is 3 at [tex]t=0[/tex], this means for some scalar constant [tex]c>0[/tex], we have

[tex]\vec v_0 =  2c\,\vec\imath + 4c\,\vec\jmath + 4c\,\vec k[/tex]

such that

[tex]\|\vec v_0\| = \sqrt{(2c)^2 + (4c)^2 + (4c)^2} = 6c = 3 \implies c = \dfrac12[/tex]

so that the initial velocity is

[tex]\vec v_0 = \vec\imath + 2\,\vec\jmath + 2\,\vec k[/tex]

Now, use the fundamental theorem of calculus to compute the velocity and position functions.

[tex]\displaystyle \vec v(t) = \vec v_0 + \int_0^t \vec a(u) \, du \\\\ ~~~~ = (\vec\imath + 2\,\vec\jmath + 2\,\vec k) + \int_0^t (2\,\vec\imath + 4\,\vec\jmath+4\,\vec k) \, du \\\\ ~~~~ = (\vec\imath + 2\,\vec\jmath + 2\,\vec k) + (2t\,\vec\imath + 4t\,\vec\jmath+4t\,\vec k) \\\\ ~~~~ = (1 +2t)\,\vec\imath + (2+4t)\,\vec\jmath + (2+4t)\,\vec k[/tex]

[tex]\displaystyle \vec r(t) = \vec r_0 + \int_0^t \vec v(u) \, du \\\\ ~~~~ = (\vec\imath - \vec k) + \int_0^t \left((1 +2u)\,\vec\imath + (2+4u)\,\vec\jmath + (2+4u)\,\vec k\right) \, du \\\\ ~~~~ = (\vec\imath - \vec k) + ((t+t^2)\,\vec\imath + (2t+2t^2)\,\vec\jmath + (2t+2t^2)\,\vec k) \\\\ ~~~~ = (t^2+t+1)\,\vec\imath + (2t^2+2t)\,\vec\jmath + (2t^2+2t-1)\,\vec k[/tex]