A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.24 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?

Answer: 2.44*10^3 km

Explanation:

NOTE: We would be solving this question in "cm" instead of the usual "m"

1 mi² = 2.59 km²

2.59 km² = 2.59*10^10 cm²

Given, area of South Dakota

A = 7.588*10^4 mi²

A = 7.588*10^4 * 2.59*10^10

A = 1.97*10^15 cm² is the area of South Dakota

S = πd²/4

S = 3.142 * 0.635² / 4

S = 3.142 * 0.4/4

S = 3.142 * 0.1

S = 0.3142 cm² is the area of 1 marshmallow

1.97*10^15 / 0.3142 = 6.27*10^15, thus is the number of marshmallows in one single layer to cover the state area

6.02*10^23 / 6.27*10^15 = 9.6*10^7, this is the number of layers

9.6*10^7 * 2.54 = 2.44*10^8, this is the height in cm

Height in km is 2.44*10^3 km

The correct result of the given operation expressed in scientific notation and with the appropriate number of significant figures should be 3.4 x 10^3. However, there are no matching options provided. A possible mistake might exist in the question.

To solve the problem, you should first add the two numbers in the brackets together. This gives us 0.82 + 0.042 = 0.862. Next, we need to multiply this result by the number outside the brackets, which is 4.4 x 103. Thus, our equation becomes 0.862 x 4.4 x 103 = 3.3928 x 103.

The result needs to be expressed in scientific notation and with the correct number of significant figures. The least precise number in our calculation is 4.4 (which has two significant figures), so our final answer should also have two significant figures. Therefore, we round 3.3928 x 103 to 3.4 x 103.

However, none of the options provided matches this result. Please double-check the question as there might be a typo.

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First, add 0.82 + 0.042 = 0.862. Second, multiply the sum 0.862 x 4.4 = 3.7928. Finally, round the result to two significant figures in scientific notation giving the answer as 3.8 x 10 or just 3.8.

To solve this problem, first, we need to add the numbers in the parentheses, then multiply the sum by 4.4 x 10⁰, which is just 4.4 since any number raised to the power of zero equals one.

Step 1: Add 0.82 + 0.042 = 0.862

Step 2: Multiply 0.862 x 4.4 = 3.7928

Lastly, we need to express the answer in scientific notation with the correct number of significant figures. As the question gives the numbers 0.82 and 0.042 with two and three significant figures respectively, we should express our final answer to two significant figures, as you always round to the fewest number of significant figures.

So, round 3.7928 to two significant figures, gives you 3.8 x 10⁰, which can alternatively be written as just 3.8, making the correct answer (a) 3.8 x 10.

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Answer:

V = 1.1658 × [tex]10^{-5}[/tex] V

Explanation:

given data

strip of copper thick = 130 µm

strip of copper wide = 4.40 mm

uniform magnetic field of magnitude B = 0.79 T

current i = 26 A

number of charge carriers per unit volume = 8.47 × [tex]10^{28}[/tex] electrons/m³

solution

we know that number density is express as

n = \frac{Bi}{Vle}      ...............1

B  is uniform magnetic field and i is current  and V is hall potential difference and l is thickness and e is electron charge 1.6 × [tex]10^{-19}[/tex]  C

so V will be as

V = \frac{iB}{nle}      .....................2

so put here value and we get V

V = [tex]\frac{26 \times 0.79}{8.47\times 10^{28}\times 130\times10^{-6}\times1.6 \times10^{-19}}[/tex]

V = 1.1658 × [tex]10^{-5}[/tex] V

Answer:

L=N/I*magnetic flux

=N/I*BA=N*mu(0)NIA/2pi*R=mu(0)N^2/2pi*R)*Pi*a^2

=mu(0)N^2a^2/2R

Explanation:

The self-inductance L of the toroid is mu(0)N^2a^2/2R.

Since

The toroidal inductor has a circular cross-section of radius a. The toroid has N turns and radius R.

The toroid is narrow ( a≪R )

So,

L=N/I*magnetic flux

=N/I*BA=N*mu(0)NIA/2pi*R=mu(0)N^2/2pi*R)*Pi*a^2

=mu(0)N^2a^2/2R

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Find the given attachment

Answer:

B) Lengths will appear shorter and time will appear to pass faster.

Explanation:

This is in line with the laws of relativity.

Answer:

A. Lengths will appear shorter and time will appear to pass slower.  

Explanation:

From theory of relativity, we know that:

L₀ = length of object, measured in stationary frame of reference

L = length of object measured from a frame moving with respect to object, called ‘relativistic length’

v = relativistic speed between observer and the object

c = speed of light

then,

L = L₀ √(1-v²/c² )

Hence, the length of the object decreases with the increase in its relativistic speed "v"

t₀ = time measured by clock at rest with respect to event.

t = time measured by clock in motion relative to the event

v = relativistic speed between observer and the object

c = speed of light

then,

t =  t₀/√(1-v²/c² )

Hence, the time increases with the increase in in relativistic speed and as a result it appears to pass slower.

Hence, the correct option is:

A. Lengths will appear shorter and time will appear to pass slower.  

Answer:

n1/p + n2/q = (n2-n1)/R

1.45/p + 1.6/1.15 = (1.6 - 1.45)/0.035 (m)

1.45/p + 1.39 = 4.28

p = 0.50 m (50 cm)

Answer with Explanation:

Let mass of one vehicle =m

Mass of other vehicle=m'

[tex]\frac{m}{m'}=\frac{1}{4}[/tex]

[tex]m'=4m[/tex]

Velocity of one vehicle=[tex]v=13 im/s[/tex]

Velocity of other vehicle=[tex]v'=13jm/s[/tex]

In x- direction

By law of conservation of momentum

[tex]mv+0=(m+m')V_x[/tex]

[tex]13m=(m+4m)V_x[/tex]

[tex]13m=5mV_x[/tex]

[tex]V_x=\frac{13}{5}[/tex]

In y- direction

By law of conservation of momentum

[tex]0+m'v'=(m+m')V_y[/tex]

[tex]4m(13)=5mV_y[/tex]

[tex]V_y=\frac{52m}{5m}=\frac{52}{5}[/tex]

Magnitude of velocity of the wreck,V=[tex]\sqrt{V^2_x+V^2_y}=\sqrt{(\frac{13}{5})^2+(\frac{52}{5})^2}=10.72 m/s[/tex]

Direction:[tex]\theta=tan^{-1}(\frac{V_y}{V_x})[/tex]

[tex]\theta=tan^{-1}(\frac{\frac{52}{5}}{\frac{13}{5}})=75.96^{\circ}[/tex]

Answer: 3.78 m

Explanation:

Volume = 1.9 * 1.9 * 1.9

Volume = 6.859 m³

Density of iron assumed to be 7870 kg/m³

Recall,

Density = mass / volume, so

Mass = density * volume

Mass = (7870 * 6.859)

Mass = 53980.33 kg

This mass we calculated, is the mass of water the new cube would have to displace, absolute minimum, so that it can float.

Now density of fresh water is assumed to be (1000 kg/m³).

Thus, the volume of water that will be displaced has to be

Volume = mass / density

Volume = (53980.33 / 1,000)

Volume = 53.98 m³.

Cube root of 53.88 = 3.78 m

Therefore, the minimum side length is 3.78 m

Answer:

3.781488032m or greater

Explanation:

General Rule of thumb is that any thing that is less dense than water will float in it.

So

density of water is 997Kg/m^3 and density of iron on the other hand is  7860kg/m^3.

so to make iron less dense it has to be in larger volume.

mass of the given cube is 53911.74Kg and we have contain it in volume such that density is less than water, mathematically this translates to .

[tex]\frac{53911.74Kg}{V}\leq \frac{997Kg}{m^3}[/tex] left side is density of iron and right side is density of water, solving v in this inequality gives us.

[tex]v \geq 54.0739m^3[/tex] which means each side is [tex]s \geq 3.7814880m[/tex]. so the minum is just greater than that.

Answer: [tex]160 \Omega[/tex]

Explanation:

According to Ohm's law:  

[tex]V=R.I[/tex]  

Where:  

[tex]V=120 V[/tex] is the voltage difference across the light bulb

[tex]R[/tex] is the resistance of the light bulb   (the value we want to find)

[tex]I=0.75 A[/tex] is the electric current

Isolating [tex]R[/tex]:  

[tex]R=\frac{V}{I}[/tex]

[tex]R=\frac{120 V}{0.75 A}[/tex]

Finally:

[tex]R=160 \Omega[/tex]  This is the resistance of the light bulb

Final answer:

To calculate the resistance of a lightbulb with a current of 0.75 amperes and a voltage difference of 120 volts, use Ohm's law. The formula is R = V / I, which yields a resistance of 160 ohms for the lightbulb.

Explanation:

Calculating the Resistance of a Lightbulb

The student has asked how to calculate the resistance of a lightbulb when a current of 0.75 amperes flows through it and the voltage difference across it is 120 volts. To find the resistance, we can use Ohm's law, which states that the resistance (R) of a circuit is equal to the voltage (V) across it divided by the current (I) flowing through it, which can be written as:

R = V / I

Plugging in the given values, we get:

R = 120 V / 0.75 A

R = 160 ohms

Therefore, the resistance of the lightbulb is 160 ohms.

Wind is driven due to the differences in air pressure around the Earth.

To find the answer, we need to know about the wind flow.

Thus, we can conclude that wind is driven due to the differences in air pressure around the Earth.

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Answer:

The magnitude of the acceleration of earth due to the gravitational pull of earth is a = Gm/r^2

Where r = the center to center distance between the earth and the moon,

m = mass of the moon, and,

G is the gravity constant.

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

[tex]a_e = \frac{Gm}{r^2}[/tex]

Explanation:

We assume that:

M to represent the mass of the earth

m to equally represent the  mass of the moon

r should be the distance between the center of the earth to the center of the moon.

Then;

the expression for the gravitational force can be written as:

[tex]F = \frac{GMm}{r^2}[/tex]

Where [tex]a_e[/tex] is the acceleration produced by the earth; then:

[tex]F =M *a_e[/tex]

Then:

[tex]M*a_e = \frac{GMm}{r^2}[/tex]

[tex]a_e = \frac{GMm}{Mr^2}[/tex]

[tex]a_e = \frac{Gm}{r^2}[/tex]

Therefore, the  magnitude of the acceleration of the earth due to  the gravitational pull of the moon  [tex]a_e = \frac{Gm}{r^2}[/tex]

Answer:

Explanation:

In a scenario where a particle of charge overrightarrow{B} enters a magnetic field with a velocity overrightarrow{V}, it experiences a force overrightarrow{F} given by: overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B}).

implies F=BqVSin\theta.

Where theta is the angle between the velocity vector of the particle and the magnetic field vector.

When a particle enters the magnetic field at an angle 90, it moves in a circular path as it experiences a centripetal force, given by F=\frac{mV^2}{R}.

Where R is the radius of the circle, V is its velocity and m is its mass

Thus, magnetic force becomes F=BqVSin90^o=\frac{mV^2{R}\implies R=\frac{mV}{Bq}.

The equation changes as below, when velocity is doubled, let us assume that the radius is given by R_1.

R_1=\frac{2mV}{Bq}=2R.

Therefore, it is obvious that the velocity of a charged particle in a circular arc is directly proportional to the radius of the arc. The radius of the circular arc doubles when the velocity of the charged particle in the circular orbit doubles only if the mass, charge and magnetic field of the particle remains constant

Hence when velocity is doubled radius of the circle also gets doubled.

When the speed of the particle doubles, the radius of the arc also doubles.

The magnetic force on the particle is calculated as follows;

[tex]F = qvB[/tex]

The centripetal force on the particle is calculated as follows;

[tex]F_c = \frac{mv^2}{R}[/tex]

The speed of the particle is calculated as follows;

[tex]\frac{mv^2}{R} = qvB\\\\mv = qBR\\\\v = \frac{qBR}{m} \\\\\frac{v_1}{R_1} = \frac{v_2}{R_2}[/tex]

when the speed of the particle doubles;

[tex]\frac{v_1}{R_1} = \frac{2v_1}{R_2} \\\\R_2v_1 = 2R_1v_1\\\\R_2 = 2R_1[/tex]

Thus, we can conclude that when the speed of the particle doubles, the radius of the arc also doubles.

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Answer:

A) μ = A.m²

B) z = 0.46m

Explanation:

A) Magnetic dipole moment of a coil is given by; μ = NIA

Where;

N is number of turns of coil

I is current in wire

A is area

We are given

N = 300 turns; I = 4A ; d =5cm = 0.05m

Area = πd²/4 = π(0.05)²/4 = 0.001963

So,

μ = 300 x 4 x 0.001963 = 2.36 A.m².

B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;

B = (μ_o•μ)/(2π•z³)

Let's make z the subject ;

z = [(μ_o•μ)/(2π•B)] ^(⅓)

Where u_o is vacuum permiability with a value of 4π x 10^(-7) H

Also, B = 5 mT = 5 x 10^(-6) T

Thus,

z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)

Solving this gives; z = 0.46m =

The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v

Explanation:

Given data,

charge = 5.5 x 10¹² C

k =9.00 x 10⁹

The electric potential V of a point charge can found by,

V= kQ / r

Assuming, r=5.00×10⁻² m

V= 5.5 x 10⁻¹²C x  9.00 x 10⁹ / 5.00×10⁻² m

V=  49.5 x 10⁻³/ 5.00×10⁻²

Electric potential V=  0.099 x 10⁻¹v

The electric potential at a point in the electric field created by a point charge is calculated by the formula V = kQ/r. The provided constants are [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex] and [tex]k = 9.00 \times 10^9[/tex] The distance r from the point charge is necessary to calculate the electric potential.

The electric potential at point A in the electric field created by a point charge can be found using the formula V = kQ/r, where V is the electric potential, k is Coulomb's constant, Q is the value of the point charge, and r is the distance from the point charge.

Given, the charge [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex], and Coulomb's constant [tex]k = 9.00 \times 10^9[/tex]

The distance r is not given in the question. Therefore, we can't calculate the electric potential at point A. However, if distance r from the point charge were provided (in meters), we could substitute k, Q, and r into the formula and solve for V.

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Due to the net electric flux that is moving outward through the closed box's surface, the box has a negative charge.

Given:-

The electric field is constant over all the faces.

From Gauss law:

[tex]\int \vec E \cdot d\vec A = \dfrac{Q}{e_o}[/tex]

[tex]\vec E =[/tex] An electric field throughout the cube.

[tex]d \vec A =[/tex] Elemental area of the cube surface.

[tex]Q =[/tex] Electric charge around the cube.

[tex]e_o =[/tex]  Electric constant.

[tex]\dfrac{Q}{e_o} = (-15-20-15+20+15+10)\\ \\ = -5A[/tex]

According to Gauss's law, the box contains a negative charge.

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The box's charge type (being positive, negative, or neutral) is determined by the net electric flux's direction. Outward flux signifies positive charge, whilst inward flux indicates a negative charge. Boxes with no net electric flux have no charge.

The determination of whether a closed box contains a positive charge, negative charge, or no charge is dependent on the direction of the net electric flux. If there is net electric flux passing outward through the surface of the box, this indicates the box contains a positive charge. Conversely, if the net electric flux is passing inward through the surface, the box contains a negative charge. Lastly, if there is no net electric flux (flux in cancelled by flux out), it suggests there is no charge in the box.

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Answer: 0.512 kgm²

Explanation:

Given

Force, F = 2*10^3 N

Angular acceleration, α = 121 rad/s²

Lever arm, r(⊥) = 3.1 cm = 3.1*10^-2 m

τ = r(⊥) * F

Also,

τ = Iα

Using the first equation, we have

τ = r(⊥) * F

τ = 0.031 * 2*10^3

τ = 62 Nm

Now we calculate for the inertia using the second equation

τ = Iα, making I subject of formula, we have

I = τ / α, on substituting, we have

I = 62 / 121

I = 0.512 kgm²

Thus, the moment of inertia of the boxers forearm is 0.512 kgm²

Answer:

The radius of curvature of the ion's orbit is 0.59 meters

Explanation:

Given that,

Mass of the 24 Mg ion, [tex]m=3.983\times 10^{-26}\ kg[/tex]

Potential difference, V = 3 kV

Magnetic field, B = 526 G

Charge on single ionized ion, [tex]q=1.6\times 10^{-19}\ C[/tex]

The radius of the the path traveled  by the charge is circular. Its radius is given by :

[tex]r=\dfrac{mv}{Bq}[/tex]

v is speed of particle.

v can be calculated using conservation of energy as :

[tex]\dfrac{1}{2}mv^2=qV\\\\v=\sqrt{\dfrac{2qV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^3}{3.983 \times 10^{-26}}} \\\\v=1.26\times 10^5\ m/s[/tex]

Radius,

[tex]r=\dfrac{3.983 \times 10^{-26}\times 1.26\times 10^5}{0.0526\times 1.6\times 10^{-19}}\\\\r=0.59\ m[/tex]

So, the radius of curvature of the ion's orbit is 0.59 meters.

The energy of a photon is the product of the planks constant and frequency. The photon energy of this light is 151 keV.

It is the product of the planks constant and frequency.

[tex]E = \dfrac {hc}\lambda[/tex]

Where,

[tex]h[/tex] -Plank's constant  = [tex]\bold {6.63\times 10^{-34}\rm \ J/s}[/tex]

[tex]c[/tex]- speed of light = [tex]\bold { 3\times 10^8 \rm \ m/s }[/tex]

[tex]\lambda[/tex][tex]\labda[/tex]- wavelength = 8.2 pm = [tex]\bold {8.2 \times 10^{-12 }\rm \ m}[/tex]

Put the values in the equation,

[tex]E = \dfrac { 6.63\times 10^{-34}\rm \ J/s\times 3\times 10^8 \rm \ m/s }{ 8.2 \times 10^{-12 }\rm \ m}\\E = 151 \rm \ keV[/tex]

Therefore, the photon energy of this light is 151 keV.

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Answer: 13 cm

Explanation:

Given

Inductance of the solenoid, L = 3.01•10⁻⁶ H

Width of the wire, x = 1.13 m

Length of the tube, z = ?

Now, we know that

L = μ₀N²A/z, where

N = x/2πr, making r subject of formula,

r = x/2πN

Also,

A = πr², on substituting for A, we have

A = πx²/4π²N²

Now finally, we substitute in the initial equation and solve

L = μ₀N²A/z

L = μ₀N²πx²/4π²N²z

L = μ₀x²/4z, making z subject of formula, we have

z = μ₀x²/4L

z = 4π*10⁻⁷ * 1.13² /4 * 3.01*10⁻⁶

z = (4π*10⁻⁷ * 1.2769) / (4 * 3.01*10⁻⁶)

z = 1.6*10^-6 / 1.2*10^-5

z = 0.13 m

Therefore, the length of the tube should be 13 cm

The length of the tube should be 469 centimeters.

To find the required length of the tube for Tia's solenoid, we can use the formula for the inductance of a solenoid:

[tex]\[ L = \frac{{\mu_0 \cdot N^2 \cdot A}}{l} \][/tex]

Where:

L = inductance of the solenoid (3.01 μH)

[tex]\( \mu_0 \)[/tex] = permeability of free space (4π × 10^-7 T*m/A)

N = number of turns of wire

A = cross-sectional area of the solenoid

[tex]\( l \)[/tex] = length of the solenoid

We're given [tex]\( L \)[/tex] and [tex]\( N \cdot l \)[/tex] (the total length of wire wound around the tube). We need to find [tex]\( l \)[/tex], the length of the tube.

First, let's rearrange the formula to solve for [tex]\( l \)[/tex]:

[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot A}}{{L}} \][/tex]

We know the cross-sectional area ( A ) can be represented as [tex]\( A = \pi r^2 \)[/tex], where ( r ) is the radius of the tube. Since the wire is uniformly wound, we can calculate ( A ) using the length of the wire and the number of turns.

[tex]\[ A = \frac{{\text{{Length of wire}}}}{{\text{{Number of turns}}}} \][/tex]

Given that the wire length is 1.13 m and there are [tex]\( N \)[/tex] turns, [tex]\( A = \frac{{1.13}}{{N}} \)[/tex].

Substituting this into the equation for [tex]\( l \)[/tex], we get:

[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot \left(\frac{{1.13}}{{N}}\right)}}{{L}} \][/tex]

[tex]\[ l = \frac{{\mu_0 \cdot 1.13}}{{L}} \][/tex]

Now, plug in the known values:

[tex]\[ l = \frac{{4\pi \times 10^{-7} \cdot 1.13}}{{3.01 \times 10^{-6}}} \][/tex]

[tex]\[ l = \frac{{4\pi \times 1.13}}{{3.01}} \][/tex]

[tex]\[ l = \frac{{4 \times 3.1416 \times 1.13}}{{3.01}} \][/tex]

[tex]\[ l = \frac{{14.1376}}{{3.01}} \][/tex]

[tex]\[ l = 4.69 \, \text{m} \][/tex]

Finally, convert this length to centimeters:

[tex]\[ l = 469 \, \text{cm} \][/tex]

So, the length of the tube should be 469 centimeters.

Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.

To find the time it takes for the current to drop to 12% of its value and the energy dissipated by the circuit, we can use the equations for the decay of current in an RL circuit and the energy in an inductor respectively.

To determine the time it takes for the current in the circuit to drop to 12% of its value at t = 0, we need to use the equation for the decay of current in an RL circuit. The equation is given by I(t) = I(0) * exp(-t/τ), where I(t) is the current at time t, I(0) is the initial current, t is the time, and τ is the time constant. In this case, we can use I(t) = I(0) * exp(-t/τ) = 0.12 * I(0). To find the time, we rearrange the equation as t = -τ * ln(0.12).

The energy dissipated by the circuit can be calculated using the equation for the energy in an inductor, which is given by E = 1/2 * L * I(0)^2, where E is the energy, L is the inductance, and I(0) is the initial current. In this case, we can substitute the values given to find the energy dissipated.

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The time needed for the current in the circuit to drop to 12% of its value at t = 0 is approximately 20.63 ms. The energy dissipated in the circuit during that time is approximately 7.22 mJ.

Given, initial current I0 can be obtained from Ohm's law, I0 = V / R = 108 / 1100 = 0.09818 A. The circuit value diminishes exponentially over time and can be expressed as I = I0 * e-Rt/2L. Therefore, to find out when the current drops to 12% of its initial value we set I / I0 = 0.12 = e-Rt/2L, and solve for t. After calculation, we find t ≈ 20.63 ms.

As for part (b), the energy dissipated in an RL circuit over time is given by W = 1/2 * L * I2, where I is the current at time t, which is given by the relation: I = I0 * e-Rt/2L. Performing the integration over the time period t = 0 to 20.63 ms, we find that the energy dissipated is approximately 7.22 mJ.

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Answer:

Explanation:

find the solution below

The work, heat, and internal energy changes are calculated for different stages of the gas expansion and cooling processes in a cylinder with a piston containing oxygen as an ideal gas.

Part A: The work done by the gas during the initial expansion can be calculated using the formula:

Work = Pressure * Change in Volume

Since the expansion is isobaric (constant pressure) and the volume doubles, the change in volume is 2 times the initial volume. Therefore, the work done is 2 times the initial pressure times the initial volume.

Part B: The heat added to the gas during the initial expansion can be calculated using the formula:

Heat = Pressure * Change in Volume, since the expansion is isobaric.

Part C: The internal energy change of the gas during the initial expansion is equal to the heat added minus the work done.

Part D: The work done during the final cooling is zero because the process is isochoric (constant volume) and no work is done.

Part E: The heat added during the final cooling can be calculated using the equation:

Heat = Change in Internal Energy, since no work is done.

Part F: The internal energy change during the final cooling is equal to the negative of the heat added.

Part G: The internal energy change during the isothermal compression is also equal to the negative of the heat added, as no work is done during an isothermal process.

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Answer:

larger, because her angular speed is larger.

Explanation:

The rotational kinetic energy is proportional to the square of the angular velocity while it is linearly proportional to the moment of inertia. So the increase of angular speed will have a larger effect of the kinetic energy than the decrease of the moment of inertia.

Answer:

Rotational kinetic energy will Increase

Explanation:

Rotational kinetic energy KE is

KE = 1/2 x I x w^2

Where I is moment of inertia,

w is angular velocity.

It can be seen that increasing angular velocity increases rotational kinetic energy.

Light is incident in air (n = 1) and refracted into a medium with an index of refraction of 2. Light incident parallel to the central axis is focused at a point 24cm from the surface

When a light  ray passes through from one medium to another medium ,due to the difference in medium index, the light deflect from its original path .

This deflection  from its path is called refraction and The phenomenon of refraction can be seen in sound and water waves.

Cornea reflect the light before light reaches to the retina. Change in the speed of light when it travels from one medium to other medium then due to this there is change in the path of light, this phenomenon is known as refraction of light.

Equation for refraction

n2/v- n1/u = n2-n1/R

= u = infinity ( parallel)

2/v= (2-1)/R

V = 2R

V= 24cm

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Answer:

[tex]6.75\mu C/m^2[/tex]

Explanation:

We are given that

Diameter,d=[tex]1\mu m=1\time 10^{-6} m[/tex]

[tex]1\mu m=10^{-6} m[/tex]

Radius,r=[tex]\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m[/tex]

Density,[tex]\rho=900kg/m^3[/tex]

Total number of electrons,n=39

Charge on electron =[tex]1.6\times 10^{-19} C[/tex]

Total charge=[tex]q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C[/tex]

Distance,s=2mm=[tex]2\times 10^{-3} m[/tex]

Mass =[tex]density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg[/tex]

Initial velocity,u=0

Final speed,v=4.5 m/s

[tex]v^2-u^2=2as[/tex]

[tex](4.5)^2-0=2a(2\times 10^{-3})[/tex]

[tex]20.25=4a\times 10^{-3}[/tex]

[tex]a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2[/tex]

Force,F=ma

[tex]qE=ma[/tex]

[tex]q(\frac{\sigma}{2\epsilon_0})=ma[/tex]

[tex]\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}[/tex]

[tex]\epsilon_0=8.85\times 10^{-12}[/tex]

[tex]\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2[/tex]

Answer: 0.666 V

Explanation:

Given

Radius of the loop, r = 14.8 cm = 0.148 m

Magnetic field present, B = 0.814 T

Rate of shrinking, dr/dt = 88.4 cm/s = 0.884 m/s

emf = dΦ/dt , where Φ = BA

emf = d(BA)/dt, where A = πr²

emf = d(Bπr²)/dt

if B is constant, then

emf = Bπ d(r²)/dt, on differentiating, we have,

emf = Bπ * 2r dr/dt

emf = 2πrB dr/dt, now if we substitute the values, we have

emf = 2 * 3.142 * 0.148 * 0.814 * 0.884

emf = 6.284 * 0.106

emf = 0.666 V

Answer:

The magnitude of the magnetic force on the particle is 67.86 N.          

Explanation:

Given that,

Charge, [tex]q=34.6\ \mu C=34.6\times 10^{-6}\ C[/tex]

Speed of particle, v = 68.4 m/s in +x direction

Magnetic field, [tex]B=(0.466 j+0.876 k)\ T[/tex]

We need to find the magnitude of the magnetic force on the particle. The magnetic force is given by :

[tex]F=q(v\times B)\\\\F=34.6\times 10^{-6}(68.4i+0+0\times (0+0.466j+0.876 k))\\\\F=\begin{pmatrix}0&-59.9184&31.8744\end{pmatrix}\\\\F=(-59.9184j+31.8744k)\ N[/tex]

The magnitude of magnetic force on the particle is :

[tex]|F|=\sqrt{(-59.9184)^2+(31.8744)^2} \\\\|F|=67.86\ N[/tex]

So, the magnitude of the magnetic force on the particle is 67.86 N.          

Answer:

R₁ = 50.77 Ω

Explanation:

Since, we know that:

Electric Power = P = VI

but from Ohm's Law:

V = IR

(or) I = V/R

Therefore,

P = V²/R

(OR) R = V²/P

where,

V = Battery Voltage

R = Resistance of combination

FOR SERIES COMBINATION:

R = Rs = (57 V)²/48 W

Rs = 67.69 Ω

but, we know that:

Rs = R₁ + R₂

R₁ + R₂ = 67.69 Ω

R₁ = 67.69 Ω - R₂  __________ eqn (1)

FOR PARALLEL COMBINATION:

R = Rp = (57 V)²/256 W

Rp = 12.69 Ω

but, we know that:

Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω

using eqn (1) and value of R₁ + R₂, we get

Rp = 12.69  = R₂(67.69 - R₂)/67.69

859.08 = 67.69 R₂ - R₂²

R₂² - 67.69 R₂ + 859.08 = 0

Solving this quadratic equation we get the answers:

Either, R₂ = 50.76 Ω

Either, R₂ = 16.92 Ω

Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,

R₂ = 16.92 Ω

using this value in eqn (1), we get:

R₁ = 67.69 Ω - 16.92 Ω

R₁ = 50.77 Ω

Final answer:

To calculate R1, first, find the total resistance for series (R1 + R2) and parallel (1/R1 + 1/R2) connections using the power formula P = V^2 / R, with provided power values for each case. Then solve the simultaneous equations.

Explanation:

The problem can be solved by using the formulas for resistances in series and parallel, as well as the formula for the power supplied by the battery.

In series, the resistances add up: Rtotal = R1 + R2. The power provided by the battery to the series circuit is given by P = V2 / Rtotal, where P is 48.0 W and V is 57.0 V. Rearranging to solve for Rtotal, we get Rtotal = V2 / P.

For parallel resistances, we have 1/Rtotal = 1/R1 + 1/R2. The power in the parallel case is P = V2 / Rtotal. With V and P given as 57.0 V and 256 W, respectively, we can solve for Rtotal.

With both Rtotal values calculated, we can set up two equations with two unknowns (R1 and R2), knowing that R1 > R2. Solving these simultaneous equations gives us the value of R1.