PHYSICS COLLEGE

The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference.
b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference.
c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference.
d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.

Answers

Answer 1

Answer:

Answer:

the answers the correct one is c

Explanation:

The diffraction pattern for a slit is

a sin θ = m λ

Where a is the width of the slit, λ the wavelength, m the order of destructive interference and θ the angle where the interference occurs.

The expression for multi-slit diffraction (diffraction grating) is

d sin θ = m λ

Where d is the distance between slits, λ the wavelength m the order of the diffraction maximums and θ the angle for these maximums.

When we compare the expressions of the answers the correct one is c

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Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).

Answers

Answer:

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

A 500 W heating coil designed to operate from 110 V is made of Nichrome 0.500 mm in diametera.Assuming the resistivity of the nichrome remains constant at is 20.0 degrees C value find the length of wire used.b. Now consider the variation of resistivity with temperature. What power is delivered to the coil of part (a) when it is warmed to 1200 degrees C.?

Answers

(a) Length of the wire is 3.162 m

(b)Power delivered to the coil is 339.7 W

Electrical Power:

The electrical power is given by

P = V² / R

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 Ω

Now the resistivity of a wire is given by

ρ= RA/L

here ρ = 1.50×10⁻⁶ Ωm

so after rearranging we get:

L = RA / ρ

Now, the radius of wirer = 0.5 / 2 mm = 0.25 mm = 2.5×10⁻⁴ m

So the cross sectional area can be calculated as follows

$A = \pi r^2\n\nA = \pi * (2.5*10^(-4))^2\n\nA = 1.96*10^(-7) m^2$

hence,

$L = (24.2 *1.96*10^(-7) / 1.50*10^(-6)) \n\nL = 3.162\; m$

(b)The dependency of resistance with temperature is as follows:

R = R₀[1 + αΔT]

α = $4*10^(-4)^\;oC^(-1)$ for Nichrome

$R' = R [1 + \alpha (1200 - 20) ]\n\nR' = R[1 + \alpha (1180) ]\n\nR' = 24.2[ 1 + 4*10^(-4) * 1180 ]\n\nR' = 24.2[1 + 0.472]\n\nR' = 24.2 * 1.472\n\nR' = 35.62 \;\Omega$

So the power generated is :

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

Learn more about electrical power:

brainly.com/question/26174188

Answer:

a) 3.162 m

b) 339.7 W

Explanation:

Assume ρ = 1.50*10^-6 Ωm, and

α = 4.000 10-4(°C)−1 for Nichrome

To solve this, we would use the formula

P = V² / R

So when we rearrange and make R subject of formula, we have

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 ohms

Recall the formula for resistivity of a wire

R = ρ.L/A

Again, in rearranging and making L subject of formula, we have

L = R.A / ρ

To make it uniform, we convert our radius from mm to m.

Diameter, D = 0.5 mm

Radius of wire = 0.5 / 2 mm = 0.25 mm = 0.00025 m

We then use this radius to find our area

A = πr²

A = π * 0.00025²

A = 1.96*10^-7 m²

And finally, we solve for L

L = (24.2 * 1.96*10^-7 / 1.50*10^-6) =

L = 3.162 m

(b)

Temperature coefficient of resistance.

R₁₂₀₀ = R₂₀[1 + α(1200 - 20.0) ]

R₁₂₀₀ = R₂₀[1 + α(1180) ]

R₁₂₀₀ = 24.2[ 1 + 4.*10^-4 * 1180 ]

R₁₂₀₀ = 24.2[1 + 0.472]

R₁₂₀₀ = 24.2 * 1.472

R₁₂₀₀ = 35.62 ohms

Putting this value of R in the first formula from part a, we have

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

A sinusoidal wave is travelling on a string under tension T = 8.0(N), having a mass per unit length of 1 = 0.0128(kg/m). It’s displacement function is D(x,t) = Acos(kx - t). It’s amplitude is 0.001m and its wavelength is 0.8m. It reaches the end of this string, and continues on to a string with 2 = 0.0512(kg/m) and the same tension as the first string. Give the values of A, k, and , for the original wave, as well as k and  the reflected wave and the transmitted wave.

Answers

Answer:

Explanation:

Please use Gauss’s law to find the electric field strength E at a distance r from the center of a sphereof radius R with volume charge density ???? = cr 3 and total charge ????. Your answer should NOT contain c. Be sure to consider regions inside and outside the sphere.

Answers

Answer:

See the explaination for the details.

Explanation:

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

Please kindly check attachment for the step by step explaination of the answer.

A Hall-effect probe to measure magnetic field strengths needs to be calibrated in a known magnetic field. Although it is not easy to do, magnetic fields can be precisely measured by measuring the cyclotron frequency of protons. A testing laboratory adjusts a magnetic field until the proton's cyclotron frequency is 9.70 MHz . At this field strength, the Hall voltage on the probe is 0.549 mV when the current through the probe is 0.146 mA . Later, when an unknown magnetic field is measured, the Hall voltage at the same current is 1.735 mV .A) What is the strength of this magnetic field?

Answers

Answer:

The value of the magnetic field is 2.01 T when Hall voltage is 1.735 mV

Explanation:

The frequency of the cyclotron can help us find the magnitude of the magnetic field, thus then we can compare the effect of increasing Hall voltage on the probe.

Magnetic field magnitude at initial Hall voltage.

The cyclotron frequency can be written in terms of the magnetic field magnitude as follows

$f = \cfrac{qB}{2\pi m}$

Solving for the magnetic field.

$B = \cfrac{2\pi mf}q$

Thus we can replace the given information but in Standard units, also remembering that the mass of a proton is $m_p=1.67 * 10^(-27) kg$ and its charge is $q_p=1.6 * 10^(-19) C$ .