PHYSICS COLLEGE

Light from a sodium vapor lamp (λ-589 nm) forms an interference pattern on a screen 0.91 m from a pair of slits in a double-slit experiment. The bright fringes near the center of the pattern are 0.19 cm apart. Determine the separation between the slits. Assume the small-angle approximation is valid here.

Answers

Answer 1

Answer:

Answer:

separation between the slits is 0.28 mm

Explanation:

given data

wave length λ = 589 nm = 589 × $10^(-9)$ m

distance between slits and the screen D = 0.91 m

fringes weight y = 0.19 cm = 0.19 × $10^(-2)$ m

solution

we find here the spacing between the two slits i.e d

so use here formula that is

y = λD ÷ d .........................1

put here value we get

0.19 × $10^(-2)$ = $(589*10^(-9)*0.91)/(d)$

solve we get

d = 0.28 mm

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Overnight a thin layer of ice forms on the surface of a 40-ft-wide river that is essentially of rectangular cross-sectional shape. Under these conditions, the flow depth is 3 ft. During the following day the sun melts the ice cover. Determine the new depth if the flowrate remains the same and the surface roughness of the ice is essentially the same as that for the bottom and sides of the river.

Answers

Answer:

the new depth is 2.3 ft

Explanation:

the solution is in the attached Word file

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Answers

Answer:

Assuming h as the height of the cylindrical tank

$F=480\pi h \,g\,\, (lb)/(ft)$

Explanation:

Assuming that the height is $h$ we can find the volume of the cylindrical tank, then:

$V=\pi*r^2*h$

The diameter is 8.00 ft then $r=4.00 ft$ the total volume of the tank is:

$V=\pi (4.00 ft)^2 h=16\pi h\,\, ft^2$

But the tank is half full of oil, then we need half of the volume. For that reason the volume of oil is:

$V_(oil)=(16\pi h)/(2)ft^2=8\pi h \,\,ft^2$

We know the density of the oil $\rho=60.0\,lb/ft^3$ , with this we can fing the mass of oil that we have because:

$\rho=(m)/(V)$ then $m=\rho V$

Then the mass of oil that we have is:

$m=(60.0(lb)/(ft^3))(8\pi h\,\,ft^2)$

$m=480\pi h (lb)/(ft)$

Note that with the value of h we have the mass in correct units.

Finally to find the force we now that $F=mg$ then we just need to multiply the mass by the gravity.

$F=480\pi h \,g\,\, (lb)/(ft)$

A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?

Answers

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

$\mu mg = (mv^2)/(R)$

here we have

$\mu g = (v^2)/(R)$

now we know that

$v = √(\mu Rg)$

here we have

$\mu = 0.600$

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

$v = √((0.600)(35)(9.81))$

$v = 14.35 m/s$

A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of 90˚, so that the normal becomes perpendicular to the magnetic field. The coil has an area of 1.5 × 10-3 m2, 50 turns, and a resistance of 180 Ω. During the time while it is rotating, a charge of 9.3 × 10-5 C flows in the coil. What is the magnitude of the magnetic field?

Answers

Answer:

3.4 x 10^-4 T

Explanation:

A = 1.5 x 10^-3 m^2

N = 50

R = 180 ohm

q = 9.3 x 106-5 c

Let B be the magnetic field.

Initially the normal of coil is parallel to the magnetic field so the magnetic flux is maximum and then it is rotated by 90 degree, it means the normal of the coil makes an angle 90 degree with the magnetic field so the flux is zero .

Let e be the induced emf and i be the induced current

e = rate of change of magnetic flux

e = dФ / dt

i / R = B x A / t

i x t / ( A x R) = B

B = q / ( A x R)

B = (9.3 x 10^-5) / (1.5 x 10^-3 x 180) = 3.4 x 10^-4 T

Final answer:

The magnitude of the magnetic field can be calculated using Faraday's Law of electromagnetic induction, by setting up and solving an equation involving the number of turns in the coil, the area of the coil, and the time it takes for the coil to rotate.

Explanation:

To calculate the magnitude of the magnetic field, we can use Faraday's Law of electromagnetic induction, which can be expressed as E = d(N∙Φ )/dt, where E represents the induced EMF, N is the number of turns, and Φ is the magnetic flux (flux equals the product of the magnetic field B, the area A through which it passes and the cosine of the angle between B and A).

Given the information in the problem, we know that E = Q/R ∙ t. Since the coil is rotated through 90 degrees, it goes from being parallel to being perpendicular to the field, resulting in a change in magnetic flux of BNA. We can set up the equation E = d(NBA)/dt = Q/R ∙ t = [(50 turns) ∙ (1.5 × 10-3 m²) ∙ B)/(t)]

We can solve this equation to determine the magnitude of the magnetic field B. Remember, always double-check your calculations to ensure their accuracy.

Learn more about Magnetic Field here:

brainly.com/question/36936308

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Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?

Answers

The measurement will be significantly affected.

Recall that the relationship between linear velocity and angular velocity is subject to the formula

$v = \omega r$ ,

Where r indicates the radius and $\omega$ the angular velocity.

As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.

A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take ????=9.81g=9.81 m/s2. angle: °