PHYSM-(e) A horizontal bar, 8.0 m in length, has a

Solution

a

First, we need to calculate the weight of the submerged wreck, which is given by the formula:

W = mg

where

m = 10^4 \, \text{kg}

and

g = 9.81 \, \text{m/s}^2

. Thus,

W = 10^4 \times 9.81 = 98100 \, \text{N}.

b

Next, we calculate the buoyant force acting on the wreck using Archimedes' principle. The volume of the wreck can be found using its mass and density:

V = \frac{m}{\rho} = \frac{10^4}{8 \times 10^3} = 1.25 \, \text{m}^3.

The buoyant force is then given by:

F_b = \rho_{water} \cdot g \cdot V = 1000 \cdot 9.81 \cdot 1.25 = 12262.5 \, \text{N}.

c

The effective force that the crane must overcome to lift the wreck is the weight minus the buoyant force:

F_{effective} = W - F_b = 98100 - 12262.5 = 85837.5 \, \text{N}.

d

Now, we can calculate the extension of the cable using Young's modulus. The formula is:

Y = \frac{F/A}{\Delta L/L}

. Rearranging gives us:

\Delta L = \frac{F \cdot L}{Y \cdot A}.

Here,

F = 85837.5 \, \text{N}

,

L = 10 \, \text{m}

,

Y = 5 \times 10^{10} \, \text{N/m}^2

, and

A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2

. Thus,

\Delta L = \frac{85837.5 \cdot 10}{5 \times 10^{10} \cdot 5 \times 10^{-4}} = \frac{858375}{2.5 \times 10^7} = 0.0343 \, \text{m} = 34.3 \, \text{mm}.

Answer

The change in the extension of the cable is approximately 34.3 mm.

Key Concept

Young's Modulus: Young's modulus is a measure of the stiffness of a material, defined as the ratio of stress (force per unit area) to strain (proportional deformation in length). The equation is given by:

Y = \frac{F/A}{\Delta L/L}.

Explanation

The extension of the cable is calculated by considering the effective force acting on it and applying Young's modulus to find how much the cable stretches under that load. The result shows how much the cable will extend when lifting the wreck.

Solution

a

To find the minimum thickness of the oil film for strong reflection of violet light, we use the condition for constructive interference in thin films. The formula is given by:

2nt = (m + \frac{1}{2})\lambda

where

n

is the refractive index of the oil,

t

is the thickness of the film,

m

is the order of interference (0, 1, 2,...), and

\lambda

is the wavelength of light in vacuum. For violet light, we take

m = 0

for the minimum thickness. Thus, we have:

2(1.5)t = \frac{1}{2}(410 \times 10^{-9})

b

Rearranging the equation to solve for

t

, we get:

t = \frac{(410 \times 10^{-9})}{2(1.5)} = \frac{410 \times 10^{-9}}{3} \approx 136.67 \times 10^{-9} \text{ m} = 136.67 \text{ nm}.

Answer

The minimum thickness of the oil film needed to observe a strong reflection of violet light is approximately 136.67 nm.

Key Concept

Thin Film Interference: The phenomenon where light waves reflected from the top and bottom surfaces of a thin film interfere with each other, leading to colorful patterns. The condition for constructive interference is given by the equation

2nt = (m + \frac{1}{2})\lambda

.

Explanation

The calculation shows how the thickness of the oil film can be determined based on the wavelength of light and the refractive index, leading to the observed colorful interference effects.

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