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Butoxors [25]
3 years ago
14

The solid rod shown has a radius of 0.75 in. if it is subjected to the force the 500lb determine the max normal stress developed

at point
Physics
1 answer:
Kipish [7]3 years ago
7 0
The normal stress follows the formula written below:

σ = F/A

There are two types of stress, axial and tangential. Since we are only given with the dimension of the radius (and not the length), the possible stress is axial. So, the area is,

A = πr² = π(0.75 in)² = 1.767 in²

So,
σ = F/A = 500 lb/1.767 in² = <em>282.94 psi</em>
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There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0
3 years ago
The most common units for expressing the density of a substance are the g/cm3.
Luden [163]
True.

Density = mass / volume,  Unit = g / cm³.

This is a common unit because of its affiliation with the SI unit and because that also our popular liquid which is water = 1 g/cm³  
6 0
3 years ago
Two blocks of masses 3 kg and 5 kg approach each other with initial velocities 4 m/s and -6 m/s respectively. The two blocks col
Kipish [7]

Answer:

Explanation:

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6 0
2 years ago
What is the speed of a wave that has a frequency of 173 Hz and a wavelength of 2.59 meters? Express your answer to the nearest w
devlian [24]

Answer:

448 m/s is the correct answer.

Explanation:

7 0
3 years ago
Read 2 more answers
A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t
const2013 [10]

Answer:

\theta=12.19^{\circ}

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

sin\theta =\dfrac{u}{v}

Now by putting the values in the above equation we get

sin\theta =\dfrac{30}{142}

sin\theta=0.21

\theta=12.19^{\circ}

Therefore the angle will be 12.19° .

 

4 0
3 years ago
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