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denis23 [38]
3 years ago
11

Amazon rectangular bar of low carbon steel A-36 is exposed to an axial strees of 150 MPa. What is the original length of the bar

in m, if it elongates 1.35 mm under these conditions?
Physics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

1.8m

Explanation:

Let the Elastics of the steel ASTM-36 E = 200000 MPa

The strain of the bar when subjected to 150 MPa is

\epsilon = \frac{\sigma}{E} = \frac{150}{200000} = 0.00075

Therefore, if the bar elongates by 1.35 mm, then the original length L would be:

\epsilon = \frac{\Delta L}{L}

L = \frac{\Delta L}{\epsilon} = \frac{1.35}{0.00075} = 1800 mm or 1.8m

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2 years ago
Unpolarized light passes through two polarizers. Find the fraction of light from the first polarizer that gets through the secon
meriva

Answer:

0.5

Explanation:

Data provided in the question:

The angle between their transmission axes, θ = 60°

Now,

We have the relation,

I₁ = I₀cos²θ

where,

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on rearranging, we get

\frac{I_1}{I_0}=cos²60°

or

\frac{I_1}{I_0}=0.5

8 0
3 years ago
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. The magnitudes of two forces are measured to be 120 ± 5 N and 60 ± 3 N. Find the sum
Tju [1.3M]

Explanation:

120+60=180

120-60=60

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3 years ago
What will be the ratio of distances between the two charges of each pair of charges (1µC, 2µC) and (2µC, -3uC) so that force her
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3 years ago
The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?
BigorU [14]

1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

where

c=3\cdot 10^{-8} m/s is the speed of light

\lambda is the wavelength of the wave

In this problem, we have light with wavelength

\lambda=425 nm=425\cdot 10^{-9} m

Substituting into the equation, we find the frequency:

f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


2. Period: 1.42 \cdot 10^{-15}s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s


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3 years ago
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