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3 years ago

13

C: 5(111 - 18) = 3(25 + 7) What’s the freaking answer

1 answer:

Minchanka [31]3 years ago

6 0

Answer:

Which of the following is the Spanish verb estar used to describe?

Explanation:

Which of the following is the Spanish verb estar used to describe?

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The greater the change in the speed of light in a different media the greater the angle of what?

valkas [14]

Answer:

Idk

Explanation:

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3 years ago

A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m

ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

$S_{el} = 414Mpa$

Yield Strain of the Specimen

$\epsilon_{el} = 0.002$

Diameter of the test-specimen

$d_0 = 12.8mm$

Gage length of the Specimen

$L_0 = 50mm$

Modulus of elasticity

$E = \frac{S_{el}}{\epsilon_{el}}$

$E = \frac{414Mpa}{0.002}$

$E = 207Gpa$

Strain energy per unit volume at the elastic limit is

$U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}$

$U'_{el} = \frac{1}{2} (414)(0.002)$

$U'_{el} = 414kN\cdot m/m^3$

Considering that the net strain energy of the sample is

$U_{el} = U_{el}' \cdot (\text{Volume of sample})$

$U_{el} = U_{el}'(\frac{\pi d_0^2}{4})(L_0)$

$U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})$

$U_{el} = 2.663N\cdot m$

Therefore the net strain energy of the sample is $2.663N\codt m$

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3 years ago

What force oppose other forces and slow objects down?

LuckyWell [14K]

Friction is a force that oppose other forces and can slow them down

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4 years ago

Jack tries to place magnets on the door of his refrigerator. He observes that the magnets don’t stick. He guesses that the door

jonny [76]

B. Construct a hypothesis

6 0

3 years ago

Read 2 more answers

In a double-slit experiment, the third-order maximum for light of wavelength 490 nm is located 15 mm from the central bright spo

jeka94

Answer:

The distance from the central bright spot are $156.8\times10^{-3}\ mm$ and $142.9\times10^{-3}\ mm$ .

Explanation:

Given that,

Wavelength = 490 nm

Distance y= 15 mm

Length L=1.6 m

New wavelength = 670

We need to calculate the distance from the central bright spot

Using formula of distance

$y= \dfrac{m\lambda L}{d}$

$d=\dfrac{m\lambda L}{y}$

Put the value into the formula

$d=\dfrac{3\times490\times10^{-9}\times1.6}{15\times10^{-3}}$

$d=156.8\times10^{-6}\ m$

$d=156.8\times10^{-3}\ mm$

We need to calculate the distance from the central bright spot for new wavelength

Using formula of distance

$d=\dfrac{m\lambda L}{y}$

Put the value into the formula

$d=\dfrac{2\times670\times10^{-9}\times1.6}{15\times10^{-3}}$

$d=142.9\times10^{-6} m$

$d=142.9\times10^{-3}\ mm$

Hence, The distance from the central bright spot are $156.8\times10^{-3}\ mm$ and $142.9\times10^{-3}\ mm$ .

8 0

3 years ago