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Alik [6]
3 years ago
6

PLEASEE HELP ME.... ​

Physics
1 answer:
andrey2020 [161]3 years ago
8 0
Wat the heck wowwwwwwwwwww
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Which of the following is characteristic of the second stage within a demographic transition?
Elan Coil [88]
The answer is A) <span>The death rate begins to fall, but birth rates remain high for a time.</span>
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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

6 0
3 years ago
20 POINTS!!!
hram777 [196]
I think its inductance. If its not then I think its none of the above
5 0
3 years ago
A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that
Marat540 [252]

Answer:

Part(a): The value of the spring constant is 3.11 \times 10^{2}~Kg~s^{-2}.

Part(b): The work done by the variable force that stretches the collagen is 1.5 \times 10^{-6}~J.

Explanation:

Part(a):

If 'k' be the force constant and if due the application of a force 'F' on the collagen '\Delta l' be it's increase in length, then from Hook's law

F = k~\Delta l....................................................................(I)

Also, Young's modulus of a material is given by

Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)

where 'A' is the area of the material and 'l' is the length.

Comparing equation (I) and (II) we can write

&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation (III), we have

k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}

Part(b):

We know the amount of work done (W) on the collagen is stored as a potential energy (U) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)

Substituting all the values, we can write

W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J

3 0
3 years ago
When an object is fully converted into energy the amount of energy liberated is
EleoNora [17]

Answer:

Mass, m = 4 kg

Explanation:

<u>Given the following data;</u>

Energy = 3.6 * 10^17 Joules

We know that the speed of light is equal to 3 * 10⁸ m/s.

To find the mass of the substance;

The theory of special relativity by Albert Einstein gave birth to one of the most famous equation in science.

The equation illustrates, energy equals mass multiplied by the square of the speed of light.

Mathematically, the theory of special relativity is given by the formula;

E = mc^{2}

Where;

  • E is the energy possessed by a substance.
  • m is the mass.
  • c is the speed of light.

Substituting into the formula, we have;

3.6 * 10^{17} = m * 300000000^{2}

3.6 * 10^{17} = m * 9*10^{16}

m = \frac {3.6 * 10^{17}}{9*10^{16}}

Mass, m = 4 kg

8 0
3 years ago
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