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Ulleksa [173]
3 years ago
7

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

m dia and has a 50-mm gage length. What is its modulus of elasticity? What is the strain energy at the elastic limit? Can you define the type of metal based on the given data?
Physics
1 answer:
ser-zykov [4K]3 years ago
6 0

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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MrRissso [65]

Answer:Explained

Explanation:

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Some Standard Precautions are

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3 0
3 years ago
A projectile is launched at an angle of 60 degrees to the horizontal at an initial speed of 10 meters per second. What is the ma
Flura [38]

Answer:

5 sq. root 3

Explanation:

theta= 60°

=> u sin theta = 10 × sin 60

= 10× sq. root 3/2

= 5 sq. root 3

3 0
2 years ago
Read 2 more answers
1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t
tiny-mole [99]

Explanation:

Charges,q_1=8\ \mu C=8\times 10^{-6}\ C

q_2=-5\ \mu C=-5\times 10^{-6}\ C

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0
3 years ago
A net force of 345 N accelerates a boy on a sled at 3.2 m/s^2 . What is combined mass of the sled
Daniel [21]

Answer:

Mass, m = 26.54kg

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

Fnet = Fapp + Fg

Where;

  • Fnet is the net force
  • Fapp is the applied force
  • Fg is the force due to gravitation

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

Net force, Fnet = 345

Acceleration, a = 3.2m/s²

<u>To find mass;</u>

Fnet = Fapp + Fg

Fnet = ma + mg

Fnet = m(a+g)

m = Fnet/(a+g)

We know that acceleration due to gravity, g = 9.8m/s²

Substituting into the equation, we have;

m = 345/(3.2 + 9.8)

m = 345/13

Mass, m = 26.54kg

6 0
2 years ago
A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc
hoa [83]

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

E V = log_2(\dfrac{N^2}{t})

N is the diameter of the aperture and t is the time of exposure

now,

log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})

\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}

inserting all the values

\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}

      N₂² = 4

      N₂ = 2 mm

hence , the correct answer is option E

4 0
3 years ago
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