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Sergeeva-Olga [200]
3 years ago
12

Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pull

ed in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.
Physics
1 answer:
bearhunter [10]3 years ago
7 0

Answer:

strain = 1.4 \times 10^{-3}

Explanation:

As we know by the formula of elasticity that

E = \frac{stress}{strain}

now we have

E = 110 GPA

E = 110 \times 10^9 Pa

Area = 15.2 mm x 19.1 mm

A = 290.3 \times 10^{-6}

now we also know that force is given as

F = 44500 N

here we have

stress = Force / Area

stress = \frac{44500}{290.3 \times 10^{-6}}

stress = 1.53 \times 10^8 N/m^2

now from above formula we have

strain = \frac{stress}{E}

strain = \frac{1.53 \times 10^8}{110 \times 10^9}

strain = 1.4 \times 10^{-3}

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A skateboarder traveling at 7.0 meters per second rolls to a stop at the top of a ramp in 3.0 seconds. What is the skateboarder’
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8 0
2 years ago
A positive and a negative charge, each with a magnitude of 1.41 x 10^-5 C, are separated by a distance of 0.44 m. Find the force
vredina [299]
For this question, we use the Coulumb's law to calculate the force on each particles. In this law, force between point charges are said to be proportional to product of each charge and is  indirectly proportional to the distance of both charges. We do as follows:

F= kq(1)q(2)/d^2 
  = (9x10^9).(1.41 x 10^-5 C).(-<span>1.41 x 10^-5 C</span><span>)/.44^2 
</span>  = 4.067 N


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3 years ago
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