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Pachacha [2.7K]
3 years ago
15

An alloy to be used for a spring application must have a modulus resilience of at least 0.79 x 106 J/m3 (0.79 x 106 Pa). What mu

st be its minimum yield strength (in MPa)? Assume that the modulus of elasticity for this alloy is 102 GPa
Using Ur=Sig^2/2E, I am getting 401.447E6 Mpa but apparently thats wrong.
Engineering
1 answer:
xxTIMURxx [149]3 years ago
3 0

Answer:

4.13 MPa

Explanation:

Given

Modulus of elasticity = E = 102 GPa

E = 102 * 10^9

Modulus resilience U = 0.79 * 106Pa

Modulus of Resilience U = y²/2E --- Make y the subject of formula

y² = 2EU

y = √(2EU)

y = √(2 * 102 * 10^9 * 0.79 * 106)

y = √1.708296E13

y = 4133153.759540044

y = 4.13E6

y = 4.13MPa

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a) 19 or select the closest answer

b) 5%

Explanation:

a)

from the voltage divide rule

V_{in} = V_0 * \frac{R_2}{R_2 + R_f}

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6 0
3 years ago
Consider a cylindrical nickel wire 2.1 mm in diameter and 3.2 × 104 mm long. Calculate its elongation when a load of 280 N is ap
pshichka [43]

Answer:

Total elongation will be 0.012 m

Explanation:

We have given diameter of the cylinder = 2.1 mm

Length of wire L=3.2\times 10^4mm

So radius r=\frac{d}{2}=\frac{2.1}{2}=1.05mm=1.05\times 10^{-3}m

Load F = 280 N

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Area of cross section A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2

We know that elongation in wire is given by \delta =\frac{FL}{AE}, here F is load, L is length, A is area and E is elastic modulus

So \delta =\frac{FL}{AE}=\frac{280\times 32}{3.461\times 10^{-6}\times 207\times 10^9}=0.012m

4 0
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