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3 years ago

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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

plied. Assume that the deformation is totally elastic and that the elastic modulus for nickel is 207 GPa (or 207 × 109 N/m2).

1 answer:

pshichka [43]3 years ago

4 0

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

Elastic modulus = 207 Gpa

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

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3 years ago

A microscope illuminator uses a transformer to step down the 120 V AC of the wall outlet to power a 12.0 V,50 W microscope bulb.

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2 years ago

Steam enters an adiabatic nozzle at 2.5 MPa and 450oC with a velocity of 55 m/s and exits at 1 MPa and 390 m/s. If the nozzle ha

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Answer:

The value of exit temperature from the nozzle = 719.02 K

Explanation:

Temperature at inlet $T_{1}$ = 450°c = 723 K

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velocity at outlet $V_{2}$ = 390 $\frac{m}{sec}$

Specific heat at constant pressure for steam $C_{p} = 18723 \frac{J}{kg k}$

Apply steady flow energy equation for the nozzle

$h_{1} + \frac{V_{1} ^{2} }{2} = h_{2} + \frac{V_{2} ^{2} }{2}$

$C_{p} T_{1} + \frac{V_{1} ^{2} }{2} = C_{p} T_{2} + \frac{V_{2} ^{2} }{2}$

Put all the values in the above formula we get,

⇒ 18723 × 723 + $\frac{55^{2} }{2}$ = $C_{p}$ $T_{2}$ + $\frac{390^{2} }{2}$

⇒ $T_{2}$ = 719.02 K

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4 years ago

If it is struck by a rigid block having a weight of 550 lblb and traveling at 2 ft/sft/s , determine the maximum stress in the c

Lelu [443]

This question is incomplete, The missing image is uploaded along this answer below;

Answer:

the maximum stress in the cylinder is 3.23 ksi

Explanation:

Given the data in the question and the diagram below;

First we determine the initial Kinetic Energy;

T = $\frac{1}{2}$ mv²

we substitute

⇒ T = $\frac{1}{2}$ × (550/32.2) × (2)²

T = 34.16149 lb.ft

T = ( 34.16149 × 12 ) lb.in

T = 409.93788 lb.in

Now, the volume will be;

V = $\frac{\pi }{4}$ d²L

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V = $\frac{\pi }{4}$ × ( 0.5 × 12 in )² × ( 1.5 × 12 in )

V = 508.938 in³

So by conservation of energy;

Initial energy per unit volume = Strain energy per volume

⇒ T/V = σ²/2E

from the image; E = 6.48(10⁶) kip

so we substitute

⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]

508.938σ² = 5,312,794,924.8

σ² = 10,438,982.5967

σ = √10,438,982.5967

σ = 3230.9414

σ = 3.2309 ksi ≈ 3.23 ksi { three significant figures }

Therefore, the maximum stress in the cylinder is 3.23 ksi

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