Magnetic micro-generators are considered for the recovery of mechanical energy from vibrations. a)-True b)- False,

Engineering

1 answer:

sleet_krkn [62]3 years ago

7 0

Answer:

False

Explanation:

Magnetic micro generators are used to generate power making use of very small generators for generation of electricity and consists of prime movers, electric generators of quite small size or conversion devices in order to convert environmental energy like heat or vibrations into electrical energy and not mechanical energy.

Therefore, magnetic micro generators are not used for the recovery of mechanical energy from vibrations but electrical energy from them.

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An aluminum bar 125 mm long with a square cross section 16 mm on an edge is pulled in tension with a load of 66,700 N and experi

AfilCa [17]

Answer: the modulus of elasticity of the aluminum is 75740.37 MPa

Explanation:

Given that;

Length of Aluminum bar L = 125 mm

square cross section s = 16 mm

so area of cross section of the aluminum bar is;

A = s² = 16² = 256 mm²

Tensile load acting the bar p = 66,700 N

elongation produced Δ = 0.43

Δ = PL / AE

we substitute

0.43 = (66,700 × 125) / (256 × E)

0.43(256 × E) = (66,700 × 125)

110.08E = 8337500

E = 8337500 / 110.08

E = 75740.37 MPa

Therefore, the modulus of elasticity of the aluminum is 75740.37 MPa

4 0

3 years ago

A 20cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter of the bar is 0.490 at this load, de

Margaret [11]

If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

<h3>What is strain?</h3>

Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

L = 20 cm d x 1 = 0.21 cm
dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
(assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
= 1561.84 MPa.