Answer:
(a) 16.27 Vpk
(b) 48.7%
Explanation:
The transformer is assumed to be an ideal 10:1 voltage divider with no internal impedance. The diode is assumed to be modeled in the forward direction by a perfect 0.7 V voltage drop with no internal impedance. That means the frequency of the supply voltage is irrelevant.
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<h3>(a)</h3>
The peak voltage will be 0.7 V less than the transformer secondary peak voltage:
((120 V)√2)/10 -0.7 V ≈ 16.27 V
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<h3>(b)</h3>
The fraction of the amplitude for which the diode is non-conducting is ...
0.7/(12√2) ≈ 0.041248
The period of conduction is symmetrical about the peak of the waveform, so it is convenient to use the arccos function to find the (half) conduction angle:
arccos(0.041248) ≈ 87.64°
As a fraction of half the cycle, this is ...
conduction fraction ≈ 87.64°/180° ≈ 48.7%
Answer:
Factor of safety against static failure = 2.097
Factor of safety against fatique failure = 1.6
Explanation:
Bending stress = 12 kpsi
Se = 40 kpsi,
Sy = 60 kpsi,
Sut = 80 kpsi
See the attached file for explanation
Answer:
d)0.991lbf
Explanation:
the property known as specific gravity is the relationship between the density of the fluid and the density of the water, so to find the density of the rock we must multiply the specific gravity by the density of the water = 1000kg / m ^ 3, when we have the density we must multiply it by volume and gravity to find the weight.
W=2.5ρVg
ρ=Density of water=1000kg / m ^ 3
V=0.00018 m^3
g=9.81m/s^2
W=(2.5)(1000)(0.00018)(9.81)
W=4.41N=0.991lbf