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AnnZ [28]
3 years ago
6

Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th

e internal bending moment is changing along the beam span. The imbalance in the bending stress resultant forces acting on a portion of the cross-section demands an internal horizontal shear force for equilibrium.
a. True
b. False
Engineering
1 answer:
jek_recluse [69]3 years ago
5 0

Answer:

False

Explanation:

When the horizontal shear forces act on the surface there is transverse shear stress at a particular point which is equal in magnitude. Pure bending is less common than a non uniform bending because the beam is not in equilibrium.

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Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a
nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer  (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

  1. Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer.  120 Vrms = 85 V and 24 Vrms = 17V = Vin
  2. Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
  3. Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

6 0
4 years ago
If i eat myself will I get twice as big or disappear completely?
Butoxors [25]
Disappear completely
5 0
3 years ago
A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate
olga2289 [7]

Answer:

1) The probability of at least 1 defective is approximately 45.621%

2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%

Explanation:

The given parameters are;

The defective rate of the device = 3%

Therefore, the probability that a selected device will be defective, p = 3/100

The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927

The probability of at least 1 = 1 - The probability of 0 defective in 20

∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621

The probability of at least 1 defective ≈ 0.45621 = 45.621%

2) The probability of at least 1 defective in a shipment, p ≈ 0.45621

Therefore, the probability of not exactly 1 defective = q = 1 - p

∴ q ≈ 1 - 0.45621 = 0.54379

The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212

Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%

4 0
3 years ago
Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th
jek_recluse [69]

Answer:

False

Explanation:

When the horizontal shear forces act on the surface there is transverse shear stress at a particular point which is equal in magnitude. Pure bending is less common than a non uniform bending because the beam is not in equilibrium.

5 0
3 years ago
Nonshielded cable with a 1.5-inch diameter should have a minimum bending radius of
Flura [38]

Answer:

c is the answer because we have to double the number

6 0
2 years ago
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