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Stells [14]
3 years ago
10

What are two factors that determine the thermal energy of a substance

Engineering
1 answer:
Sav [38]3 years ago
6 0
Mass and chemical composition
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Elements of parallel computing
Leya [2.2K]

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<h3><u>E</u><u>lements of parallel </u><u>computing:</u></h3>

  • Computer systems organization.
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6 0
2 years ago
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
What is differentiation​
oksian1 [2.3K]

Answer:

the action or process of differentiating or distinguishing between two or more things or people.

7 0
3 years ago
Someone should help me and explain stress strain curve​
dimulka [17.4K]

Answer:

In engineering and materials science, a stress–strain curve for a material gives the relationship between stress and strain. It is obtained by gradually applying load to a test coupon and measuring the deformation, from which the stress and strain can be determined (see tensile testing).

Explanation:

please mark brainliest

4 0
3 years ago
A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12
nydimaria [60]

Answer:

The value of Modulus of elasticity E = 85.33 × 10^{6} \frac{lbm}{in^{2} }

Beam deflection is = 0.15 in

Explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

Load = 125 × 32 = 4000\frac{ft lbm}{s^{2} }

We know that moment of inertia is given as

I = \frac{bt^{3} }{12}

I = \frac{5 (1.5^{3} )}{12}

I = 1.40625 in^{4}

Deflection = 0.15 in

We know that deflection of the beam in this case is given as

Δ = \frac{PL^{3} }{48EI}

0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}

E = 85.33 × 10^{6} \frac{lbm}{in^{2} }

This is the value of Modulus of elasticity.

Beam deflection is = 0.15 in

6 0
3 years ago
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