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ICE Princess25 [194]
3 years ago
14

Solve the inequality. Then graph your solution. -9v – 10 < 7y +6

Engineering
1 answer:
Cerrena [4.2K]3 years ago
5 0
16

if you add 9+10 you get 18 - 7+6
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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
A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto
vazorg [7]

Answer:

(M_t)_{rated}=61.11lb-in

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = 4 \times 0.7457 =2.9828 KW

service factor(k)= 2.75

now,

KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }

M_t= 18,989.09 \ N-mm= 168.06 lb-in

torque rating

(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in

4 0
4 years ago
According to the
zysi [14]

Answer:

The part of the system that is considered the resistance force is;

B

Explanation:

The simple machine is a system of pulley  that has two pulleys

The effort, which is the input force at A gives the value of the tension at C and  D which are used to lift the load B

Therefore, we have;

A = C = D

B = C + D = C + C = 2·C

∴ C = B/2

We have;

C = B/2 = A

Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B

The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done

It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force

Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.

3 0
3 years ago
Is a street the same as a avenue
-BARSIC- [3]

they're essentially the same thing so i'd say yes

5 0
3 years ago
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A "type 30" brake chamber has a diaphragm diameter of 8" which gives it
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The piston makes four strokes in the crankshaft makes two revolutions between combustion firings. The diameter of the piston, and the inside diameter of the cylinder, is called a bore, so the area of the head of the piston is pi times the diameter squared divided by four.
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3 years ago
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