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Tom [10]
3 years ago
11

What boolean operation does the circuit compute?

Physics
1 answer:
vagabundo [1.1K]3 years ago
8 0

A circuit computes various types of Boolean operations. As in the question no specific circuit has been shown,so we have to know the fundamental logic operations.

Out of the various types of logic operations in Boolean algebra,the simple operations are- addition,subtractions,multiplications,division through logic gates.

Various logic or Boolean operations are  NOT GATE,OR GATE ,AND GATE .

There are other operations also like NAND GATE,NOR GATE and XOR GATE etc

NOT gate is the simple logic gate which inverts one quantity.it makes 1[high] to 0[low] and vice versa.

OR gate gives 1 [high] when one of the input is 1 and low when all the inputs are low.

AND operation gives 1 when all the inputs are high and low when one of the input is low

NAND operation is AND gate followed by a NOT gate

NOR operation is the OR gate followed by a NOT gate.

Remembering these simple rules we can get the idea of the operations involved in the circuit. Each circuit has different symbols and gives different results.

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Calculate the total resistance in a parallel circuit made up of resistances of 2Ω, 3Ω, and 4Ω. (Hint: When converting to a decim
natta225 [31]

Answer:

Total resistance = 0.92Ω

Explanation:

For parallel connected resistors we have effective resistance

                 \frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+.....

Here parallel circuit made up of resistances of 2Ω, 3Ω, and 4Ω.

That is

             R₁ = 2Ω

             R₂ = 3Ω

             R₃ = 4Ω

Substituting

             \frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\\\frac{1}{R_{eff}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\\\\frac{1}{R_{eff}}=\frac{6+4+3}{12}\\\\R_{eff}=\frac{12}{13}=0.92ohm

Total resistance = 0.92Ω

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3 years ago
In Which senecio is an animal doing work
Arisa [49]

Answer:

A horse pulls a wagon along a road

4 0
3 years ago
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likoan [24]

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difference

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3 years ago
10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time perio
Afina-wow [57]

Answer:

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)

V2R+h=g(R2(R+h)2)

V=√g(R2R+h)

V= sqrt(9.8 × (6371000)^2/(6371000+360000)

V= sqrt(9.8× (4.059×10^13/6731000)

V=sqrt(65648789.18)

V= 8102.39m/s

Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)

T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)

T=sqrt(3.40×10^21)/ (3.99×10^14)

T= sqrt(0.862×10^7)

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