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3 years ago

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Consider the two level (AND-OR) function f1(A, B, C, D) = AC0D + AB0D0 and two level (OR-AND) function f2(A, B, C, D) = (A+B +D0

)(B0 +C +D0 ). Find the following two level circuits expressions for f1 and f2. (Hint: use both SOP and POS form of f1 and f2

1 answer:

Soloha48 [4]3 years ago

3 0

Answer:

Answer for the question :

"Consider the two level (AND-OR) function f1(A, B, C, D) = AC0D + AB0D0 and two level (OR-AND) function f2(A, B, C, D) = (A+B +D0 )(B0 +C +D0 ). Find the following two level circuits expressions for f1 and f2. (Hint: use both SOP and POS form of f1 and f2"

is explained in the attachment.

Explanation:

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A. The water

Biotic factors mean living this such as frogs, grasshoppers, snakes, etc.

Aboriginal factors means non-living such as water, temperature, sunlight, etc.
Abiotic factors are the non-living parts of the environment that have a major influence on living organisms which makes water an abiotic factor.

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4 0

3 years ago

Read 2 more answers

A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0

Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

Explanation:

Given that,

Electric field $E=-6.00\times10^{5}i\ N/C$

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

$F=F_{e}$

$ma=qE$

$a=\dfrac{Eq}{m}$

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

$a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}$

$a=-5.74\times10^{13}\ m/s62$

The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). We need to calculate the initial peed

Using equation of motion

$v^2-u^2=2as$

Where, s = distance

Put the value into the formula

$0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}$

$u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}$

$u=2834783.94$

$u=2.83\times10^{6}\ m/s$

The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

$t=\dfrac{u}{a}$

Where, u = initial velocity

a = acceleration

Put the value into the formula

$t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}$

$t=0.493\times10^{-7}\ sec$

Hence, (a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

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2 years ago

On a cold day, snakes usually lie very still and eat little or nothing while birds usually move around and eat a lot of food. Wh

RoseWind [281]

Birds are warm-blooded, which means they need to move and eat more in the cold to stay warm; snakes are cold-blooded, so they don't have to.

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3 years ago

A toy car having mass m = 1.50 kg collides inelastically with a toy train of mass M = 3.60 kg. Before the collision, the toy tra

baherus [9]

Answer:

$v_{f}$ = 3,126 m / s

Explanation:

In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.

the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s

Before the crash

p₀ = m v₁₀ + M v₂₀

After the inelastic shock

$p_{f}$ = m $v_{1f}$ + M $v_{2f}$

p₀ = $p_{f}$

m v₀ + M v₂₀ = m $v_{1f}$ + M $v_{2f}$

We cleared the end of the train

M $v_{2f}$ = m (v₁₀ - v1f) + M v₂₀

Let's calculate

3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45

$v_{2f}$ = (-3.9 + 8.82) /3.60

$v_{2f}$ = 1.36 m / s

As we can see, this speed is lower than the speed of the car, so the two bodies are joined

set speed must be

m v₁₀ + M v₂₀ = (m + M) $v_{f}$

$v_{f}$ = (m v₁₀ + M v₂₀) / (m + M)

$v_{f}$ = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)

$v_{f}$ = 3,126 m / s

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3 years ago

Until 1883, every city and town in the United States kept its own local time. Today, travelers reset their watches only when the

inna [77]

Answer:

135° (degrees)

Explanation:

The Earth rotates 360° in about 24 h

How many degrees in 1 hour

$360^{\circ}=24\ h\\\Rightarrow 1\ h=\frac{360}{24}\\\Rightarrow 1\ h=15^{\circ}$

So, in 1 hour the Earth rotates 15 degrees. This means that a person would have to reset their watch by 1 hour after travelling 15 degrees

In 9 hours

$9\ h=9\times 15^{\circ}=135^{\circ}\\\Rightarrow 9\ h=135^{\circ}$

So, a person would have to travel 135 degrees in order to reset their watch by 9 hours.

7 0

3 years ago