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Artemon [7]
2 years ago
12

7. The binary addition 1 + 1 + 1 gives ​

Engineering
2 answers:
scoray [572]2 years ago
7 0
1+1+1=3 should be the answer right?
kenny6666 [7]2 years ago
4 0

11 [2-bit]

011 [3-bit]

0011 [4-bit]

________

1 + 1 + 1 = 3

________

3 = 2 + 1

2¹ 2⁰

3 = (.. × 0) + (2¹ × 1) + (2⁰ × 1)

3 = ..011

Since 2³, 2⁴, 2⁵, .. are not used, they are represented as 0.

[ 2⁷ 2⁶ 2⁵ 2⁴ 2³ 2² 2¹ 2⁰ ]

[ 128 64 32 16 8 4 2 1 ]

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Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly fr
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Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

n_{collision}=\frac{9947.62m}{0.05m}\approx198952\  times

c) By using the equation of the ideal gases you obtain:

PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules

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Consider a N-channel enhancement MOSFET with VGS = 3V, Vt = 1 V, VDS = 10 V, and lambda =0 (channel length modulation parameter)
AveGali [126]

The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region.

<u>Explanation:</u>

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