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Sergeu [11.5K]
3 years ago
5

Draw the logic circuit for each of the following. For each gate, determine if it generates either EVEN or ODD parity bit and fin

ds the output for the given input data: (Remember: A XOR generates EVEN parity bit. A XNOR generates Odd Parity bit, whatever how many inputs they have.)
Data Inputs Which kind parity bit can it generate?
4-input XOR, input data-1001 Even Parity Bit ODD Parity Bit
5-input XOR, input data-10010
6-input XOR, input data-101001
7-input XOR, input data 1011011
Engineering
1 answer:
omeli [17]3 years ago
8 0

Answer:

a) 4-input XOR, input data-1001  = 0 Even parity Bit

b)  5-input XOR, input data-10010 = 0 Even parity Bit  

c) 6-input XOR, input data-101001 = 1 Even parity Bit

d) 7-input XOR, input data 1011011 = 1 Even parity Bit

Explanation:

a) 4-input XOR, input data-1001  ;  generates 0 Even parity Bit

b)  5-input XOR, input data-10010 ; generates 0 Even parity Bit  

c) 6-input XOR, input data-101001 ; generates 1 Even parity Bit

d) 7-input XOR, input data 1011011 ; generates 1 Even parity Bit

Attached below is the Logic circuits of the data inputs

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The 1000-lb elevator is hoisted by the pulley system and motor M. If the motor exerts a constant force of 500 lb on the cable, d
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Answer:

\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{32965.5}{0.65}\\ p_{in}=50716.1538 lb.ft/s\\

In hp:

p_{in}=\frac{50716.1538}{500}\\ p_{in}=101.432 hp

Part B:

\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{50356.2}{0.65}\\ p_{in}=77471.07692 lb.ft/s\\

In hp:

p_{in}=\frac{77471.07692}{500}\\ p_{in}=154.94215 hp

Explanation:

Weight of elevator=1000-lb

Force=500 lb

s=15 ft

Force on pulley=F=3*500=1500 lb

g=32.2 ft/s^2

According to Newton's Second law:

\sum F_y=ma_y

According to attachment:

F-W=ma_y

1500-1000=\frac{1000}{32.2}a_y

a_y=16.1 ft/s^2

According to third equation of motion:

v^2=v_o^2+2a_y(S-So)\\

Where:

Vo is initial velocity

V is final velocity

S is final distance

So is starting distance

v^2=(0)^2+2*16.1*(15)\\v^2=483\\v=21.977 ft/s

Output power:

P_{out}=F.v\\P_{out}=1500*21.977\\P_{out}=32965.5 lb.ft/s

\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{32965.5}{0.65}\\ p_{in}=50716.1538 lb.ft/s\\

In hp:

p_{in}=\frac{50716.1538}{500}\\ p_{in}=101.432 hp

Part B:

When S=35 ft

v^2=(0)^2+2*16.1*(35)\\v^2=1127\\v=33.5708 ft/s

Output power:

P_{out}=F.v\\ P_{out}=1500*33.5708 \\ P_{out}=50356.2 lb.ft/s

\epsilon=\frac{p_{out}}{P_{in}} \\p_{in}=\frac{p_{out}}{\epsilon} \\p_{in}=\frac{50356.2}{0.65}\\ p_{in}=77471.07692 lb.ft/s\\

In hp:

p_{in}=\frac{77471.07692}{500}\\ p_{in}=154.94215 hp

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Answer:

See attachment below

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Answer:

program:

#include<stdio.h>

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