Design the necessary circuit using Logisim to implement the situation described above. Use Kmaps for simplification. Be VERY car
eful to get the correct functions for your four outputs before simplifying and designing the circuit with Logisim. You should minimize the entire circuit. This means you should look at each function AND its complement to see which combination will give you the minimal circuit. For example, using the functions that represent UL, UR (the complement of UR), LR, and LL (with NOT gates after UR and LL of course) might give the minimal circuit overall.Your circuit should have four inputs and four LED outputs. Your outputs should be such that the LEDs are arranged to light up the dots in their actual places (stack the 4 LEDs into a square.)All inputs should be labeled (in Logisim, not by hand). You can label outputs if you wish, but since you are placing the LEDs in the correct position, that isn't necessary. Please use the labeling for input (and output if you choose) as given above so everyone's circuit is labeled the same.You should also include your name as a label on the circuit.Test your circuit to be sure it is working correctly.Name the file using your first initial and last name followed by 4-2, for example lnull4-2.circ.
2 answers:
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Answer:
,
Explanation:
Since man and river report constant speeds and velocities are mutually perpendicullar, the absolute speed of the man is calculated by the Pythagorean Theorem:
The required time to make the crossing is:
The omitted part of the question shown in bold format
The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume μs = 0.18 and μk = 0.16. What couple must be applied to the shaft to exert a 30-lb force on the clamped object?
Answer:
C = 0.967 in. lb
Explanation:
Given that:
The lead of the threaded shaft of the C-clamp is 0.05 in.
∴ the pitch of the screw = 0.05 in
the mean radius of the thread is r = 0.15 in.
Assuming:
(μs)= 0.18 which implies the coefficient of the static friction
(μk) = 0.16 (coefficient of kinetic friction)
Force = 30-lb
What couple must be applied to the shaft to exert a 30-lb force on the clamped object?
To determine the Couple (C) that must be applied; we use the expression:
C = Fr tan ( +
)
where; F = force
r = mean radius
= angle of kinetic friction
= pitch angle
NOW, let's take then one after the other.
From the coefficient of the static friction and the kinetic friction; we can solve for their respective angles, so we have:
Angle of static friction ()
() =
() = 10.204°
Angle of kinetic friction ()
= 9.0903°
To determine the pitch angle(); we apply the expression:
() =
() =
() =
() = 3.0368°
Have gotten our parameters to solve for Couple (C); then we have:
C = Fr tan ( +
)
substituting our values; we have:
C = (30 × 0.15) tan ( 9.0903 + 3.0368)
C = 4.5 × tan ( 12.1271)
C = 4.5 × 0.2148761968
C = 0.96669428854 in.lb
C = 0.967 in. lb
Therefore, 0.967 in. lb couple must be applied to the shaft to exert a 30-lb force on the clamped object.
Answer:
mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Explanation:
The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
