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Rainbow [258]
3 years ago
5

Do the following addition exercises by translating the numbers into 8-bit 2's complement binary numbers, performing the arithmet

ic, and translating the result back into a decimal number. Indicate where overflow occurs and why, based on the binary arithmetic:
(a) 47 + 38
(b) 47 - 38
(c) -47 - 38
(d) 47 + 88
(e) -47 + 88
(f) 47 - 88
Engineering
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

I am attaching a file with the solution and explanation as the number character limit is exceeding.

Explanation:

Download docx
You might be interested in
1. A thin plate of a ceramic material with E = 225 GPa is loaded in tension, developing a stress of 450 MPa. Is the specimen lik
mina [271]

Answer:

fracture will occur as the value is less than E/10 (= 22.5)

Explanation:

If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.

\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}

=  2\times 750 (\frac{\frac{0.2mm}{2}}{0.001 mm}})^{1/2}

                 = 15 GPa

fracture will occur as the value is less than E/10 = 22.5

7 0
3 years ago
Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

4 0
3 years ago
Determine the static pressure to stagnation pressure ratio associated with the following motion in standard air: (a) a runner mo
vredina [299]

Answer: a) 0.00017

b) 0.0013

c) 0.0022

d) 0.017

Explanation:

3 0
3 years ago
What information you would gain by watching the video. Plz Help due today.
Archy [21]

Answer: It doesnt let me watch the link im sorry :(

Explanation:

3 0
2 years ago
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are current
postnew [5]

Answer:

D=0.016m

Explanation:

From the question we are told that:

Discharge Rate F_r=0.5kgls

Pressure P=15Kpa

Temperature T=25=>298K

Ambient pressure is 1 atm.

Generally the equation for Density is mathematically given by

\rho=\frac{PM}{RT}

\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}

\rho=16.958kg/m^2

Generally the equation for Flow rate is mathematically given by

F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}

Where

Q=Heat coefficient\ ratio\ of\ Nitrogen

Q=1.4

\mu= Discharge\ coefficient

\mu=0.68

Therefore

0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}

A=2.129*10^{-4}

Where

A=\frac{\pi}{4}D^2

\frac{\pi}{4}D^2=2.129*10^{-4}

D=0.016m

8 0
3 years ago
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