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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
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Explain what a margin of safety is in driving as well as how it can help minimize risk.
Yakvenalex [24]

Answer:

A safety margin is the space left between your vehicle and the next to provide room, time and visibility at every instant

Explanation:

A safety margin is defined as an allowance given between your vehicle and the next vehicle in front to provide enough room, visibility and time to move in a safe manner to prevent the occurrence of an accident at anytime the frontal vehicle suddenly stops or slows down

Safety margins help minimize risks in the following way

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6 0
3 years ago
If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical
vampirchik [111]

Answer:

critical stress required for the propagation is 27.396615 ×10^{6} N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 ×10^{9} N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}    .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm  = 0.3 ×10^{-3} m

so now put value in equation 1 we get

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}

( σc ) = \sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}

( σc ) = 27.396615 ×10^{6} N/m²

so critical stress required for the propagation is 27.396615 ×10^{6} N/m²

6 0
3 years ago
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