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3 years ago

What is the range of a 32-bit unsigned integer?

Engineering

1 answer:

Katyanochek1 [597]3 years ago

4 0

Answer:

0 to 4294967295

Explanation:

Unsigned integers have only positive numbers and zero. Their range goes from zero to (2^n)-1.

In the case of a 32-bit unsigned integer this would be.

(2^32) - 1 = 4294967296 - 1 = 4294967295

So the range goes from 0 to 4294967295 or form zero to about 4.3 billion.

The minus one term is because the zero takes one of the values.

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Compare the output of full-wave rectifier with and without filter

lara31 [8.8K]

Answer:

Full wave rectification flips the negative half cycle of the sine wave to positive so the result is two positive half cycles.

Explanation:

hope it helps a lil

4 0

3 years ago

Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distanc

Andrei [34K]

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

$\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}$

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

$\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m$

b) f = 1 GHz = 10⁹ Hz.

$\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm$

4 0

3 years ago

Jack has been concerned about the rapidly changing green regulations in his state and his ability as a mechanical engineer to ke

uysha [10]

Answer:

Option A, B and D

Explanation:

Jack can easily convince boss if he focus around two major aspects of the company

a) Revenue enhancement - Jack must outline the benefits of his research that can be used to improvise customer offerings and hence can be further used to devise more energy-efficient options to customer

b) Reduction in mistakes - Issues such as poor implementation can be avoided with better approach and understanding.

Hence, option A, B and D are correct

3 0

3 years ago

What is measurement in term of electrical engineering

laila [671]

Explanation:

voltage, current and resistance are the Volt [ V ], Ampere [ A ] and Ohm [ Ω ]

3 0

3 years ago

Technician A states that air tools generally produce more noise than electric tools, so wear ear protection when using air tools

dusya [7]

Answer:

Both Technician A and Technician B are correct

Explanation:

Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.

Technician B is also correct because it is always advisable to use impact sockets while using impact guns due to the ability of the impact sockets to withstand the force caused by operating impact guns and make work neater when nuts and bolts are being loosened or tightened.

5 0

3 years ago