What character string does the binary ASCII code 1010100 1101000 1101001 1110011 0100000 1101001 1110011 0100000 1000101 1000001 1010011 1011001 0100001?

Answers

Answer 1

Answer:

Answer: This is EASY!

Explanation:

To make it easy, you would convert those binary numbers and to denary. And this gives:

84 104 105 115 32 105 115 32 69 65 83 89 33

Then, the denary numbers generated can be converted to ASCII code using this ASCII Table in the attachment below. And the result is: This is EASY!

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Which property of real numbers is shown below?3 + ((-5) + 6) = (3 + (-5)) + 6
commutative property of addition
identity property of multiplication
associative property of addition
commutative property of multiplication

Answers

The property of realnumbers is shown below is associative property of addition. The correct option is C.

What is associative property of addition?

According to the associativeproperty of addition, you can arrange the addends in several ways without changing the result.

According to the commutative property of addition, you can rearrange the addends without altering the result.

When more than two numbers are added together or multiplied, the outcome is always the same, regardless of how the numbers are arranged.

This is known as the associativeproperty. As an illustration, 2 (7 6) = (2 7) 6. 2 + (7 + 6) = (2 + 7) + 6.

Thus, the correct option is C.

For more details regarding associative property, visit:

brainly.com/question/30111262

#SPJ1

Answer:

Explanation:

What is 29.95 inHg in kPa?

Answers

Answer:

101.42235 kPa

Explanation:

The unit inHg means "inches of mercury", Its a pressure unit commonly used by the US aviators.

The conversion value to KPa (kilopascal) is

1 inHg= 3.386389 kPa

So now we only have to multiply:

29.95 inHg * 3.386389 kPa/in Hg =101.42235 kPa

Have a nice day and Good Luck!

Compute the electrical resistivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in length in which a current of 0.25 A passes in an axial direction. A voltage of 24 V is measured across two probes that are separated by 45 mm (1.75 in.).

Answers

No no no no no no no no no no

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

Answers

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

$(d_(2) )/(d_(1) ) =(1)/(2) (\sqrt{1+8F^(2) } -1)$

$10*2=\sqrt{1+8F^(2) } -1$

Solving for F:

F = 7.4162

$v=F\sqrt{gd_(1) }$

Here, g = gravity = 32.2 ft/s²

$v=7.4162*√(32.2*1) =42.0833ft/s$

b) The flow rate:

$q=v*L*d_(1) =42.0833*80*1=3366.664ft^(3) /s$

c) The Froude number:

$v_(2) =(q)/(L*d_(2) ) =(3366.664)/(80*10) =4.2083ft/s$

$F=\frac{v_(2)}{\sqrt{gd_(2) } } =(4.2083)/(√(32.2*10) ) =0.2345$

d) The flow energy dissipated:

$E=((d_(2)-d_(1))^(3) )/(4d_(1)d_(2)) =((10-1)^(3) )/(4*1*10) =18.225ft$

e) The critical depth:

$d_(c) =((((q)/(L))^(2) )/(g) )^(1/3) =((((3366.664)/(80))^(2) )/(32.2) )^(1/3) =3.8030ft$

For a p-n-p BJT with NE 7 NB 7 NC, show the dominant current components, with proper arrows, for directions in the normal active mode. If IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA, calculate the base transport factor, emitter injection efficiency, common-base current gain, common-emitter current gain, and ICBO. If the minority stored base charge is 4.9 * 10-11 C, calculate the base transit time and lifetime.

Answers

Answer:

=> base transport factor = 0.98.

=> emitter injection efficiency = 0.99.

=> common-base current gain = 0.97.

=> common-emitter current gain = 32.34.

=> ICBO = 1 × 10^-6 A.

=> base transit time = 0.325.

=> lifetime = 1.875.

Explanation:

(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).

The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.

(1). The base transport factor = ICp/IEp=9.8/10 = 0.98.

(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 = 0.99.

(3).common-base current gain = 0.98 × 0.99 = 0.9702.

(4).common-emitter current gain =0.97 / 1- 0.97 = 32.34.

(5). Icbo = Ico = 1 × 10^-6 A.

(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.

(7).lifetime;

= > 2 = √0.325 + √ lifetime.

= Lifetime = 2.875.

The answer to the question mark the park in a

Answers

The correct answer is that being a

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