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pochemuha
3 years ago
8

Might give brainliest

Engineering
1 answer:
zepelin [54]3 years ago
7 0

Answer:

Blissymbols

Explanation:

Blissymbols is a constructed language with hundreds of basic symbols that represent words.

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Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the
Sunny_sXe [5.5K]

Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C

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3 years ago
While discussing VIN numbers, Technician A says that the first digit of the VIN identifies the country where the vehicle was man
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Usually the first digit of the vin id’s the country it was built. So technician A would be correct. That’s usually how it is. Hope this helps. Please let me know if this is incorrect
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3 years ago
When were dresses made
klio [65]

Answer:

The world's oldest dress called the Tarkhan Dress is at 5,100 to 5,500 years of age.

Does that help? Or do you need something else? I can change my answer if this is not what you need! :D

Explanation:

6 0
3 years ago
A wire is 0.92 m long and 1.2 mm2 in cross-sectional area. It carries a current of 5.0 A when a 2.2 V potential difference is ap
Flura [38]

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3 0
3 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
DochEvi [55]

Answer:

m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]

Q_{water} = 3599.435\,kJ

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Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

m_{LP}=0.45\,kg

6 0
3 years ago
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