?Why the efficiency of Class A amplifier is very poor

what are the characteristics of an ideal fluid the general relation between shear stress and velocity gradient

Dafna11 [192]

Answer:

ideal fluid follow Newtonian law

that is, shear stress is directly proportional to rate change of shear strain.

watch handwritten explanation

6 0

3 years ago

Please help asap! I will give brainlist!

dexar [7]

Answer:

i did this this is b

Explanation:

7 0

3 years ago

Read 2 more answers

Software, such as a word processor, search engine, or mobile interface, typically includes plug-in support specific to a languag

bearhunter [10]

Answer:

Check the explanation

Explanation:

class LanguageHelper:

language=set()

#Constructor

def __init__(self, words):

for w in words:

self.language.add(w)

def __contains__(self,query):

return query in self.language

def getSuggestionns(self,query):

matches = []

for string in self.language:

if string.lower().startswith(query) or query.lower().startswith(string) or query.lower() in string.lower():

matches.append(string)

return matches

lh = LanguageHelper(["how","Hi","What","Hisa"])

print('how' in lh)

print(lh.getSuggestionns('hi'))

===========================================

OUTPUT:-

==================

True

['Hisa', 'Hi']

====

7 0

3 years ago

Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is

VARVARA [1.3K]

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

$\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}$

Solve the following for the real stress and pressure for the stable. $\sigma_{r1}=K(\varepsilon_{r1})^{n}$

$K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}$

Solve the following for the true state stress and stress2.

$\sigma_{r2}=K(\varepsilon_{r2})^n$

$=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa$