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3 years ago

Text that is located between <title >and </title > appears in the browser's_____

Engineering

1 answer:

Vinil7 [7]3 years ago

8 0

Answer:

Tab title

Explanation:

The tab title can be defined as the title that shows up in a browser tab known as page title. The title only has texts, other tags within the texts are not considered.

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Now, suppose that you have a balanced stereo signal in which the left and right channels have the same voltage amplitude, 500 mV

eimsori [14]

Answer:

R₁ = 32kΩ

Explanation:

See attached image

6 0

3 years ago

python Write a program that takes a date as input and outputs the date's season. The input is a string to represent the month an

kupik [55]

Answer:

month = input("Input the month (e.g. January, February etc.): ")

day = int(input("Input the day: "))

if month in ('January', 'February', 'March'):

season = 'winter'

elif month in ('April', 'May', 'June'):

season = 'spring'

elif month in ('July', 'August', 'September'):

season = 'summer'

else:

season = 'autumn'

if (month == 'March') and (day > 19):

season = 'spring'

elif (month == 'June') and (day > 20):

season = 'summer'

elif (month == 'September') and (day > 21):

season = 'autumn'

elif (month == 'December') and (day > 20):

season = 'winter'

print("Season is",season)

Explanation:

4 0

3 years ago

a metal rod 24mm diameter and 2m long is subjected to an axial pull of 40 kN. If the rod is 0.5mm, then find the stress-induced

ozzi

Answer:

i dont know but i will take the points tho hahah

Explanation:

8 0

2 years ago

(25%) A well-insulated compressor operating at steady state takes in air at 70 oF and 15 psi, with a volumetric flow rate of 500

lubasha [3.4K]

Answer:

You can look it up

Explanation: if you don't know what it is look it up on .

6 0

3 years ago

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature

Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

$P_{1}$ = 5 Mpa

$T_{1}$ = 1623°C

= 1896 K

$V_{1}$ = 0.05 $m^{3}$

Also given $\frac{V_{2}}{V_{1}} = 20$

Therefore, $V_{2}$ = 1 $m^{3}$

R = 0.27 kJ / kg-K

$C_{V}$ = 0.8 kJ / kg-K

Also given : $P_{1}V_{1}^{1.25}=C$

Therefore, $P_{1}V_{1}^{1.25}$ = $P_{2}V_{2}^{1.25}$

$5\times 0.05^{1.25}=P_{2}\times 1^{1.25}$

$P_{2}$ = 0.1182 MPa

a). Work transfer, δW = $\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

$\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}$

= 527200 J

= 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

= $\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

= $\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W$

= $\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200$

= 197.7 kJ

6 0

3 years ago

Text that is located between <title >and </title > appears in the browser's_____​

Text that is located between <title >and </title > appears in the browser's_____