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Svet_ta [14]
3 years ago
10

What is the resulting value for the Hubble constant?Write your answer in scientific notation

Physics
1 answer:
KonstantinChe [14]3 years ago
5 0

\bf{Given : Hubble\;Constant(H_0) = 73(\frac{km}{s \times Mpc})}

\bf{Given : Conversion\;Factor = (\frac{1Mpc}{3.086 \times 10^1^9\;km})}

\bf{\implies Resulting\;Hubble\;Constant = 73(\frac{km}{s \times Mpc})(\frac{1Mpc}{3.086 \times 10^1^9\;km})}

\bf{\implies Resulting\;Hubble\;Constant = 73(\frac{1}{s \times 3.086 \times 10^1^9})}

\bf{\implies Resulting\;Hubble\;Constant = (\frac{23.65}{s \times 10^1^9})}

\bf{\implies Resulting\;Hubble\;Constant = 2.365 \times 10^-^1^8(1/s)

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3 years ago
Kindly Don't Spàm!<br>Thank uh !:)<br><img src="https://tex.z-dn.net/?f=%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20" id="Te
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\:  \underline{ \boxed{  { \color{gray}\frak{\huge \: Answer : }}}}

<h3> Collector Current at Saturation :</h3>

  • \sf \large {I_{C(SAT)} = \frac{ V_{CC} }{R_C} }

\:  \:

  • \sf \large {I_{C(SAT)} = \frac{12}{2.2 \times 10 ^{3} } }

\:  \:

  • \bold{ \bf \large {I_{C(SAT)} =5.45mA }}

\:  \:

________________________________

<h3> Value Of Cut - off Voltage : </h3>

  • \sf \large \: V_{CS(cut-off)} = V_{CC}

\:  \:

Therefore ,

\:  \:

  • \bf \large \: V_{CS(cut-off)} = \: 12v

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________________________________

{ \bf\large \: Base \:  Current , I_B =  \frac{V_{CC}}{R_C} }

  • \sf \large \: I_B ={ \frac{12}{2.2 \times 10 ^{3} } }

\:  \:

  • \bf \large \: I_B = 50μA

\:  \:

_______________________________

<h3>Collector Current ,</h3>

  • \sf \large \: I_C = β * I_B

\:  \:

  • \sf \large \: I_C = 50 * 50 * 10^{-6}

\:  \:

  • \bf \large \: I_C = 2.5mA

\:  \:

________________________________

<h3> Collector to emitter Voltage : </h3>

  • \sf \large \: V_{CE} = V_{CC} - ( I_C * R_C )

\:  \:  \:

  • \sf \large \: V_{CE} = 12 - ( 2.5 * 10-³ * 2.2 * 10³ )

\:  \:

  • \bf \large \: V_{CE} = 6.5v

________________________________

<h3>Q - point are :</h3>

  • \sf \large \: I_{CEQ} = 2.5mA

\:  \:

  • \sf \large \: V_{CEQ} = 6.5v

________________________________

<h3>Q - point located on the DC load line as shown in fig ~</h3><h3 /><h3 /><h3 /><h3 />

________________________________

Hope Helps!:)

\:  \:  \:  \:

5 0
2 years ago
A rolling ball has an initial velocity of 5 meters per second. 30 seconds later, its velocity it 2 meters per second. What is th
sergeinik [125]
The answer is -0.2 hope this helps. :)
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3 years ago
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The angle of elevation to the top of a very tall Building is found to be 6° from the ground at a distance of 1 mi from the base
DanielleElmas [232]

Answer:

555

Explanation:

The scenario is shown in the image attached below. A Right Angled Triangle is formed. We have an angle of 6 degrees, the side adjacent to angle which measures 1 mile and we need to find the side opposite to the angle.

Since 1 mile = 5280 feet, the side adjacent to the angle has a measure of 5280 feet.

Tangent ratio relates the opposite and adjacent side by following formula:

tan(\theta)=\frac{Opposite}{Adjacent}

Using the given values, we get:

tan(6)=\frac{x}{5280}\\\\ x=tan(6) \times 5280\\\\ x = 555

Thus, rounded to nearest foot, the height of the building is 555 feet

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