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julsineya [31]
3 years ago
8

Light from the star Betelgeuse takes 640 years to reach Earth. How far away is Betelgeuse in units of light-years? Name any hist

orical event that was occurring on Earth at about the time the light left Betelgeuse. Is the distance to Betelgeuse unusual compared with other stars?
Physics
2 answers:
Eduardwww [97]3 years ago
6 0

Answer:

Betelgeuse is 640 light years away from earth.

Explanation:

A light-year is an astronomical unit to measure the distance the light travels in a calendar year.

If the light from a star takes 640 years to reach us, then that star its 640ly away from us.

Betelgeuse has been labeled as a Variant Star, which means that its brightness can fluctuate over the course of years, this has made difficult for astronomers to measure the exact distance of the star. Right now the star is estimated to be around 613 and 881ly away from earth, although, for the sake of your second question, we will take 640 years as our estimated value.

In 1380 (640 years ago) the Battle of Kulikovo took place. A battle of remarkable importance to Russian history, in which the Russian army, led by Prince Dmitry of Moscow, defeated the Mongol army, defining a turning point in the Mongol dominance, and setting the bases for what Russia is today.

galina1969 [7]3 years ago
6 0

Answer:

The distance is 641.8207 light years, and the star Betelgeuse is further away when compared to other stars

Historical event: Benedict XI succeeds Boniface VIII as pope (1302)

Explanation:

the solution is in the attached Word file

Download docx
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With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons
Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

m=\frac{Y-Y_{1}}{X-X_{1}}  (1)

Where:

m=5.7 is the slope of the line

Y_{1}=2390.7pounds is the airplane weight with  51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

X_{1}=51gallons is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

(X_{1},Y_{1})=(51,2390.7)

Rewritting (1):

Y=m(X-X_{1})+Y_{1}  (2)

As Y is a function of X:

Y=f_{(X)}=m(X-X_{1})+Y_{1}  (3)

Substituting the known values:

f_{(X)}=5.7(X-51)+2390.7  (4)

f_{(X)}=5.7X-290.7+2390.7  (5)

f_{(X)}=5.7X+2100  (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

f_{(81)}=5.7(81)+2100  (7)

f_{(81)}=2561.7  (8)   This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0
3 years ago
1. ability to cause change * Pick One PLEASE HELP ME
kkurt [141]

Answer:

electromagnetic wave  i think

Explanation:

8 0
4 years ago
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vlada-n [284]

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2 years ago
Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr
lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time  of computer A is CT_{A} =200 ps

Clock time  of computer B is CT_{B} =250 ps

Effective CPI of computer A is CPI_{A} =1.5

Effective CPI of computer B isCPI_{B} =1.7

CPU time of A is

CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec

CPU time of B is

CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec

Hence Computer A is Faster by \frac{425}{300} =1.41

Computer A is 1.41 times faster than the Computer B

4 0
3 years ago
A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
Keith_Richards [23]

Answer

given,

length of slender rod =80 cm = 0.8 m

mass of rod = 0.39 Kg

mass of small sphere = 0.0200 kg

mass of another sphere weld = 0.0500 Kg

calculating the moment of inertia of the system

I = \dfrac{ML^2}{12}+\dfrac{m_1L^2}{4}+\dfrac{mL^2}{4}

I = \dfrac{0.39\times 0.8^2}{12}+\dfrac{0.02\times 0.8^2}{4}+\dfrac{0.05\times 0.8^2}{4}

I =0.032\ kg.m^2

using conservation of energy

\dfrac{1}{2}I\omega^2 = (m_1-m_2)g\dfrac{L}{2}

\omega=\sqrt{\dfrac{(m_1-m_2)gL}{I}}

\omega=\sqrt{\dfrac{(0.05-0.02)\times 9.8 \times 0.8}{0.032}}

\omega=2.71 \rad/s

we know,

v = r ω

v = \dfrac{L}{2} \times 2.71

v = \dfrac{0.8}{2} \times 2.71

v = 1.084 m/s

3 0
4 years ago
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