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Margaret [11]
3 years ago
12

Sir William Herschel counted the number of stars in different directions, and getting similar numbers in each direction along th

e disk, concluded the Sun was near the center of a disk like collection of stars. Herschel came to the wrong conclusion about the Sun's location because(A) He was predisposed to believe the Sun is the center of the Milky Way, and ignored the data which disagreed with that conclusion. (B) He did not know that interstellar dust made it hard from him to see a large part of the Milky Way's disk. (C) He only counted globular cluster, and not regular stars. (D) The Sun's position in the Milky Way at that particular time was very unusual and skewed in his results.
Physics
1 answer:
AVprozaik [17]3 years ago
8 0

Answer: the correct answer is (B) He did not know that interstellar dust made it hard from him to see a large part of the Milky Way's disk.

Explanation:

We live in a dusty Galaxy. Because interstellar dust absorbs the light from stars, Herschel could see only those stars within about 6000 light-years of the Sun.

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Difference between of echo and reverberation​
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A difference between of echo and reverberation​  is described below in details.

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A van der Graaff generator belt takes 250 electrons up to the ball at the top every second. Assuming that there is no loss, how
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2 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
3 years ago
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