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Vladimir79 [104]

3 years ago

15

Radar, the first link in the cosmic distance chain, is used to establish the baseline distance necessary for the second link, pa

rallax. What baseline distance must we know before we can measure parallax?

1 answer:

Elena L [17]3 years ago

7 0

Answer:

the Earth-Sun distance.

Explanation:

To measure parallax the base line distance that we need to know is the Earth-Sun distance. This distance is called 1 astronomical unit. This distance is 1.496×10^8 Km. So, this base line distance is necessary for the second link parallax.

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Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

alekssr [168]

<u>Answer:</u> The angle of diffraction is 0.498°

<u>Explanation:</u>

To calculate the angle of diffraction, we use the equation given by Bragg, which is:

$n\lambda =2d\sin \theta$

where,

n = order of diffraction = 3

$\lambda$ = wavelength of the light = $580nm=5.80\times 10^{-7}m$ (Conversion factor: $1m=10^{9}nm$ )

d = spacing between the crystal planes = 0.100 mm = $1.0\times 10^{-4}$ (Conversion factor: 1 m = 1000 mm)

$\theta$ = angle of diffraction = ?

Putting values in above equation:

$3\times 5.80\times 10^{-7}=2\times 1.00\times 10^{-4}\sin \theta\\\\\sin \theta = 0.0087\\\\\theta=\sin ^{-1} (0.0087)=0.498$

Hence, the angle of diffraction is 0.498°

6 0

3 years ago

A right triangle ABC has sides /AB/ =9cm, /BC/=12cm find /AC/ and angles ACB and ABC if angle BAC= 90°

dezoksy [38]

fdi9bn 08989089089008i uuuuuuri9ij

8 0

3 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

$F = 150000 lb$

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

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3 years ago

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the number of neutrons may b 21

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The magnetosphere protects Earth from _____. a.solar radiation b. the solar wind c. the cold of space d.asteroid hits e.gamma ra

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2 years ago

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