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crimeas [40]
3 years ago
9

Astronomers know that the distance between the Earth and the Sun averages 1.50 x108 km. How can astronomers use the observed ste

llar parallax and a little knowledge of geometry to measure the distance to nearby stars?
Physics
1 answer:
rodikova [14]3 years ago
3 0

Answer:

The distance of stars and the earth can be averagely measured by using the knowledge of geometry to estimate the stellar parallax angle(p).

From the equation below, the stars distances can be calculated.

D = 1/p

Distance = 1/(parallax angle)

Stellar parallax can be used to determine the distance of stars from an observer, on the surface of the earth due to the motion of the observer. It is the relative or apparent angular displacement of the star, due to the displacement of the observer.

Explanation:

Parallax is the observed apparent change in the position of an object resulting from a change in the position of the observer. Specifically, in the case of astronomy it refers to the apparent displacement of a nearby star as seen from an observer on Earth.

The parallax of an object can be used to approximate the distance to an object using the formula:

D = 1/p

Where p is the parallax angle observed using geometry and D is the actual distance measured in parsecs. A parsec is defined as the distance at which an object has a parallax of 1 arcsecond. This distance is approximately 3.26 light years

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Classify each possible hypothesis about a medicinal aloe vera plant as falsifiable or non-falsifiable.
pychu [463]

Answer:

Aloe vera is a succulent plant species ( with fleshy water storing tissue) belonging to the genus Aloe.

It is used in many products ( including beverages, cosmetic products as gel, medicinal ointments). Its juice is quite bitter in taste. It is known to be the best skin moisturizer and can be used for healing wounds by enhancing the renewal of damaged cells. It is also used as an anticarcinogen, thus reduces the chances of cancer.

Thus, the given hypothesis can be classified as  falsifiable ( which is proven wrong) or non-falsifiable ( which is proven right) in the following way-

Falsifiable-

Aloe vera juice tastes  better than carrot juice.

Non Falsifiable-

Aloe vera gel is the best  natural skin moisturizer.

Aloe vera gel can heal  wounds by boosting  cell renewal.

Drinking aloe juice can  reduce the risk of lung  cancer.

7 0
3 years ago
what is the best term to describe information sent as patterns and codes in the controlled flow of electrons through a circuit
Vlad1618 [11]
But I was told it was A. Electronic signal
4 0
3 years ago
What component of a longitudinal sound wave is analogous to a trough of a transverse wave?
anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.  

6 0
3 years ago
Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),
yuradex [85]

Answer:

v=3.564\ m.s^{-1}

\Delta v =2.16\ m.s^{-1}

Explanation:

Given:

  • mass of John, m_J=30\ kg
  • mass of William, m_W=30\ kg
  • length of slide, l=3\ m

(A)

height between John and William, h=1.8\ m

<u>Using the equation of motion:</u>

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3

v_J=5.94\ m.s^{-1}

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

30\times 5.94+0=(30+20)v

v=3.564\ m.s^{-1} is the velocity with which they leave the slide.

(B)

  • frictional force due to mud, f=105\ N

<u>Now we find the force along the slide due to the body weight:</u>

F=m_J.g.sin\theta

F=30\times 9.8\times \frac{1.8}{3}

F=176.4\ N

<em><u>Hence the net force along the slide:</u></em>

F_R=71.4\ N

<em>Now the acceleration of John:</em>

a_j=\frac{F_R}{m_J}

a_j=\frac{71.4}{30}

a_j=2.38\ m.s^{-2}

<u>Now the new velocity:</u>

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2\times 2.38\times 3

v_J_n=3.78\ m.s^{-1}

Hence the new velocity is slower by

\Delta v =(v_J-v_J_n)

\Delta v =5.94-3.78= 2.16\ m.s^{-1}

8 0
3 years ago
A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?
evablogger [386]
20m away

the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds
4 0
3 years ago
Read 2 more answers
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