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Katyanochek1 [597]

3 years ago

7

Suppose you observe a binary system containing a main-sequence star and a brown dwarf. The orbital period of the system is 1 yea

r, and the average separation of the system is 1 AUAU . You then measure the Doppler shifts of the spectral lines from the main-sequence star and the brown dwarf, finding that the orbital speed of the brown dwarf in the system is 22 times greater than that of the main-sequence star.

1 answer:

Agata [3.3K]3 years ago

6 0

Complete Question:

How massive is the brown dwarf

Answer:

Mass of the brown dwarf, M₂ = 4.132 * 10²⁷ kg

Explanation:

Let M₁ = Mass of the main-sequence star

M₂ = Mass of the brown dwarf

v₁ = speed of the main-sequence star

v₂ = speed of the brown dwarf

The average separation of the system r₁ = r₂ = 1 AU

Orbital speed of the brown dwarf is 22 times that of the main - sequence star

That is, v₂ = 22 v₁

Since the sun is a main sequence star

Mass of the main - sequence star = 2 * 10³⁰ kg

Centripetal force, F = Mv²/r

M₁v₁²/r₁² = M₂v₂²/r²

M₁v₁² = M₂v₂²

M₂/M₁ = v₁²/v₂²

M₂/M₁ = (1/22)²

M₂/M₁ = 0.002066

M₂ = 0.002066M₁

M₂ = 0.002066 * 2 * 10³⁰

M₂ = 4.132 * 10²⁷ kg

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Answer:

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Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

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$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

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This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

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Let's calculate now the height for t = 0.555 s

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A bullet of mass 0.017 kg traveling horizontally at a high speed of 290 m/s embeds itself in a block of mass 5 kg that is sittin

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Answer:

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Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

where

p:Linear momentum

m: mass

v:velocity

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For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

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Pf : Final linear momentum quantity

Data

m₁ = 0.017 kg : mass of the bullet

m₂ = 5 kg : mass of the block

v₀₁ = 290 m/s : initial velocity of the bullet

v₀₂ = 0 : initial velocity of the block₂

(a) Speed of the block after the bullet embeds itself in the block

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = (m₁+ m₂)vf

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( 0.017)*( 290) + (5)*(0) = ( 0.017 + 5 )*vf

4.93+ 0 = ( 5.017 )*vf

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b) Total translational kinetic energy before (K₁) and after the collision(K₂).

K₁ = 1/2(m₁*v₀₁² + m₂*v₀₂²)

K₁ = 1/2(0.017*(290)² + 5*(0)²) = 714.85 J

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K₂= 1/2(0.017+ 5)*(0.98)² = 2.41 J

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