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Eva8 [605]
4 years ago
13

Two stars in a faraway part of the Milky Way are orbiting each other as a binary star system. By

Physics
1 answer:
natita [175]4 years ago
3 0

Answer: M_{total}= 1.85

Explanation: Estimate the total mass of a binary system is done by a reformulation of <em>Kepler's Third Law</em>, which states that the square of the period of a planet's orbit is proportional to the cube of its semimajor axis, i.e.:

a^{3}=(M_{1}+M_{2})P^{2}

where

a is semimajor axis in astronomical units (AU);

P is period measured in years;

M_{1}+M_{2} is total mass of the two-stars system;

For the two stars faraway in the Milky Way:

1 year is equivalent of 365 days, so period in years:

P=\frac{594}{365}

P = 1.63 years

Calculating total mass:

a^{3}=(M_{total})P^{2}

M_{total}=\frac{a^{3}}{P^{2}}

M_{total}=\frac{1.7^{3}}{1.63^{2}}

M_{total}= 1.85

<u>The total mass of the two-object system is 1.85 mass units.</u>

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2 kg and 3 kg objects slide together, and then they break apart. If the final velocity of 2 kg is 10 m/s,
lara [203]

Answer:

P = (2 + 3) * V       where V is their initial speed  (total momentum)

P = 2 * 10 + 3 * Vx     where Vx here would be V3

If the initial momentum is not known how can one determine the final velocity of the 3 kg obj.

Also work depends on the sum of the velocities

W (initial) = 1/2 (2 + 3) V^2     the initial kinetic energy

W (final) = 1/2 * 2 * V2^2 + 1/2 * 3 * V3^2

It appears that more information is required for this problem

8 0
3 years ago
Dimas reads that it takes 270 N to lift a firefighter's equipment. If a firefighter does 5400 J of work carrying the equipment u
Alex787 [66]

Answer:

h = 20 m

Explanation:

Given that,

Force to lift a firefighter's equipment, F = 270 N

Work done, W = 5400 J

We need to find the height climbed by the firefighter. We know that,

Work done, W = Fd

So,

d=\dfrac{W}{F}\\\\d=\dfrac{5400}{270}\\\\d=20\ m

So, the firefighter climb from a height of 20 m.

7 0
3 years ago
Read 2 more answers
Finding the work done in pulling a stranded climber to safety. In an unfortunate accident, a rock climber finds herself stuck 17
Gnesinka [82]

Answer:

-9446.22 J

Explanation:

Parameters given:

Mass of climber and harness = 56kg

Mass of cable = 0.7kg

Distance between climber and top of rock face = 17m

The work done in pulling the climber is given as:

W = F * d

F is the force applied on the rope. It is opposite the force of gravity pulling the climber, hence, it is given as:

F = -Fg

Fg = force of gravity

Fg = m * g

g = acceleration due to gravity.

The mass of the climber, harness and cable = 56 + 0.7 = 56.7kg

=> Fg = 56.7 * 9.8

Fg = 555.66 N

Therefore, the work done will be:

W = - 555.66 * 17

W = -9446.22 J

The negative value of work means that the work done is opposite the value of the force acting on the climber.

4 0
4 years ago
Which of the following is a sign of Anorexia Nervosa?
Alla [95]
B is the correct answer because anorexia nervosa is an eating disorder in which the victim refrains from eating due to body image insecurities.
4 0
3 years ago
Consider the two-body situation at the right. A 3.50x103-kg crate (m1) rests on an inclined plane and is connected by a cable to
4vir4ik [10]

Answer:a=2.42 m/s^2

Explanation:

Given

mass m_1=3.50\times 10^{3} kg

m_2=1.00\times 10^{3} kg

\theta(inclination)=30^{\circ}

\mu =0.21

Let T be the tension in the rope

From Diagram

m_1g\sin \theta -T-f_r=m_1a-----------------1

where f_r=friction\ force

f_r=\mu m_g\cos \theta

For block m_2

T=m_2a-----------2

From 1 & 2

m_1g\sin \theta -m_2a-\mu m_1g\cos \theta =m_1a

m_1g(\sin \theta -\mu \cos \theta )=(m_1+m_2)a

\frac{9.8\times 3.5\times 10^3}{4.5\times 10^3}(0.5-\0.21\cos 30)=4.5a

a=2.42 m/s^2

4 0
4 years ago
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