1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
AnnyKZ [126]
3 years ago
12

Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This

statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is |v with arrow| = 160 km/s and the orbital period of each is 13.7 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 1030 kg.)
Physics
1 answer:
zepelin [54]3 years ago
6 0

Answer:

 M = 4.61 10³¹ kg

Explanation:

For this exercise we will use Newton's Second Law where the force is the gravitational outside

       F = ma

       G M₁ M₂ / r² = m a

The acceleration is centripetal

       a = v² / r’

Let's replace

The mass of the two stars in it, the distance between them is r and the distance around the center of mass is

       r’= r / 2

       G M² / r² = M v² / (r / 2)

       G M / r = 2 v²

The linear velocity module is constant, so we can use the kinematic relationship

       v = d / t

The distance of a circle of radius r ’is

      d = 2π r ’= 2π (r / 2)

      d = π r

We replace

      v = π r / T

Let's write the two equations

      v² = ½ G M / r

      v = π r / T

      r = v T /π

      v² = ½ G M π/ vT

      M = 2 v³ T / π G

Let's reduce the magnitudes to the SI system

     v = 160 Km / s = 1.60 10⁵ m / s

     T = 13.7 days (24 h / 1 day) (3600s / 1 h) = 1.18 10⁶ s

Let's calculate

     M = 2 (1.60 10⁵)³ 1.18 10⁶ / (π 6.67 10⁻¹¹)

     M = 4.61 10³¹ kg

You might be interested in
A person starts at a position of 16 meters and finishes at a position of
DochEvi [55]

Answer:

38 is a good girl and a great place to work for u and I miss you y and I miss you much love your love and love to

7 0
3 years ago
If gravity on the earth increased, what affect would it have on the moon
Rufina [12.5K]

Answer:

If gravity on Earth is increased, this gravitational tugging would have influenced the moon's rotation rate. If it was spinning more than once per orbit, Earth would pull at a slight angle against the moon's direction of rotation, slowing its spin. If the moon was spinning less than once per orbit, Earth would have pulled the other way, speeding its rotation.

6 0
3 years ago
What happened to the kelp forest when the otter was hunted to near extinction?
never [62]

Answer: Sea otter is the pioneer species in the kelp forest as it regulates and controls the population of other species in the kelp forest.

Explanation:

If sea otters are hunted and their population is brought to extinction then this will cause major harm the ecosystem of the kelp forest and it will disturb the ecological balance in the kelp forest. The herbivorous animals consumed by the sea otters will increase in population and they will consume a lot of vegetation in the forest. The kelp forest which forms the coastline will not remain effective in providing protection against the storms to the neighboring areas.

7 0
3 years ago
A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk
Murrr4er [49]

Answer:

\omega_2=5.1rad/s

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

L_1=L_2

Or:

I_1\omega_1=I_2\omega_2

And we want to calculate:

\omega_2=\frac{I_1\omega_1}{I_2}

The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

which for our values is:

\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

6 0
3 years ago
Thirty beats are heard in one minute when two notes are played together. If the higher note has a frequency of 740 Hz, what is t
Tatiana [17]
30 beats/1min = 30 beats/60 sec = .5 beat/sec = 1/2 Hz
740 Hz - 1/2 Hz = 739.5 Hz

Answer - c. 739.5 Hz
6 0
3 years ago
Other questions:
  • A person sees a purple flower at a florist shop. What's true about the purple flower?
    11·2 answers
  • A temperature measuring device should be placed in the __________ area of the refrigerated holding unit.
    11·1 answer
  • Why evaporation are called surface phenomenon?
    11·2 answers
  • A person jumps forward off the front of a stationary boat floating on a lake. Identify the reaction force in this case, between
    6·1 answer
  • A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a 100 V battery. A Teflon slab (dielectric c
    13·1 answer
  • The electrical and magnetic components of an EM wave are _________________.
    11·1 answer
  • What are two examples of population distribution?
    14·1 answer
  • (b) A cylinder of cross-sectional area 0.65m2 and
    5·1 answer
  • A 95-kg astronaut is stranded from his space shuttle. He throws a 2-kg hammer away from the shuttle with a velocity of 19 m/s .
    11·1 answer
  • All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!