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Grace [21]
3 years ago
7

Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the pla

nets are determined to be 45.6 km/s and 55.1 km/s. The slower planet's orbital period is 8.68 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?
Physics
1 answer:
Taya2010 [7]3 years ago
5 0

Answer:

6.19744\times 10^{31}\ kg

4.91 years

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of star

v = Oribital velocity of star

T = Oribital time period

R = Radius of orbit

M=\frac{4\pi^2R^3}{GT}

v=\frac{2\pi r}{T}\\\Rightarrow R=\frac{vT}{2\pi}

\\\Rightarrow M=\frac{4\pi^2\left(\frac{vT}{2\pi}\right)^3}{GT}\\\Rightarrow M=\frac{Tv^3}{2\pi G}

\\\Rightarrow M=\frac{8.68\times 365.25\times 24\times 3600\times 45600^3}{2\pi 6.67\times 10^{-11}}\\\Rightarrow M=6.19744\times 10^{31}\ kg

Mass of the star is 6.19744\times 10^{31}\ kg

M=\frac{T_1v_1^3}{2\pi G}=\frac{T_2v_2^3}{2\pi G}\\\Rightarrow M=T_2v_2^3

\\\Rightarrow T_2=T_1\left(\frac{v_1}{v_2}\right)^3\\\Rightarrow T_2=8.68\left(\frac{45.6}{55.1}\right)^3\\\Rightarrow T_2=4.91\ years

Orbital period of the faster planet is 4.91 years

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An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0km/hour is blowing southward.
qaws [65]

Answer:

a)  correct answer is C , b) 14º  from the west to the north, c)   v_{1g} = 300.79 km / h

Explanation:

This is a relative speed exercise using the addition of speeds.

1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C

2 and 3) In this case we must compose the speed using the Pythagorean Theorem.

     v_{1a}² = v_{1g}² + v_{ag}²

where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth

in this case let's clear the speed of the airplane with respect to the Earth

  v_{1g} = √(v_{1a}² - v_{ag}²)

 v_{1g} = √ (310² - 75²)

 v_{1g} = 300.79 km / h

we find the direction of the airplane using trigonometry

   sin θ = v_{ag} / v_{1a}

   θ = sin⁻¹ (v_{ag} /v_{1a})

   θ = sin⁻¹ (75/310)

   θ= 14º

the pilot must direct the aircraft at an angle of 14º from the west to the north

7 0
3 years ago
You pull with a force of 77 N on a piece of luggage of mass 23 kg, but it does
Vinvika [58]

Answer:

The force of static friction acting on the luggage is, Fₓ = 180.32 N

Explanation:

Given data,

The mass of the luggage, m = 23 kg

You pulled the luggage with a force of, F = 77 N

The coefficient of static friction of luggage and floor, μₓ = 0.8

The formula for static frictional force is,

                                      Fₓ = μₓ · η

Where,

                                  η - normal force acting on the luggage 'mg'

Substituting the values in the above equation,

                                   Fₓ = 0.8 x 23 x 9.8

                                        = 180.32 N

Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N

5 0
3 years ago
Read 2 more answers
A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w
myrzilka [38]
The centripetal acceleration is given by
a_c =  \frac{v^2}{r}
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, a_c = 15.625 m/s^2 and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s
8 0
3 years ago
Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 75.9 N to the north. Jenny’s foot exerts
Lady_Fox [76]

Answer:

Fr^2 = 75.9N+105.8N=181.7

<u><em>Fr = </em></u><u><em>181.7N.</em></u>

6 0
3 years ago
Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
katrin [286]

Answer:

a= 4.4×10 m/s^2

Explanation:

pressure P  = E/c

Where, E = 100 W/m^2 intensity of light

c= speed of light  = 3×10^8 m/s

P = 1000/ 3×10^8

P = 3.33×10^(-6) Pa

Force F = P×A

  • P is the pressure and c= speed of light

F = 3.33×10^{-6}×6.65×10(-29)

= 2.22×10^{-6}

acceleration a  = F/m = 2.22×10^{-6}/ 5.10×10^{-27}

a= 4.4×10 m/s^2

4 0
3 years ago
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