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Sauron [17]
3 years ago
14

A planet or butting a distant star has been observed to have an orbital period of 0.76 earth years at a distance of 1.2 au. What

is the mass of the star the planet is orbiting
Physics
1 answer:
NNADVOKAT [17]3 years ago
6 0

Answer:

The mass of the star is, M = 5.9567x10³⁰ Kg

Explanation:

Given

The orbital period of the planet, T = 0.76 year

                                                        = 2.3967x10⁷ seconds

The distance between planet and sun, R+h = 1.2 a.u

                                                                        = 1.795 x 10¹¹ meters

The orbital period of the planet is given by the formula

                                 T={2\pi\sqrt{\frac{(R+h)^{2}}{GM}}}

Squaring and solving for M

                                   M=\frac{4\pi ^{2} (R+h)^{3}}{GT^{2} }

Substituting the given values in the above equation

                          M=\frac{4\pi ^{2}(1.795X10^{11} )^{3} }{6.673X10^{-11}X(2.3967X10^{7})^{2}}    

                                     M = 5.9567 x 10³⁰ Kg

Hence, the mass of the star the planet is orbiting, M = 5.9567 x 10³⁰ Kg

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In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten
sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

V = iR

V = (2 mA)(10 kohm)

V = 20 volts

so voltage across the capacitor + voltage across resistor = V

V_c + 20 = 50

V_c = 30 V

Now we know that

U = \frac{q^2}{2C}

here rate of change in energy of the capacitor is given as

\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}

\frac{dU}{dt} = (30)(2 mA)

\frac{dU}{dt} = 0.06 W

3 0
3 years ago
In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second
wolverine [178]

Answer:

 f_{beat} = 1.64\ Hz

Explanation:

given,

frequency of tuba.f = 64 Hz

Speed of train approaching, v = 8.50 m/s

beat frequency = ?

using Doppler's effect formula

 f' = f(\dfrac{v}{v-v_s})

v_s is the velocity of the source

v is the speed of sound, v = 340 m/s

now,

 f' = 64\times (\dfrac{340}{340 - 8.50})

       f' = 65.64 Hz

now, beat frequency is equal to

 f_{beat} = f' - f

 f_{beat} = 65.64 - 64

 f_{beat} = 1.64\ Hz

hence, beat frequency is equal to 1.64 Hz

3 0
3 years ago
Is distance traveled by a moving train qualitative discrete continuous or na?
butalik [34]
Continuous. Discrete values are values like 1, 2, 3, 4, etc. - they're values that are <em>distinct</em>, and typically there's some idea of a <em>next </em>and a <em>previous </em>value. When we're counting whole numbers, there's a definitive answer to which number comes after, and which number comes before. With continuous values, there's no real "next" or "last" value.

Motion is measured with <em>continuous </em>values; a train might move 300 yards in 1 minute, but we can look at smaller and smaller chunks of time to keep getting shorter and shorter distances. There is no <em />"next" distance the train moves after those 300 yards - it just doesn't make sense for there to be.

It's also measured <em>quantitatively</em>, not <em>qualitatively</em>. This just means that we can use numerical values to measure it, rather than other descriptors like color, smell, or taste.
8 0
3 years ago
hich one of the following statements could be an operational definition of electric current?View Available Hint(s)Which one of t
klemol [59]

Answer:

it is True as the operational definition of electric current.

Explanation:

The definition of electric current is

         I = dQ / dt

By convention the direction of the current is the direction in which a positive charge flows.

The initial expression is the derivative that is the change of the load in the unit of time and this occurs in a given cross-sectional cable.

The proposed definition is the same as this, so it is True as the operational definition of electric current.

8 0
3 years ago
Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s
Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375  m / s

3 0
4 years ago
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