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grin007 [14]
2 years ago
8

If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet a

nd the star?
Physics
1 answer:
Levart [38]2 years ago
5 0

Answer:

9.36*10^11 m

Explanation

Orbital velocity  v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

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A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass
Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

L=2\pi R

where R is the radius of the wheel.

The period of revolution is:

T=120 s

Therefore, the tangential speed of the book is:

v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R

8 0
3 years ago
What is Gravitational force?​
mote1985 [20]

Answer:

the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface

Explanation:

7 0
2 years ago
Read 2 more answers
The electric field of a sinusoidal electromagnetic wave obeys the equation E=(375V/m)cos[(1.99×107rad/m)x+(5.97×1015rad/s)t].a.
andrey2020 [161]

Answer:

A) max = Eo = 375 V/m, B)  B = 125 10⁻⁸ T , C) f = 9.50 101⁴ Hz, D) λ = 3.158 10⁻⁷ m, E)  T = 1.05 10⁻¹⁵ s , F) invisible for humans   G)  v = c = 3 10⁸ m/s

Explanation:

The expression given for the electric field is

       E = 375 cos (1.99 107x + 5.97 105t)

The general formula for the electric field of a transverse traveler is

     E = Eo cos (kx-wt)

Where k is the wave number and w the angular velocity

A) The amplitudes as electric is

     Emax = Eo = 375 V / m

B) the electric and magnetic field are related

     E / B = c

     B = E / c

     B = 375/3 108

     B = 125 10⁻⁸ T

C) angular velocity and frequency is related

    .w = 2π f

     f = w / 2π

     f = 5.97 10¹⁵ / 2π

     f = 9.50 101⁴ Hz

D) the speed of light has the formula

      c = λ f

      λ = c / f

      λ = 3 10⁸ / 9.50 10¹⁴

      λ = 3.158 10⁻⁷ m

E) The period

     T = 1 / f

     T = 1 / 9.5 10¹⁴

     T = 1.05 10⁻¹⁵ s

F) let's reduce the wavelength nm

   λ = 3.158 10⁻⁷ m (10⁹nm / 1m)

   λ = 3.158 10²nm = 315.3 nm

The visible radiation range is between 400nm and 700nm. This radiation is ultraviolet and is invisible humans

G) All electromagnetic radiation has a speed at the speed of light (c)

      v = c = 3 10⁸ m/s

5 0
3 years ago
The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e
Maksim231197 [3]

Answer:

\fbox{strength  \: of \:  the \:  other \:  charge  =  - 0.0196 Ke \: Coulomb}

Explanation:

Given:

Force between pair of charges= 900 newtons

The distance between the charges = 0.01 meters

Strength of Charge first q1 = 2e-10 Coulomb

To find:

Strength of Charge second q2 = ____ Coulomb?

Solution:

We know that,

Force between two charges separate by distance r is given by the equation,

|F| =  K_e \frac{q1 \cdot \: q2}{ {r}^{2} }  \\ 900 =K_e  \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times  {10}^{ - 4}  =  K_e {(2e - 10)\cdot \: q2} \\ q2 =   \frac{9 \times  {10}^{ - 2} }{(2e - 10) K_e}  \\  \\  \fbox{We \:  know \:  that \:  e = 2.71 } \\  substituting \: the \: value \: \\ q2 =  \frac{9 \times  {10}^{ - 2} }{(2 \times 2.71 - 10)K_e}  \\ q2 =  \frac{0.09}{ - 4.58 K_e}  \\ q2 =  \frac{-0.0196}{K_e}\: coulomb

\fbox{strength  \: of \:  the \:  other \:  charge  =  - 0.0196 Ke \: Coulomb}

<em><u>Thanks for joining brainly community!</u></em>

3 0
2 years ago
Make the curvature radius 0.6 m, the refractive index 1.5, and the diameter 0.6 m. place the lamp so that the source of light is
vodka [1.7K]
We are given
r = 0.6 m
n = 1.5
D = 0.6 m, R1 = 30 cm
R2 = 120 cm

We are asked to get the focal length and the distance of the focal plane from the lens

We use the formula
1 / f = ( n - 1) (1/R1 - 1/R2)
Substituting and solving for f
1/ f = (1.5 - 1) (1/30 - 1/120)
f = 80 cm

The focal length is 80 cm and the distance of the focal plane from the lesn is 80 cm - 30 cm = 50 cm.<span />
5 0
2 years ago
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