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3 years ago

10

In an extrasolar planetary system containing a single planet, the parent star is measured to move about its center of mass every

24 years. Given this, what is the orbital period of the planet?

1 answer:

Mademuasel [1]3 years ago

4 0

Answer:

Orbital Time Period is 24 years

Explanation:

This can be explained by the definition of time period.

Time period can be defined as the time taken by an object to complete one cycle, here, time taken to complete one revolution.

Also, we know that an extra solar planet which is also called as an exo planet is that planet which is outside our solar system and orbits any star other than our sun. The system in consideration is extra solar system with a single planet.

Therefore, the time taken by the parent star to move about its mass center is the orbital time period that is 24 years.

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AysviL [449]

I think it’s 150 because if u multiply 3 x 50 it’s 150 but so it’s ore simple you do 3 x5 equals 15 then add the zero from the 50 so you’ll get 150 hope I helped

4 0

3 years ago

Which of these scenarios describes circular motion?

Anika [276]

A)

The moon orbiting the Earth

6 0

3 years ago

A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the high

m_a_m_a [10]

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to = 4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω² .................1

and kinetic energy is directly proportional to square of the amplitude.

so

$\frac{KE2}{KE1} = \frac{A2^2}{A1^2}$ .............2

$\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}$

$\frac{KE2}{KE1}$ = 0.52284

so factor that bug maximum KE change is 0.52284

5 0

3 years ago

What are three kinds of forces that affect an object’s motion? What effects do these forces have?

Harman [31]

Answer:

Different forces (including magnetism, gravity, and friction) can affect motion

4 0

2 years ago

A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio

Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

$\sum F_{x}=0$

we can see there are only two forces in x, which are the weight on x and the static friction, so:

$-W_{x}+f_{s}=0$

when solving for the static friction we get:

$f_{s}=W_{x}$

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

$W_{x}=mg sin \alpha$

we can substitute this on our sum of forces equation:

$f_{s}=mg sin \alpha$

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

$f_{s}=N\mu_{s}$

so we can substitute this on our equation:

$N\mu_{s}=mg sin \alpha$

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

$\sum F_{y}=0$

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

$N-W_{y}=0$

when solving for N we get:

$N=W_{y}$

When seeing the free-body diagram we can determine that:

$W_{y}=mg cos \alpha$

so we can substitute that in the sum of y-forces equation, so we get:

$N=mg cos \alpha$

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

$mg cos \alpha \mu_{s}=mg sin \alpha$

we can divide both sides of the equation into mg so we get:

$cos \alpha \mu_{s}=sin \alpha$

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for $\mu_{s}$ we get:

$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

and we can substitute the given value so we get:

$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

8 0

4 years ago