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Alexus [3.1K]
4 years ago
13

An exoplanetary system has two known planets. Planet X orbits in 290 days and Planet Y orbits in 145 days. Which planet is close

st to its host star? If the star has the same mass as the Sun, what is the semi-major axis of the orbits for Planets X and Y?
Physics
1 answer:
satela [25.4K]4 years ago
6 0

Answer:

Planet Y

Rx = 1.587 Ry

Explanation:

Tx = 290 days

Ty = 145 days

Let the semi major axis of planet X is Rx and of plant Y is Ry.

According to the Kepler's third law of planetary motion

\frac{T_{x}^{2}}{T_{y}^{2}}=\frac{R_{x}^{3}}{R_{y}^{3}}

\frac{290^{2}}{145^{2}}=\frac{R_{x}^{3}}{R_{y}^{3}}

4=\frac{R_{x}^{3}}{R_{y}^{3}}

1.587=\frac{R_{x}}{R_{y}}

Rx = 1.587 Ry

So, planet Y is closest to star.

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driving down the highway , you find yourself behind a heavily loaded tomato truck. you follow close behind the truck keeping the
prohojiy [21]

Answer:

The tomato won't hit the car

Explanation:

According to the statement, the car moves at constant speed behind the truck fully loaded with tomatoes, and in the same direction. When a tomato falls from the top of the truck, it should not hit the car as the tomato falls due to the force of gravity, while horizontally has the same speed and in the same direction as the truck.  So we assume that the tomato will fall to the road without touching the car.

Have a nice day!

4 0
4 years ago
A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su
madreJ [45]

maximum static friction acting on the object will be

F_s = \mu_s mg

plug in all values

F_s = 0.40 \times 15 \times 9.8 = 58.8 N

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force

So here it is given that applied force is 20 N

so here object will not move due to this force and it will remain at rest always

due to this applied force

6 0
3 years ago
Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.
Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

f=\frac{R}{2}

For this mirror, R = 20 cm, so its focal length is

f=\frac{20}{2}=+10 cm (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm

b)

We can solve this part by using the magnification equation:

M=-\frac{y'}{y}=\frac{q}{p}

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0
3 years ago
Why is it better to use the metric system, rather than the English system, in scientific measurement?
fiasKO [112]

A. The English system uses one unit for each category of measurement.

4 0
3 years ago
Read 2 more answers
An ice skater has a moment of inertia of 5.0 kg · m2when her arms are outstretched, and at this time she is spinning at 3.0 rev/
KatRina [158]

Answer:

Explanation:

Given

Initial Moment of Inertia I_1=5 kg-m^2

initial Spin N_1=3 rev/s

\omega _1=2\pi N_1=2\pi 3=6\pi rad/s

Final Moment Moment of Inertia I_2=2 kg-m^2

Conserving Angular momentum

L_1=L_2

I_1\omega _1=I_2\omega _2

5\times 6\pi=2\times \omega _2

\omega _2=15\pi rad/s

N_2=\frac{\omega _2}{2\pi }

N_2=\frac{15\pi }{2\pi}=7.5 rev/s

8 0
4 years ago
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