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3 years ago

5

Dont decorate a tree in front of a cat lol we finished decorating the tree last night and the cat climbed up the tree and knocke

d it over XD

1 answer:

vladimir2022 [97]3 years ago

8 0

You should’ve ate the cat , that’s the only reasonable way to get revenge :)

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These 2 processes cause watersheds to change.

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What does K stand for in Hookes law?

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3 years ago

Fan object moves in uniform circular motion in a circle of radius R=200 meters, and the objectes 5.00 seconds to

Leviafan [203]

Answer:

The centripetal acceleration of the object is $31550.72\ m/s^2$ .

Explanation:

We have,

Radius of a circular path is 200 m

It takes 5 seconds to complete 10 revolutions. The angular velocity of the object is given by the rate of change of angular displacement per unit time :

$\omega=\dfrac{\Delta \theta}{\Delta t}\\\\\omega=\dfrac{2\pi \times 10}{5}\\\\\omega=12.56\ rad/s$

The centripetal acceleration of the object is given by :

$a=\omega^2 r\\\\a=(12.56)^2 \times 200\\\\a=31550.72\ m/s^2$

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3 years ago

A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s

kondaur [170]

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

$T^2=\frac{4\pi ^2}{GM} a^2$

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

$M = 3.5M_{sun} = 6.96*10^{30} kg$

$a=4.2AU = 6.28*10^{11} meters$

$G = 6.67*10^{-11}m^3kg^{-1}s^{-2}$

We put this into Kepler's law and get:

$T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.$

This when converted to years is 4.6 years.

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