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olchik [2.2K]
1 year ago
8

In recent years, astronomers have found planets orbiting nearby stars that are quite different from planets in our solar system.

Kepler-12b, has a diameter that is 1.7 times that of Jupiter (RJupiter = 6.99 x 107 m), but a mass that is only 0.43 that of Jupiter (MJupiter = 1.90 x 1027 kg). Part A What is the value of g on this large, but low-density, world? Express your answer with the appropriate units. НА ? g= Value Units Submit Request Answer Provide Feedback
Physics
1 answer:
Marizza181 [45]1 year ago
6 0

3.86 m/s^2  is the value of gravity  on this large, but low-density, world.

given :

Kepler-12b

diameter= 1.7 times of Jupiter (R_Jupiter = 6.99 × 10^7 m),

mass = 0.43  Jupiter (M_Jupiter = 1.90 × 10^27 kg ).

g = GM/r^2

g = 6.67×10^-11 × 0.43×1.9×10^27/( 1.7×6.99×10^7)^2

g = 3.859 ~ 3.86 m/s^2

Gravity, also referred to as gravitation, is the unchanging force of attraction that binds all matter together in mechanics. It is by far the weakest known force in nature, so it has no effect on determining the internal properties of common matter.

On Earth, everything has weight, or a gravitational pull that is imposed by the planet's mass and proportional to the object's mass. A measure of the force of gravity is the acceleration that freely falling objects experience. At the surface of the Earth, gravity accelerates at a rate of about 9.8 meters per second. As a result, an object's speed increases during free fall by about 9.8 meters per second. At the Moon's surface, a freely falling body accelerates to about 1.6 m/s2.

To know more about  gravity visit : brainly.com/question/14428640

#SPJ4

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if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? *
SSSSS [86.1K]

Answer:

0.013N

Explanation:

F=\frac{mv^{2} }{r}

m=0.04, v=4.4m/s, r=60

F=0.013 N

4 0
3 years ago
If a block of wood dropped from a tall building has attained a velocity of 78.4 m/s how long has it been falling
ryzh [129]

Gravity adds 9.8 m/s to the speed of a falling object every second.

An object dropped from 'rest' (v = 0) reaches the speed of 78.4 m/s after  falling for (78.4 / 9.8) = <em>8.0 seconds</em> .

<u>Note:</u>

In order to test this, you'd have to drop the object from a really high cell- tower, building, or helicopter.  After falling for 8 seconds and reaching a speed of 78.4 m/s, it has fallen 313.6 meters (1,029 feet) straight down.

The flat roof of the Aon Center . . . the 3rd highest building in Chicago, where I used to work when it was the Amoco Corporation Building . . . is 1,076 feet above the street.

5 0
3 years ago
calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2​
igor_vitrenko [27]

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0
3 years ago
Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 time
skad [1K]

Answer:

g' = 13.5 m/s²

Explanation:

The acceleration due to gravity on surface of earth is given by the formula:

g = GMe/Re²   --------------- euation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

Me = Mass of Earth

Re = Radius of Earth

Now, the the acceleration due to gravity on the surface of Kepler-62e is:

g' = GM'/R'²   --------------- euation 1

where,

g' = acceleration due to gravity on surface of Kepler-62e

G = Universal Gravitational Constant

M' = Mass of Kepler-62e = 3.57 Me

R' = Radius of Kepler-62e = 1.61 Re

Therefore,

g' = G(3.57 Me)/(1.61 Re)²

g' = 1.38 GMe/Re²

using equation 1:

g' = 1.38 g

where,

g = 9.8 m/s²

Therefore,

g' = 1.38(9.8 m/s²)

<u>g' = 13.5 m/s²</u>

6 0
3 years ago
3
Elanso [62]

Answer:

ive got no idea just ask your teacher

7 0
3 years ago
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