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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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A transverse wave is traveling through a canal. If the distance between two successive crests is 3.17 m and four crests of the w
Thepotemich [5.8K]

Answer:

a. 0.18Hz

b. 0.56m/s

Explanation:

From the question we can deduct the following parameters

The wavelength, λ is define as the distance between two successful crest or trough and from the question we conclude that wavelength is 3.17m.

Also the period of the wave T can be computed as

T=22.6/4

T=5.65secs.

a. To compute the frequency, recall that frequency, F=1/period.

Hence,

F=1/5.65

F=0.18Hz

b. Next we compute the wave speed.

Wave speed=frequency *wavelength

Wave speed =0.18*3.17

Wave speed =0.56m/s

5 0
3 years ago
Two long wires hang vertically. wire 1 carries an upward current of 1.60
charle [14.2K]
3.00 is my correct answer this is what my teacher told me.thanks
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3 years ago
One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤
OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

P(t)=60 W, 0\leq t

P(t)=25 W, 5.0\leq t

Aplying the values to the equation we have:

Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}

Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}

Power_{avg} = 42.5W

5 0
2 years ago
Gardeners understand that plants have specific pH requirements. Most plants thrive in soil with a pH that is slightly acidic, ne
Delvig [45]

Answer:

D) Add a base like limestone or calcium carbonate.

Explanation:

-To reduce the soil's pH, you need to add an alkaline or base solution.

-Calcium carbonate has an alkanizing property and forms a carbonic acid and calcium hydroxide when dissolved in water.

-The solution will reduces the soil's pH.

8 0
3 years ago
An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu
Doss [256]

Answer:

400000\ \text{N/C}

Explanation:

q_1 = Charge at 3000 m = 40 C

q_2 = Charge at 1000 m = -40 C

r_1 = 3000 m

r_2 = 1000 m

k = Coulomb constant = 9\times10^9\ \text{Nm}^2/\text{C}^2

Electric field due to the charge at 3000 m

E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}

Electric field due to the charge at 1000 m

E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}

Electric field at the aircraft is E_1+E_2=40000+360000=400000\ \text{N/C}.

7 0
3 years ago
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