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algol [13]
3 years ago
11

Imagine that a new planet is discovered with two moons of equal mass: Moon A and Moon B. The mass of the new planet is greater t

han the combined mass of its moons. Moon A is farther away from the new planet than Moon B. What is the planet's gravitational pull on Moon A compared to the planet's gravitational pull on Moon B?
The planet's gravity repels Moon A with a greater force than it repels Moon B, which is why Moon A is farther away.

The gravitational pull on Moon B is greater than on Moon A because Moon B is closer to the new planet than Moon A.

The gravitational pull on Moon B is greater than on Moon A because Moon B is farther away from the new planet than Moon A.

The gravitational pull on Moon A is the same as the gravitational pull on Moon B because distance does not affect the planet's gravity.
Chemistry
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer: The gravitational pull on Moon B is greater than on Moon A because Moon B is closer to the new planet than Moon A.

Explanation:

The gravitational force exerted by the two objects on each other is inversely proportional to the square of the distance between the objects.

F=G\frac{mM}{r^2}

F = gravitational force or pull

G = gravitational constant on that planet

M =  mass of the object-1

m = mass of object-2(Mass of Moon-A or Moon-B)

r = distance between two objects

Force\propto \frac{1}{r^2}

With decease in distance 'r' the force between the object increases or vice versa.So, from this we can say that the gravitational pull on Moon-B is more than the gravitational pull on Moon-A because the Moon B is closer than the Moon-A from the new planet.

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Answer: Finding the [H3O+] and pH of Strong and Weak Acid Solutions The larger the Ka, the stronger the acid and the higher the H+ concentration at equilibrium. hydronium ion, H3O+, 1.0, 0.00, H2O, 1.0×10−14, 14.00.

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In the science fiction movie, The Abyss, a diver is able to breathe while his head is immersed in a specially prepared liquid pu
Klio2033 [76]

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5 0
4 years ago
If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera
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Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

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a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0
3 years ago
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