The period of the orbit would increase as well
Explanation:
We can answer this question by applying Kepler's third law, which states that:
"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"
Mathematically,
Where
T is the orbital period
a is the semi-major axis of the orbit
In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, 
: but as we saw from the previous equation, the orbital period of the Earth is proportional to , therefore as increases, T increases as well.
Therefore, the period of the orbit would increase.
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Answer:
this situation would not be physically possible
It's b, because the more force an object it is given the harder it will be for it to slow down.
Answer:
As the mass of an object increases, its gravitational force increases.
As an object's distance to other objects increases, its gravitational force on those objects increases.
Explanation:
The gravitational force of one object on another is calculated with the equation
F = (G*m1*m2)/(r²),
where G is the gravitational constant,
M1 and M2 are the masses of the two objects, and
r is the distance between them
We can see that the force has a direct relationship with both of the mass values, and an inverse square relationship with the distance between them.
Hope this helped!
Answer:
The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is 
Explanation:
Here we have
Initial temperature of air T₁ = 20 °C = 293.15 K
Final temperature of air T₁ = 120 °C = 393.15 K
Initial pressure P₁ = 1 atm = 101325 Pa
Final pressure P₂ = Required
Area = A
Therefore we have for the pressure cooker, the volume is constant that is does not change
By Chales law
P₁/T₁ = P₂/T₂
P₂ = T₂×P₁/T₁ = 393.15 K× (101325 Pa/293.15 K) = 135,889.22 Pa
∴ P₂ = 135.88922 KPa = 135.9 kPa
Where Force = 
we have
Force = 
.
