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.
Learn more about Kepler's third law:
brainly.com/question/11168300
#LearnwithBrainly
Answer:
angular speed of both the children will be same
Explanation:
Rate of revolution of the merry go round is given as
f = 4.04 rev/min
so here we have
here we know that angular frequency is given as
now this is the angular speed of the disc and this speed will remain same for all points lying on the disc
Angular speed do not depends on the distance from the center but it will be same for all positions of the disc
Answer:
Diffusing the gradient ensures that most of the molecules in high concentration zone will wind up in the previously low concentration by the spontaneous movement of small molecules.
Explanation:
A gradient of concentration is the difference between in concentration of one place / area substance to different area. Having a molecule flow down its concentration gradient means moving the molecules from hypotonic areas to the concentration hypertonic areas
Diffusing the gradient ensures that most of the molecules in high concentration zone will wind up in the previously low concentration by the spontaneous movement of small molecules.
The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '. D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .--
They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.
