Answer:
The answer is below
Explanation:
a) The initial velocity (u) = 24 m/s
We can solve this problem using the formula:
v² = u² - 2gh
where v = final velocity, g= acceleration due to gravity = 9.8 m/s², h = height.
At maximum height, the final velocity = 0 m/s
v² = u² - 2gh
0² = 24² - 2(9.8)h
2(9.8)h = 24²
2(9.8)h = 576
19.6h = 576
h = 29.4 m
b) The time taken to reach the maximum height is given as:
v = u - gt
0 = 24 - 9.8t
9.8t = 24
t = 2.45 s
The total time needed for the apple to return to its original position = 2t = 2 * 2.45 = 4.9 s
Is the variable you change, independent, I, something I change.
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
'What is the magnitude of the force needed to stop the horses and bring the box into equilibrium?' ≈42N; according to the vectors rules.
'Where would you locate the rope to apply the force?' - in point D.
PS. zoom out the attached picture.
