Answer:
t = 0.24 s
Explanation:
As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:
Translation: ΣF = ma
Rotation: ΣM = Iα ; where α = angular acceleration
Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:
ΣM = I(a/R)
Now we are going to resolve and combine these equations.
For translation: Fx - Ffr = ma
We know that Fx = mgSin27°, so we substitute:
(1) mgSin27° - Ffr = ma
For rotation: (Ffr)(R) = (2/3mR²)(a/R)
The radius cancel each other:
(2) Ffr = 2/3 ma
We substitute equation (2) in equation (1):
mgSin27° - 2/3 ma = ma
mgSin27° = ma + 2/3 ma
The mass gets cancelled:
gSin27° = 5/3 a
a = (3/5)(gSin27°)
a = (3/5)(9.8 m/s²(Sin27°))
a = 2.67 m/s²
If we assume that the acceleration is a constant we can use the next equation to find the velocity:
V = √2ad; where d = 0.327m
V = √2(2.67 m/s²)(0.327m)
V = 1.32 m/s
Because V = d/t
t = d/V
t = 0.327m/1.32 m/s
t = 0.24 s
Answer: 
Explanation:
Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period 
of a planet is proportional to the cube of the semi-major axis 
of its orbit”:
(1)
Now, if is measured in years (Earth years), and is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 
), equation (1) becomes:
(2)
So, knowing 
and isolating from (2) we have:
![a=\sqrt[3]{T^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7BT%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{(619.36 years)^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%28619.36%20years%29%5E%7B2%7D%7D)
(4)
Finally:
T
his is the distance between the dwarf planet and the Sun in astronomical units
Converting this to kilometers, we have:
Answer:
<em>The end of the ramp is 38.416 m high</em>
Explanation:
<u>Horizontal Motion
</u>
When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.
The maximum horizontal distance traveled by the object can be calculated as follows:
If the maximum horizontal distance is known, we can solve the above equation for h:
The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:
h= 38.416 m
The end of the ramp is 38.416 m high
