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:
Q1: B.2 Q2: B.Waxing crescent Q3: A.Waxing Gibbous
Explanation:
Answer:
A. Repeat the experiment to be sure the results are valid.
Answer:
50 m
Explanation:
F = ma
10 N = (10 kg) a
a = 1 m/s²
Given:
v₀ = 0 m/s
a = 1 m/s²
t = 10 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²
Δx = 50 m
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