The tension in the string and the acceleration must be equal for both masses. (See the free body diagrams)
Answer:
In this case reflection occurs.
Explanation:
This happens when they bounce on objects that they can not pass through.
Hope it helps
okay.
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:
going to work my way down
Troposphere
contains weather
contains life forms
stratosphere
contains ozone layer
mesosphere
where asteroids burn up
very low temps
thermosphere
widely varying temps
exosphere
almost no molecules
Explanation:
Answer:
1.63 N
Explanation:
F = GMm/r^2
= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2
= 1.63 N ( 3 sig. fig.)