Answer:
B.the same as Lily's.
Explanation:
When a disc is rotating on its axis uniformly with some angular velocity , every particle on it rotates with the same angular velocity , because every particle on it completes a rotation simultaneously . So in one rotation they take same time . Hence particles whether on the outer edge or near the axis , all have same angular velocity .
They but differ due to their different linear velocity . Linear velocity of particle near outer edge have greater magnitude. .
Answer:
7.89 7.91
Explanation:
The ranges of measurement lie between 7.92-0.05 and 7.92+0.05
7.87g and 7.97g
Answer: hello some part of your question is missing attached below is the missing detail
answer :
<em>w</em>f = M( v cos∅ )D / I
Explanation:
The Angular speed <em>wf </em>of the system after collision in terms of the system parameters and I can be expressed as
considering angular momentum conservation
Li = Lf
M( v cos∅ ) D = ( ML^2 / 3 + mD^2 ) <em>w</em>f
where ; ( ML^2 / 3 + mD^2 ) = I ( Inertia )
In terms of system parameters and I
<em>w</em>f = M( v cos∅ )D / I
Answer:
No
Explanation:
The vertical component of Jack's initial velocity is:
5.0
⋅
sin
30
∘
=
5.0
⋅
1
2
=
2.5
m/s
With gravitational acceleration
9.8
m/s
2
, he will reach the highest point of his trajectory after:
2.5
9.8
≈
0.255
s
The average vertical component of his velocity in that
0.255
s
will be:
1
2
⋅
2.5
=
1.25
m/s
So the highest point of his trajectory will be:
0.255
⋅
1.25
≈
0.32
m
So he will pass approximately
7
cm
above the top of the candle.
The horizontal component of his velocity will be a constant:
5.0
⋅
cos
30
∘
=
5.0
⋅
√
3
2
≈
4.33
m/s
So Jack's trajectory will be substantially longer than it is high and he will spend little time anywhere near above the candle.
