A uniform circular disk of moment of inertia 8.0 kg m2 is rotating at 4.0 rad/s. A small lump of mass 1.0 kg is dropped on the d
1 answer:
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The velocity when function p(t)=11 is 8 .
According to the question
The position of a car at time t represented by function :
Now,
When function p(t) = 11 , t will be
11 = t²+2t-4
0 = t² + 2t - 15
or
t² +2t-15 = 0
t² +(5-3)t-15 = 0
t² +5t-3t-15 = 0
t(t+5)-3(t+5) = 0
(t-3)(t+5) = 0
t = 3 , -5
as t cannot be -ve as given ( t≥0)
so,
t = 3
Now,
the velocity when p(t)=11
As we know velocity =
therefore to get the value of velocity from function p(t)
we have to differentiate the function with respect to time
v(t) = 2t + 2
where v(t) = velocity at that time
as t = 3 for p(t)=11
so ,
v(t) = 2t + 2
v(t) = 2*3 + 2
v(t) = 8
Hence, the velocity when function p(t)=11 is 8 .
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Answer:
Its period if its length is increased by a factor of four is 5 s.
Explanation:
The period of a simple pendulum is given by;
Given;
initial period, T₁ = 2.5
initial length, = L₁
new length, L₂ = 4L₁
the new period, T₂ = ?
Therefore, its period if its length is increased by a factor of four is 5 s.