Answer:
charcoal is the correct answer
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The minimum number of tickets that could admit all of them is six (6).
This thing is impossible to explain in words, so I shall attempt it with a diagram:
Here are the six ladies:
( A ) ( B )
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( C ) ( D )
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( E ) ( F )
-- 'E' and 'F' are the daughters of 'C' and 'D' .
-- 'C' and 'D' are the daughters of 'A' and 'B' .
So look what we have now:
-- 'A' and 'B' are the mothers of 'C' and 'D' .
There's 2 of the mothers.
-- 'C' and 'D' are the mothers of 'E' and 'F' .
There's the OTHER 2 mothers.
-- 'A' and 'B' are the grandmothers of 'E' and 'F' .
There's the 2 grandmothers.
-- 'E' and 'F' are the daughters of 'C' and 'D' .
There's 2 of the daughters.
-- 'C' and 'D' are the daughters of 'A' and 'B' .
There's the OTHER 2 daughters.
You want to know what ? !
The group is even bigger than THAT.
There are also 2 GRAND-daughters in the family ... 'E' and 'F' .
So now you have a list of 12 people ! ... 4 mothers, 2 grandmothers,
4 daughters, and 2 grand-daughters ... and they all get in to the
Christmas Market with only six tickets. Legally !
Such a deal !
Don't forget : Christmas this year is also the first day of Chanukah !
All for the same price !
Newton's first law of motion says something like "An object remains
in constant, uniform motion until acted on by an external force".
Constant uniform motion means no change in speed or direction.
If an object changes from rest to motion, that's definitely a change
of speed. So it doesn't remain in the state of constant uniform
motion (none) that it had when it was at rest, and that tells us
that an external force must have acted on it.
The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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