How many atoms are in a molecule of C6H12O6?
1 answer:
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Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω = (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω
+ α
t
α= (ω
- ω
)/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ = ω
t + 1/2α
t²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω
+ α
t
t= (ω - ω
)/α
=> (0- 3.333)/-1.25
t= 2.67s
(a) 4.06 cm
In a simple harmonic motion, the displacement is written as
(1)
where
A is the amplitude
is the angular frequency
is the phase
t is the time
The displacement of the piston in the problem is given by
(2)
By putting t=0 in the formula, we find the position of the piston at t=0:
(b) -14.69 cm/s
In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:
(3)
Differentiating eq.(2), we find
And substituting t=0, we find the velocity at time t=0:
(c) -101.13 cm/s^2
In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:
Differentiating eq.(3), we find
And substituting t=0, we find the acceleration at time t=0:
(d) 5.00 cm, 1.26 s
By comparing eq.(1) and (2), we notice immediately that the amplitude is
A = 5.00 cm
For the period, we have to start from the relationship between angular frequency and period T:
Using and solving for T, we find