Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
<u>α = -0.16 rad/s²</u>
<u>Negative sign shows deceleration</u>
<u></u>
(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
<u>t = 33.2 s</u>
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Below are the choices the can be found elsewhere:
A.) 14 newtons upward
<span>B.) 45 newtons upward </span>
<span>C.) 67 newtons upward </span>
<span>D.) 130 newtons upward </span>
<span>E.) 150 newtons upward
</span>
The answer is A.) 14 newtons upward
Answer:
4200 Joules
Explanation:
Work done =force x distance
From the question , we’re given f =350N ,
d = 12m
Using the above formula, we have
Workdone = 350 x 12
= 4200 Joules
Answer: The riders are subjected to 11.5 revolutions per minute
Explanation: Please see the attachments below
