Answer:
0.1 m
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
Time period:
Explanation:
Mass of the wheel, m = 49 kg
Radius of the hoop, r = 0.73 m
Initial angular speed of the wheel, 
Final angular speed of the wheel, 
Time, t = 22 s
(a) If I is the moment of inertia of the hoop. It is equal to,



We know that the work done is equal to change in kinetic energy.



W = -1858.05 Joules
(b) Let P is the average power. It is given by :


P =84.45 watts
Hence, this is the required solution.
Answer:
a. The angular frequency is doubled.
e. The period is reduced to one-half of what it was.
Explanation:
Angular frequency is given as;
ω = 2πf

when the frequency is doubled

Thus, the angular frequency will be doubled.
Amplitude in simple harmonic motion is the maximum displacement.
Frequency is related to period in simple harmonic motion as given in the equation below;

when the frequency is doubled;

Thus, the period will be reduced to one-half of what it was.
<u><em>In accordance with the International Regulation for the prevention of collisions at sea</em></u><u>:
</u>
<u>1.- A sailing boat has a passing preference over a motorized boat, </u><u>except when the motor boat is limited by its draft</u><u>.
</u>
<u>2.- The sailboat must maintain its course and speed.
</u>
<u>3.- </u><em><u>If it is evident that the PWC does not respond</u></em><u>, the sailboat must sound the warning signal, and change its course to starboard.
</u>
<u>4.- </u><u><em>All actions must be taken as soon as possible</em></u><u>.
</u>
<u>5.- If a sailboat is using its engine, the situation changes, and in that case, both ships must alter to starboard.</u>