Take the derivative to find the velocity of the object:

The object stops when
:

so the answer is E.
Answer:
a)32.34 N/m
b)10cm
c)1.6 Hz
Explanation:
Let 'k' represent spring constant
'm' mass of the object= 330g =>0.33kg
a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.
ΣF=kx-mg=0
k=mg / x
k= (0.33 x 9.8)/ 0.1
k= 32.34 N/m
b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.
Therefore, amplitude of the oscillation is 10cm
c)frequency of the oscillation can be determined by,
f= 1/2π 
f= 1/2π 
f= 1.57
f≈ 1.6 Hz
Therefore, the frequency of the oscillation is 1.6 Hz
155Ω
Explanation:
R = R ref ( 1 + ∝ ( T - Tref)
where R = conduction resistance at temperature T
R ref = conductor resistance at reference temperature
∝ = temperature coefficient of resistance for conductor
T = conduction temperature in degrees Celsius
T ref = reference temperature that ∝ is specified at for the conductor material
T = 600 k - 273 k = 327 °C
Tref = 300 - 273 K = 27 °C
R = 50 Ω ( 1 + 0.007 ( 327 - 27) )
R = 155Ω
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