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
Needs are less than resources, and the population increases.
The velocity of the body is zero; option A
<h3>What is the motion of an oscillating body?</h3>
The motion of an oscillating body is known as simple harmonic motion.
Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.
For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.
In conclusion, oscillating bodies undergo simple harmonic motion.
Learn more about simple harmonic motion at: brainly.com/question/24646514
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The change in temperature here corresponds to a sensible heat. The amount of energy required can be calculated by multiplying the specific heat capacity, the amount of the substance and the corresponding change in temperature.
Heat required = mCΔT
Heat required = 0.368 kg (0.0920 cal/g°C) (60 - 23)°C
Heat required = 1.25 cal
