A. Increase is correct because the statement said the moons gravitational "PULL"
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
Answer:
Mumbai
Explanation:
Because Mumbai is a coastal city
due to high humidity,vigorous rusting damages all the iron structures
∴ rusting is a major problem in Mumbai
Answer:
A) The resultant force is 43.4 [N]
B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis
Explanation:
We need to make a sketch of the different forces acting on the heavy crate.
In the attached image we can see the forces and the sum of the vector with their respective angles.
Forces in the X-axis

Forces in the y-axis
![FDiony=0[N]\\Fshirley= 16.5*sin(30)=8.25[N]\\Fjoany=19.5*sin(60)=16.88 [N]\\\\Forcesy=0+8.25-16.88= -8.63[N]](https://tex.z-dn.net/?f=FDiony%3D0%5BN%5D%5C%5CFshirley%3D%2016.5%2Asin%2830%29%3D8.25%5BN%5D%5C%5CFjoany%3D19.5%2Asin%2860%29%3D16.88%20%5BN%5D%5C%5C%5C%5CForcesy%3D0%2B8.25-16.88%3D%20-8.63%5BN%5D)
Using the Pythagorean theorem

The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.