A compass has a magnitized needle that follows magnetic fields in the earth to lead you to the South I believe because of how strong the magnetic pull is.
Answer:
The stiffness of one of the individual spring is 390 N/m.
Explanation:
It is given that 43 identical springs are placed side-by-side and connected to a large massive block.
The stiffness of the 43 spring combination is 16770 N/m
We need to find the stiffness of one of the individual springs. Let k is the stiffness of one spring. The effective spring stiffness of this width spring is given by :
![K=N\times k](https://tex.z-dn.net/?f=K%3DN%5Ctimes%20k)
![k=\dfrac{K}{N}](https://tex.z-dn.net/?f=k%3D%5Cdfrac%7BK%7D%7BN%7D)
![k=\dfrac{16770\ N/m}{43}](https://tex.z-dn.net/?f=k%3D%5Cdfrac%7B16770%5C%20N%2Fm%7D%7B43%7D)
k = 390 N/m
So, the stiffness of one of the individual spring is 390 N/m. Hence, this is the required solution.
Answer:
FT = 103.15 N and T = 1.75 s
Explanation:
There are 3 forces at play here: gravity acting in vertical direction, tension from the string that makes 18 degrees with the vertical, and centripetal force generated by the circular motion which is horizontal.
Since the bob is moving at the constant speed, according to Newton 's 1st law the net force is 0. To find the tension force that makes a fixed angle of 18 degrees with the vertical, we can calculate that tension in the vertical direction and balance it with gravity force. Let g = 9.81m/s2
![F_T*cos18^o = mg = 10*9.81 = 98.1](https://tex.z-dn.net/?f=F_T%2Acos18%5Eo%20%3D%20mg%20%3D%2010%2A9.81%20%3D%2098.1)
![F_T = 98.1/cos18^o = 103.15 N](https://tex.z-dn.net/?f=F_T%20%3D%2098.1%2Fcos18%5Eo%20%3D%20103.15%20N)
To calculate the centripetal force in the horizontal direction, we can calculate tension in the horizontal direction and balance it with the centripetal force
![F_Tsin18^o = F_c](https://tex.z-dn.net/?f=F_Tsin18%5Eo%20%3D%20F_c)
![F_c = 103.15*sin18^o = 31.87 N](https://tex.z-dn.net/?f=F_c%20%3D%20103.15%2Asin18%5Eo%20%3D%2031.87%20N)
The centripetal acceleration a is
![a_c = F_c / m = 31.87 /10 = 3.187 m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%20F_c%20%2F%20m%20%3D%2031.87%20%2F10%20%3D%203.187%20m%2Fs%5E2)
The radius of circular motion is:
![8sin18^o = 2.47 m](https://tex.z-dn.net/?f=8sin18%5Eo%20%3D%202.47%20m)
The angular speed is
![\omega^2 R = a_c](https://tex.z-dn.net/?f=%5Comega%5E2%20R%20%3D%20a_c)
![\omega = \sqrt{\frac{a_c}{R}} = \sqrt{3.187 / 2.47} = 3.6 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Ba_c%7D%7BR%7D%7D%20%3D%20%5Csqrt%7B3.187%20%2F%202.47%7D%20%3D%203.6%20rad%2Fs)
The time it would take to sweep 1 period, or 2π rad is
![T = 2\pi/\omega = 2\pi/3.6 = 1.75 s](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%2F%5Comega%20%3D%202%5Cpi%2F3.6%20%3D%201.75%20s)