Answer:
0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz
Explanation:
The fundamental frequency of a standing wave on a string is given by
![f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B2L%7D%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%7D)
where
L is the length of the string
T is the tension in the string
is the mass per unit length
For the string in the problem,
L = 30.0 m
![\mu=9.00\cdot 10^{-3} kg/m](https://tex.z-dn.net/?f=%5Cmu%3D9.00%5Ccdot%2010%5E%7B-3%7D%20kg%2Fm)
T = 20.0 N
Substituting into the equation, we find the fundamental frequency:
![f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B2%2830.0%29%7D%5Csqrt%7B%5Cfrac%7B20.0%7D%7B%289.00%5Ccdot%2010%5E%7B-3%7D%7D%7D%3D0.786%20Hz)
The next frequencies (harmonics) are given by
![f_n = nf](https://tex.z-dn.net/?f=f_n%20%3D%20nf)
with n being an integer number and f being the fundamental frequency.
So we get:
![f_2 = 2 (0.786 Hz)=1.572 Hz](https://tex.z-dn.net/?f=f_2%20%3D%202%20%280.786%20Hz%29%3D1.572%20Hz)
![f_3 = 3 (0.786 Hz)=2.358 Hz](https://tex.z-dn.net/?f=f_3%20%3D%203%20%280.786%20Hz%29%3D2.358%20Hz)
![f_4 = 4 (0.786 Hz)=3.144 Hz](https://tex.z-dn.net/?f=f_4%20%3D%204%20%280.786%20Hz%29%3D3.144%20Hz)
Answer:
The hiker followed a road heading north for 2 miles in 30 minutes.
Explanation:
In order to describe the motion of an object, distance covered and time taken must be required. The total path covered by an object is called the distance travelled.
The hiker followed a road heading north for 2 miles in 30 minutes. This describes the motion of hiker. The motion shows how fast the hiker is moving.
Distance, d = 2 miles = 3218.6 m
times, t = 30 minutes = 1800 seconds
So, we can say that the hiker is moving with a speed of 1.78 m/s in north direction.
Hence, this is the required solution.
<span>Answer:
Spherical Distribution
Feedback: Correct
The stars in the halo component have highly-inclined random orbits that orbit the center of our Galaxy. The stars within the halo would therefore make up a spherical distribution of stars surrounding the center of the Galaxy. In comparison, the disk stars move in elliptical orbits, which are nearly circular and are confined to the disk of the Galaxy. Disk stars therefore have very small inclinations and do not move above or below the plane of the Galactic disk.</span>
N2O is nitrous oxide becuase of the nitrogen and oxygen
Answer:
Fc = 19.2 N
Explanation:
In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:
![F_c=m\frac{v^2}{r}](https://tex.z-dn.net/?f=F_c%3Dm%5Cfrac%7Bv%5E2%7D%7Br%7D)
m: mass of the rock = 600g = 0.6 kg
v: tangential velocity of the Honda = 4m/s
r: radius of the Honda = 50cm = 0.5m
You replace the values of m, r and v in the equation for Fc:
![F_c=(0.6kg)\frac{(4m/s)^2}{0.5m}\\\\F_c=19.2N](https://tex.z-dn.net/?f=F_c%3D%280.6kg%29%5Cfrac%7B%284m%2Fs%29%5E2%7D%7B0.5m%7D%5C%5C%5C%5CF_c%3D19.2N)
hence, the force has a magnitude of 19.2 N
If the rock would have more mass the centripetal force would be higher