Answer:
The length of the beam increasing is 9.64 ft/s.
Explanation:
Given that,
Height = 210 ft
Distance =290 ft
According to figure,
We need to calculate the angle
....(I)
Put the value of x in the equation
![\cos\theta=\dfrac{210}{290}](https://tex.z-dn.net/?f=%5Ccos%5Ctheta%3D%5Cdfrac%7B210%7D%7B290%7D)
![\cos\theta=\dfrac{21}{29}=0.72](https://tex.z-dn.net/?f=%5Ccos%5Ctheta%3D%5Cdfrac%7B21%7D%7B29%7D%3D0.72)
Now, ![\sin\theta=\dfrac{20}{29}](https://tex.z-dn.net/?f=%5Csin%5Ctheta%3D%5Cdfrac%7B20%7D%7B29%7D)
On differentiate of equation (I)
![-\sin\theta\dfrac{d\theta}{dt}=-\dfrac{-210}{x^2}\dfrac{dx}{dt}](https://tex.z-dn.net/?f=-%5Csin%5Ctheta%5Cdfrac%7Bd%5Ctheta%7D%7Bdt%7D%3D-%5Cdfrac%7B-210%7D%7Bx%5E2%7D%5Cdfrac%7Bdx%7D%7Bdt%7D)
![\sin\theta=\dfrac{210}{x^2}\dfrac{dx}{dt}](https://tex.z-dn.net/?f=%5Csin%5Ctheta%3D%5Cdfrac%7B210%7D%7Bx%5E2%7D%5Cdfrac%7Bdx%7D%7Bdt%7D)
Put the value in the equation
![\sin\dfrac{20}{29}\times2.0=\dfrac{210}{(290)^2}\dfrac{dx}{dt}](https://tex.z-dn.net/?f=%5Csin%5Cdfrac%7B20%7D%7B29%7D%5Ctimes2.0%3D%5Cdfrac%7B210%7D%7B%28290%29%5E2%7D%5Cdfrac%7Bdx%7D%7Bdt%7D)
![\dfrac{dx}{dt}=\sin\dfrac{20}{29}\times2.0\times\dfrac{290^2}{210}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%3D%5Csin%5Cdfrac%7B20%7D%7B29%7D%5Ctimes2.0%5Ctimes%5Cdfrac%7B290%5E2%7D%7B210%7D)
![\dfrac{dx}{dt}=9.64\ ft/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%3D9.64%5C%20ft%2Fs)
Hence, The length of the beam increasing is 9.64 ft/s.
<span>The lagrangian center of mass between two particles of equal mass can be calculated using the following equation: L=12MV2â’Uext.+Nâ‘i=112mi~v2iâ’Uint. M=â‘mi is the total mass, V is the speed of the center of mass, Uext.=â‘Uext.,i, and ~vi is the speed of the ith particle relative to the center of mass.</span>
0.4 x 18 = 7.2 kg m/s
The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s
7.2 - 5 = 2.2 kg m/s is the motmentum of the ball now
the velocity is 2.2/0.4 = 5.5 m/s
I’m sorry i haven’t found the answer to this
Answer:
7.1 Hz
Explanation:
In a generator, the maximum induced emf is given by
![\epsilon= 2\pi NAB f](https://tex.z-dn.net/?f=%5Cepsilon%3D%202%5Cpi%20NAB%20f)
where
N is the number of turns in the coil
A is the area of the coil
B is the magnetic field strength
f is the frequency
In this problem, we have
N = 200
![A=0.030 m^2](https://tex.z-dn.net/?f=A%3D0.030%20m%5E2)
![\epsilon=8.0 V](https://tex.z-dn.net/?f=%5Cepsilon%3D8.0%20V)
B = 0.030 T
So we can re-arrange the equation to find the frequency of the generator:
![f=\frac{\epsilon}{2\pi NAB}=\frac{8.0 V}{2\pi (200)(0.030 m^2)(0.030 T)}=7.1 Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B%5Cepsilon%7D%7B2%5Cpi%20NAB%7D%3D%5Cfrac%7B8.0%20V%7D%7B2%5Cpi%20%28200%29%280.030%20m%5E2%29%280.030%20T%29%7D%3D7.1%20Hz)