Answer:
Acceleration is the change in velocity divided by time
Explanation:
This is the correct answer because distance divided by time is the position. Speed multiplied by time is the distance. And acceleration is not just velocity, but the change in velocity over time.
Answer:
On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means
![\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m](https://tex.z-dn.net/?f=%5Cfrac%7B%5Clambda%7D%7B4%7D%20%3D%200.275~m%5C%5C%5Clambda%20%3D%201.1~m)
This means that the relation between the wavelength and the length of the string is
![3\lambda/2 = L](https://tex.z-dn.net/?f=3%5Clambda%2F2%20%3D%20L)
By definition, this standing wave is at the third harmonic, n = 3.
Furthermore, the standing wave equation is as follows:
![y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)](https://tex.z-dn.net/?f=y%28x%2Ct%29%20%3D%20%28A%5Csin%28kx%29%29%5Csin%28%5Comega%20t%29%20%3D%20A%5Csin%28%5Cfrac%7B%5Comega%7D%7Bv%7Dx%29%5Csin%28%5Comega%20t%29%20%3D%20A%5Csin%28%5Cfrac%7B2%5Cpi%20f%7D%7Bv%7Dx%29%5Csin%282%5Cpi%20ft%29%20%3D%20A%5Csin%28%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7Dx%29%5Csin%28%5Cfrac%7B2%5Cpi%20v%7D%7B%5Clambda%7Dt%29%20%3D%20%282.45%5Ctimes%2010%5E%7B-3%7D%29%5Csin%285.7x%29%5Csin%2859.94t%29)
The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.
![v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)](https://tex.z-dn.net/?f=v_y%28x%2Ct%29%20%3D%20%5Cfrac%7Bdy%28x%2Ct%29%7D%7Bdt%7D%20%3D%20%5Comega%20A%5Csin%28kx%29%5Ccos%28%5Comega%20t%29%5C%5Ca_y%28x%2Ct%29%20%3D%20%5Cfrac%7Bdv%28x%2Ct%29%7D%7Bdt%7D%20%3D%20-%5Comega%5E2A%5Csin%28kx%29%5Csin%28%5Comega%20t%29)
![a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0](https://tex.z-dn.net/?f=a_y%28x%2Ct%29%20%3D%20-%2859.94%29%5E2%282.45%5Ctimes%2010%5E%7B-3%7D%29%5Csin%28%285.7%29%280.138%29%29%5Csin%2859.94t%29%20%3D%200)
For this equation to be equal to zero, sin(59.94t) = 0. So,
![59.94t = \pi\\t = \pi/59.94 = 0.0524~s](https://tex.z-dn.net/?f=59.94t%20%3D%20%5Cpi%5C%5Ct%20%3D%20%5Cpi%2F59.94%20%3D%200.0524~s)
This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:
![v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s](https://tex.z-dn.net/?f=v_y%28x%3D0.138%2Ct%3D0.0524%29%20%3D%20%2859.94%29%282.45%5Ctimes%2010%5E%7B-3%7D%29%5Csin%28%285.7%29%280.138%29%29%5Ccos%28%2859.94%29%280.0524%29%29%20%3D%200.002~m%2Fs)
Answer:
55%
Explanation:
take efficiency=power output/power input multiply by 100%
Answer:
wave frequency
Explanation:
i took the test, trust me
Answer: 0.333 h
Explanation:
This problem can be solved using the <u>Radioactive Half Life Formula</u>:
(1)
Where:
is the final amount of the material
is the initial amount of the material
is the time elapsed
is the half life of the material (the quantity we are asked to find)
Knowing this, let's substitute the values and find
from (1):
(2)
(3)
Applying natural logarithm in both sides:
(4)
(5)
Clearing
:
(6)
Finally:
This is the half-life of the Bismuth-218 isotope