Answer:
![a=4.8m/s^2](https://tex.z-dn.net/?f=a%3D4.8m%2Fs%5E2)
Explanation:
Hello,
In this case, since the acceleration in terms of position is defined as its second derivative:
![a=\frac{d^2x(t)}{dt^2}=\frac{d^2}{dt^2}(2.9+8.8t+2.4t^2)](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bd%5E2x%28t%29%7D%7Bdt%5E2%7D%3D%5Cfrac%7Bd%5E2%7D%7Bdt%5E2%7D%282.9%2B8.8t%2B2.4t%5E2%29)
The purpose here is derive x(t) twice as follows:
![a=\frac{d^2x(t)}{dt^2}=\frac{d}{dt}(8.8+2*2.4*t)\\ \\a=4.8m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bd%5E2x%28t%29%7D%7Bdt%5E2%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%288.8%2B2%2A2.4%2At%29%5C%5C%20%5C%5Ca%3D4.8m%2Fs%5E2)
Thus, the acceleration turns out 4.8 meters per squared seconds.
Best regards.
Answer:
IT and science courses
Explanation:
As Health Information Management department requires a person who should have fair knowledge about science and medical so that he/she can deal with the data extracted from diseases. He/she should also have fair knowledge about IT so that the person can manage the files and data very well. The data should be managed accurately so different IT programs help in completing this goal. IT knowledge will help to keep data safe, accurate and complete. Due to IT knowledge they can upgrade and redesign programs according to their need.
Answer:
0.25A
1.0A
2.5A
Explanation:
this is only for the calculated current column on edge
Answer:
210.3 degrees
Explanation:
The net force exerted on charge A = 59.5 N
Use the x and y coordinates of net force to get the direction
arctan (y/x)
Answer:
Kinetic energy of the electron ![E_k=1.23\ eV](https://tex.z-dn.net/?f=E_k%3D1.23%5C%20eV)
Explanation:
It is given that,
Initial speed of an electron, u = 660000 m/s
Final speed of the electron, v = 0 (at rest)
The kinetic energy of an electron is possessed due to the motion of an electron. The mathematical formula for the kinetic energy is given by :
![E_k=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=E_k%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
m is the mass of electron
![E_k=\dfrac{1}{2}\times 9.1\times 10^{-31}\times (660000)^2](https://tex.z-dn.net/?f=E_k%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%209.1%5Ctimes%2010%5E%7B-31%7D%5Ctimes%20%28660000%29%5E2)
![E_k=1.98\times 10^{-19}\ J](https://tex.z-dn.net/?f=E_k%3D1.98%5Ctimes%2010%5E%7B-19%7D%5C%20J)
Since, ![1\ eV=1.602\times 10^{-19}\ J](https://tex.z-dn.net/?f=1%5C%20eV%3D1.602%5Ctimes%2010%5E%7B-19%7D%5C%20J)
![E_k=\dfrac{1.98\times 10^{-19}}{1.602\times 10^{-19}}](https://tex.z-dn.net/?f=E_k%3D%5Cdfrac%7B1.98%5Ctimes%2010%5E%7B-19%7D%7D%7B1.602%5Ctimes%2010%5E%7B-19%7D%7D)
![E_k=1.23\ eV](https://tex.z-dn.net/?f=E_k%3D1.23%5C%20eV)
So, the initial kinetic energy of the electron is 1.23 eV. Hence, this is the required solution.