Answer:
The electrical potential energy is 0.027 Joules.
Explanation:
The values from the question are
charge (q) = ![4.5 \times 10^{-5} C](https://tex.z-dn.net/?f=4.5%20%5Ctimes%2010%5E%7B-5%7D%20C)
Electric Field strength (E) = ![2.0 \times 10^{4} N/C](https://tex.z-dn.net/?f=2.0%20%5Ctimes%2010%5E%7B4%7D%20N%2FC)
Distance from source (d) = 0.030 m
Now the formula for the electrical potential energy (U) is given by
![U = q \times E \times d](https://tex.z-dn.net/?f=U%20%3D%20q%20%5Ctimes%20E%20%5Ctimes%20d)
So now insert the values to find the answer
![U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m](https://tex.z-dn.net/?f=U%20%3D%204.5%20%5Ctimes%2010%5E%7B-5%7D%20C%20%5Ctimes%202.0%20%5Ctimes%2010%5E%7B4%7D%20N%2FC%20%5Ctimes%200.030%20m)
On further solving
![U = 0.027 J](https://tex.z-dn.net/?f=U%20%3D%200.027%20J)
Answer:
![v_f=9,07~m/s](https://tex.z-dn.net/?f=v_f%3D9%2C07~m%2Fs)
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes uniformly in time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
![v_f=v_o+at](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat)
The car initially travels at vo=7.35 m/s and accelerates at a rate of
during t=2.09 s.
The final velocity is:
![v_f=7.35+0.824*2.09](https://tex.z-dn.net/?f=v_f%3D7.35%2B0.824%2A2.09)
![\mathbf{v_f=9,07~m/s}](https://tex.z-dn.net/?f=%5Cmathbf%7Bv_f%3D9%2C07~m%2Fs%7D)
To be able to do the experiment, and like that being able to find the correct answer to your hypothesis by testing it.