Answer:
0.175 m/s
Explanation:
32.8105 cm = 0.328105 m
According to the law of conservation in energy, as the spring oscillates, its kinetic energy is converted to spring elastic energy and vice versa. So the maximum speed of the bananas occur when spring elastic energy is 0 and vice versa, spring elastic energy is maximum (at amplitude) with kinetic energy, and speed, is 0
![E_e = E_k](https://tex.z-dn.net/?f=E_e%20%3D%20E_k)
![kx^2/2 = mv^2/2](https://tex.z-dn.net/?f=kx%5E2%2F2%20%3D%20mv%5E2%2F2)
![13.9602*(0.328105)^2/2 = 49.0943v^2/2](https://tex.z-dn.net/?f=13.9602%2A%280.328105%29%5E2%2F2%20%3D%2049.0943v%5E2%2F2)
![v^2 = \frac{13.9602*(0.328105)^2}{49.0943} = 0.0306](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7B13.9602%2A%280.328105%29%5E2%7D%7B49.0943%7D%20%3D%200.0306)
![v = \sqrt{0.0306} = 0.175 m/s](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B0.0306%7D%20%3D%200.175%20m%2Fs)
30 30 30 30 30 30 30 30 30 30
Answer:
the force between the two charges is 198.37 N
Explanation:
Given;
charge of the first particle, Q₁ = -3 x 10⁻⁵ C
charge of the second particle, Q₂ = 9.0 x 10⁻⁵ C
distance between the two charges, r = 0.35 m
The force between the two charges is calculated from Coulomb's law;
![F = \frac{kQ_1Q_2}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BkQ_1Q_2%7D%7Br%5E2%7D)
where k is Coulomb's constant = 9 x 10⁹ Nm²/C²
![F = \frac{(9\times 10^9)(3.0\times 10^{-5})(9\times10^{-5} )}{0.35^2} \\\\F = 198.37 \ N](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%289%5Ctimes%2010%5E9%29%283.0%5Ctimes%2010%5E%7B-5%7D%29%289%5Ctimes10%5E%7B-5%7D%20%29%7D%7B0.35%5E2%7D%20%5C%5C%5C%5CF%20%3D%20198.37%20%5C%20N)
Therefore, the force between the two charges is 198.37 N
Explanation:
It is given that,
Electric field experienced by an electron, ![E=1200\ N/C](https://tex.z-dn.net/?f=E%3D1200%5C%20N%2FC)
Charge on electron, ![q=-1.6\times 10^{-19}\ C](https://tex.z-dn.net/?f=q%3D-1.6%5Ctimes%2010%5E%7B-19%7D%5C%20C)
Mass of electron, ![m=9.1\times 10^{-31}\ C](https://tex.z-dn.net/?f=m%3D9.1%5Ctimes%2010%5E%7B-31%7D%5C%20C)
The electric force is balanced by the force acting on electron as,
ma = qE
![a=\dfrac{qE}{m}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BqE%7D%7Bm%7D)
![a=\dfrac{-1.6\times 10^{-19}\times 1200}{9.1\times 10^{-31}}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B-1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%201200%7D%7B9.1%5Ctimes%2010%5E%7B-31%7D%7D)
![a=-2.1\times 10^{14}\ m/s^2](https://tex.z-dn.net/?f=a%3D-2.1%5Ctimes%2010%5E%7B14%7D%5C%20m%2Fs%5E2)
So, the acceleration of the electron is
. As the electron is a negatively charged particle, the field lines is from negative to positive charge.