(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.
(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
<h3>
Potential energy of the proton</h3>
U = qΔV
where;
- q is charge of the proton
- ΔV is potential difference
U = q(Ed)
U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)
U = 3.6 x 10⁻¹⁸ J
<h3>Potential difference between the negative plate and a point midway</h3>
ΔV = E(0.5d)
ΔV = 0.5Ed
ΔV = 0.5 (1500)(1.5 x 10⁻²)
ΔV = 11.25 V
<h3>Speed of the proton </h3>
U = ¹/₂mv²
U = mv²
v² = 2U/m
where;
- m is mass of proton = 1.67 x 10⁻²⁷ kg
v² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)
v² = 4.311 x 10⁹
v = √(4.311 x 10⁹)
v = 6.57 x 10⁴ m/s
Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
The potential difference between the negative plate and a point midway between the plates is 11.25 V.
The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
Learn more about potential difference here: brainly.com/question/24142403
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Sometimes in the same direction but most of them go in the opposite direction
Answer:
a = 0,1[m/s^2]
Explanation:
First we need to indentify the initial data.
And using this kinematic equation we have:
![v = 4[m/s]\\v_{0}= 2 [m/s] \\t = 20[s]\\\\v= v_{0}+a*t\\a=\frac{v-v_{0}}{t} \\a= \frac{4-2}{20} \\a=0.1[m/s^{2}]](https://tex.z-dn.net/?f=v%20%3D%204%5Bm%2Fs%5D%5C%5Cv_%7B0%7D%3D%202%20%5Bm%2Fs%5D%20%5C%5Ct%20%3D%2020%5Bs%5D%5C%5C%5C%5Cv%3D%20v_%7B0%7D%2Ba%2At%5C%5Ca%3D%5Cfrac%7Bv-v_%7B0%7D%7D%7Bt%7D%20%5C%5Ca%3D%20%5Cfrac%7B4-2%7D%7B20%7D%20%5C%5Ca%3D0.1%5Bm%2Fs%5E%7B2%7D%5D)
The kinetic energy of the tomato is :
K.E = 1/2 mv^2
K.E = 1/2 x 0.18 kg x 11 m/S^2
K.E = 0.99
Hope this helps
