(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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Answer:
Explanation:
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The frequency of a wave is just the reciprocal of the period.For this one . . .
Frequency = 1/(period) = 1/(5 sec) = 0.2 per sec (0.2 Hz) .
Answer:
The answer is e.
Explanation:
We take:
Taking the gas as an ideal gas, we can use the ideal gas law:
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Then:
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Taking R=0.08205atmL/molK.
