(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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Calculate the pressure due to sea water as density*depth.
That is,
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa
Atmospheric pressure is 101.3 kPa
Total pressure is 94423 + 101.3 = 94524 kPa (approx)
The area of the window is π(0.44 m)^2 = 0.6082 m^2
The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx
Answer:
A. 98,000 J
Explanation:
The gravitational potential energy of an object is given by
U = mgh
where
m is the mass of the object
g is the gravitational acceleration
h is the heigth above the ground
In this problem,
m = 2000 kg
g = 9.8 m/s^2
h = 5.0 m
Substituting into the equation, we find
Answer:The percentage of the work input that becomes work output is the efficiency of a machine. Because there is always some friction, the efficiency of any machine is always less than 100 percent.
Explanation:
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