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zavuch27 [327]
3 years ago
10

A proton and an electron are held in place on the x axis. The proton is at x = -d, while the electron is at x = +d. They are rel

eased simultaneously, and the only force that affects their motions is the electrostatic force of attraction that applies to the other. Which particle reaches the origin first? Give your reasoning.
Physics
1 answer:
Over [174]3 years ago
6 0
The protons and electrons are held in place on the x axis.
The proton is at x = -d and the electron is at x = +d. They are released at the same time and the only force that affects movement is the electrostatic force that is applied on both subatomic particles. According to Newton's third law, the force Fpe exerted on protons by the electron is opposite in magnitude and direction to the force Fep exerted on the electron by the proton. That is, Fpe = - Fep. According to Newton's second law, this equation can be written as
                               Mp * ap = -Me * ae
where Mp and Me are the masses, and ap and ae are the accelerations of the proton and the electron, respectively. Since the mass of the electron is much smaller than the mass of the proton, in order for the equation above to hold, the acceleration of the electron at that moment must be considerably larger than the acceleration of the proton at that moment. Since electrons have much greater acceleration than protons, they achieve a faster rate than protons and therefore first reach the origin.
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Kisachek [45]

Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

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W_s = \frac{1}{2}kc^2

Work lost to friction:

W_f =\mu mg(c-x)

Energy:

E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2

(a) Solve for v:

v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}

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\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}

\frac{dv}{dx}=0 if:

\mu g-\frac{k}{m}x = 0

x_{max} = \frac{\mu gm}{k}

v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}

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3 years ago
What happens to the position of an object as an unbalanced force acts on?
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Answer:

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I need an answer for this plzz!!<br>number 2 <br>anybody can help ??
Anuta_ua [19.1K]
2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a \:  =  \:  \frac{dv}{dt}  \:  =  \:  \frac{25 \:  \frac{m}{s} }{50 \: s} \:  =  \: 0.5 \:  \frac{m}{ {s}^{2} }
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2ax \:  =  \:  {v}^{2}  \:  -  \:  {u}^{2}
x \:  =  \:  \frac{ {v}^{2}  \:   -  \:  {u}^{2} }{2a}  \:  =  \:   \frac{{(25 \:  \frac{m}{s})}^{2}  \:  -  \:  {(0 \:  \frac{m}{s} )}^{2} }{2 \:  \times  \: 0.5 \:  \frac{m}{ {s}^{2} } } \:  =  \: 625 \: m
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
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