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ANTONII [103]
3 years ago
12

What is the strength of an electric field that will balance the weight of a proton?

Physics
1 answer:
alukav5142 [94]3 years ago
3 0

The weight (force of gravity) on an object near the surface of the earth is given by

F = mg

where m is the object's mass and g is the acceleration of objects near the surface of the earth.

The electric force on a charged object is given by

F = Eq

where E is the electric field at the point where the object is located and q is the charge of the object.

The weight (force of gravity) of a proton will act downwards (toward the surface of the earth), so in order for this force to be balanced, there must be an electric field pointing upward such that the electric force is equal and opposite to the weight. Set the magnitudes of the weight and electric force equal to each other:

mg = Eq

E = mg/q

Known values:

m = 1.67×10⁻27kg (a quick Google search will yield the mass of a proton)

g = 9.81m/s²

q = 1.60×10⁻19C

Plug in the known values and solve for E:

E = 1.67×10⁻27×9.81/(1.6×10⁻19)

E = 1.02×10⁻⁷N/C

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The formula for the acceleration due to gravity is:

a = Gm/r²
where
G is the universal gravitational constant = 6.6726 x 10⁻¹¹ N-m²/kg²
m is the mass of planet
r is the radius of planet

So, if they have the same a:

m₁/r₁² = m₂/r₂²
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How can you use graphs to calculate the displacement of an object?
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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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Answer: B. I hope you get this right.

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