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alexandr1967 [171]
2 years ago
12

Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha

rge q3 = +6.40 × 10−19 C is moved gradually along the x axis from x = 0 to x = +5.0 m. At what values of x will the magnitude of the electrostatic force on the third particle from the other two particles be (a) minimum and (b) maximum? What are the (c) minimum and (d) maximum magnitudes?
Physics
1 answer:
mel-nik [20]2 years ago
4 0

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × 10^{-26} N

(d) 7.37 ×10^{-29} N

Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force=     F=\frac{kq1q2}{x^{2} }

F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

F= 7.37×10^{-29 N

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tatuchka [14]

We have that the values for F north, F east, F up are

  • F_N=1.09090909*10^{-5}
  • F_E=5.18181818*10^{-6}
  • F_E=2*10^{-6}

From the Question we are told that

electric force F_1 = 1.2 x 10^{-3} N(N)

electric force , F_2=5.7 x 10^{-4} N(E)

electric force , F_3=2.2 x 10^{-4} N (U)

charge on this ball one q_1= 110 nC.

charge on this ball two q_2= -50 nC.

Generally the equation for the F north  is mathematically given as

F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}

F_N=1.09090909*10^{-5}

For F East

F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}

F_E=5.18181818*10^{-6}

For F UP

F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}

F_E=2*10^{-6}

For more information on this visit

brainly.com/question/21811998

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1. Find the charge if the number of electrons is 4 x 10-18​
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22. The answer is 22.
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As mass increases what happens to the kinetic energy
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As mass increases kinetic energy also increases; kinetic energy is directly proportional to mass so whatever is done to either affects the other one the same. i hope this helps :)
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How to find the total displacement of an object ?
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3 years ago
Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De
My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

3 0
3 years ago
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