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grigory [225]
3 years ago
13

There are two balloons of charges +3.37 x 10-6 C and –8.21 x 10-6 C. The distance between the two balloons is 2.00 m. Determine

the force between the two balloons.
Physics
1 answer:
andrew-mc [135]3 years ago
4 0

F = 0.06N. Since the charges has different signs the force of atraction between them is 0.06N.

In order to solve this exercise we have to use Coulomb's Law equation which says that the magnitude of each of the electric forces with which two point charges at rest interact is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them. Given by the equation:

F = k\frac{|q_{1}q_{2}|}{d^{2}}. Where k is the Coulomb's Constant k = 9x10^{9}\frac{Nm^{2} }{C^{2} }, q1 and q2 are the charges value in Coulomb (C), and d is the distance between charges in meters (m).

There are two balloons of charges +3.37 x 10-6 C and –8.21 x 10-6 C. The distance between the two balloons is 2.00 m. Calculate the force between the two ballons.

F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|(3.37x10^{-6}C)(-8.21x10^{-6}C)|}{(2.00m)^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|-2.77x10^{-11}C^{2} )|}{4.00m^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} }(6.92x10^{-12})\frac{C^{2} }{m^{2}}\\F = 0.06N

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g_M=\frac{6.67*10^{-11}*8.93*10^{22}}{(1821*10^3)^2m^2} \\g_M=1.7962 m/s^2

According to law of conservation of energy

Initial Energy=Final Energy

K.E_i+mgh_i=K.E_f+mgh_f

\frac{1}{2}m(v_0)^2+mgh_o= \frac{1}{2}m(v_f)^2+mgh_f\\At\ maximum\ height\ v_f=0\\\frac{1}{2}m(v_0)^2+0=mgh_f\\v_0=\sqrt{2gh_f}

For Jupiter's moon Io:

Velocity is given by:

v_0_M=\sqrt{2g_Mh_f_M}

For Earth Velocity is given by:

v_0_E=\sqrt{2g_Eh_f_E}

Now:

v_o_M=v_o_E

\sqrt{2g_Mh_f_M}=\sqrt{2g_Eh_f_E}\\h_f_E=\frac{g_Mh_f_M}{g_E}

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h_f_E=\frac{1.7962*500*10^3m}{9.8} \\h_f_E=91642.85 m\\h_f_E=91.64Km

91.64 km high material would go on earth if it were ejected with the same speed as on Io.

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Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
3 years ago
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