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Dmitrij [34]
3 years ago
8

Two charged spheres on a frictionless horizontal surface are attached to opposite ends of a string & are in static equilibri

um. The 18 kg red sphere has more charge than the 11 kg green sphere. The total charge magnitude on the spheres is 10 μC & they have the same polarity. As a result the tension is 2.5 N & the centers of the spheres are 0.06 m apart. Determine the charge magnitude on each sphere.
Physics
1 answer:
deff fn [24]3 years ago
5 0

Answer:

the red sphere has a charge of 9.899 10⁻⁶ C and

the green sphere 0.101 10⁻⁶ C

Explanation:

For this exercise we will use coulomb's law

          F = k q₁ q₂ / r²

where they indicate the value of the force F = 2.5 N, the distance between them r = 0.06 m

they also indicate the value of the total load

         q₁ + q₂ = 10 10⁻⁶ C

we substitute the values

         2.5 = 9 10⁹ q₁ q₂ / 0.06²

         2.5 0.06 2/9 10⁹ = q₁ q₂

         1 10⁻¹² = q₁ q₂

         10 10⁻⁶= q₁ + q₂

we have two equations with two i unknowns, so the system can be solved.

         q₁ = 1 10⁻¹² / q₂

we substitute in the other equation

         10 10⁻⁶ = 10⁻¹² / q₂ + q₂

let's solve the equation

         10 10⁻⁶ q₂ = 10⁻¹² + q₂²

         q₂² - 10⁻⁵ q₂ + 10⁻¹² = 0

          q₂ = [10⁻⁵ ± √(10⁻¹⁰ - 4 10⁻¹²)] / 2

          q₂ = [10⁻⁵ + - 0.9798 10⁻⁵] / 2 = 10⁻⁵ (1 + - 0.9798) / 2

          q₂ = 0.9899 10⁻⁵ C

          q₂ = 0.0101 10⁻⁵ C

let's find the charge of the other sphere

         q₂ = 0.9899 10⁻⁵ C

         q₁ = 10⁻⁵ - q₂

          q₁ = 10⁻⁵ - 0.9899 10⁻⁵

          q₁ = 0.0101   10⁻⁵    C

therefore the red sphere has a charge of 9.899 10⁻⁶ C and

the green sphere 0.101 10⁻⁶ C

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The x-coordinates of two objects moving along the x-axis are given as a function of time (t). x1= (4m/s)t x2= -(161m) + (48m/s)t
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The ball lands 154.3 ft from the origin at an angle of 13.6° from the eastern direction toward the south.

Explanation:

Hi there!

The position vector of the ball is described by the following equation:

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Where:

r =  poisition vector of the ball at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity (eastward).

t = time.

ax = horizontal acceleration (eastward).

y0 = initial horizontal position.

v0y = initial horizontal velocity (southward).

ay = horizontal acceleration (southward)

z0 = initial vertical position.

v0z = initial vertical velocity.

g = acceleration due to gravity.

We have to find at which time the vertical component of the position vector is zero (the ball is on the ground) and then we can calculate the horizontal distance traveled by the ball at that time, using the equations of the horizontal components of the position vector.

Let´s place the origin of the system of reference at the throwing point so that x0 and y0 and z0 = 0.

y =  z0 + v0z · t + 1/2 · g · t²            (z0 = 0)

0 = 48 ft/s · t - 1/2 · 32 ft/s² · t²

0 = t (48 ft/s - 16 ft / s² · t)                 (t= 0, the origin point)

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- 48 ft/s / -16 f/s² = t

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Now, we can calculate how much distance the ball traveled in that time.

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x = 150 ft

And now let´s calculate the distance traveled in southward direction:

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y =  1/2 · ay · t²

y = 1/2 · (-8 ft/s²) · (3 s)²

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Then, the final position vector will be:

r = (150 ft, -36 ft, 0)

The traveled distance is the magnitude of the position vector:

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To calculate the angle, we have to use trigonometry (see attached figure):

cos angle  = adjacent side / hypotenuse

cos α = x/r

cos α = 150 ft / 154.3 ft

α = 13.6°

The ball lands 154.3 ft from the origin at an angle of 13.5° from the eastern direction toward the south.

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3 years ago
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