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Juliette [100K]
3 years ago
5

The two hot-air balloons in the drawing are 48.2m and 61.0 m above the ground.A person in the left balloon observes that the rig

ht balloon is 13.3 degrees above the horizontal.What is the horizontal distance x between the two balloons?

Physics
1 answer:
mafiozo [28]3 years ago
4 0

Answer:

The horizontal distance x between the two balloons is 54.15 m

Explanation:

The diagram described as obtained online is presented in the image attached to this solution.

Let the horizontal distance between the two balloons be x

Difference in height (vertical distance) between the two balloons = 61 - 48.2 = 12.8 m

Using trigonometric relations, it is evident that

Tan 13.3° = 12.8/x

x = 12.8/tan 13.3° = 12.8/0.2364 = 54.15 m

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Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv
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Answer:

 electric field E = - k Q (1 /r(r-a)), force    F = - k Q qo / r (r-a) and force for r>>a    F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

       

           E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

          λ = q / x

As it is constant, we can write it based on differentials

         λ = dq / dx

         dq = λ dx

We already have all the terms, let's  integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

         E = k ∫ λ dx / x²

         E = k la (- 1 / x)

Let's get the negative sign from the parentheses

         E = - k λ (1 / x)

         E = - k λ (1 /(r-a)  -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

         E = - k Q/a  [a / r (r-a)]

         E = - k Q (1 /r(r-a))

b) We calculate the force.  

         F = E qo

         F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

         F = - k Qqo / r² (1 -  a/r)

         F ≈ - k Q qo / r²

6 0
3 years ago
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