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IceJOKER [234]
3 years ago
5

A hunter points a rifle directly at a coconut that he intends to shoot off a tree. It so happens that the coconut falls from the

tree at the exact instant the hunter pulls the trigger. What happens to the bullet?
Show your motion diagram and complete solution.

Show the solution to prove your answer.

It passes above the coconut

It passes beneath coconut

It hits the coconut
Physics
1 answer:
yan [13]3 years ago
4 0

Answer:

It hits the coconut

Explanation:

The coconut fell immediately after the trigger was pulled as a result of the impact made by the bullet on the coconut. The Third Law of  motion states that : For every action, there is an equal and opposite reaction. The action was performed when the bullet struck the coconut, while the coconut reacted by falling of the tree.

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Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
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Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

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3 years ago
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Answer:

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Explanation:

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