Question

76. [0/20 Points]10 kvDETAILSMY NOTESASK YOUR TEACHERA 5.87-kg rocket with a charge of 3.20 mC rests on a horizontal plate. Another horizontal plate above the first has a hole directlyabove the rocket. A potential difference of 10.0 kV is established between the two plates, causing the rocket to rise into the air.What is the maximum height the rocket reaches before it comes crashing back to earth?

Answer 1

0.5563 m

Explanation:

Given:

Mass (m) = 5.87 kg

Voltage (V) = 10 kV = 10000 V

Charge (q) = 3.2 mC = 3.2 x 10^-3 C

First, we calculate the kinetic energy, as follows:

`\begin{gathered} KE=q\cdot V \\ \\ KE=3.2\cdot10^(-3)\cdot10000 \\ \\ KE=32\text{ N} \end{gathered}`

This is the kinetic energy of the rocket when it comes out of the hole, it is equal to the potential energy, therefore we can calculate the maximum height to which the rocket rises as follows:

`\begin{gathered} KE=PE \\ \\ PE=mgh \\ \\ \text{ we replacing:} \\ \\ 32=5.87\cdot9.8\cdot h \\ \\ h=(32)/(5.87\cdot9.8) \\ \\ h=0.5563\text{ m} \end{gathered}`

The maximum height of the rocket is 0.5563 meters