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Karo-lina-s [1.5K]
1 year ago
6

At the Earth's surface, a projectile is launched straight up at a speed of 10.0km / s. To what height will it rise? Ignore air r

esistance and the rotation of the Earth.
Physics
1 answer:
icang [17]1 year ago
6 0

At h = 5199.28 km  it will rise. Kinetic energy is a type of energy that is present in a particle or object in motion.

<h3>What is kinetic energy?</h3>

Kinetic energy is a type of energy that is present in a particle or object in motion. When work, which entails the transfer of energy, is done on an object by applying a net force, that object acquires kinetic energy.

There are five types of kinetic energy: mechanical, electrical, acoustic, thermal, and radiant. Kinetic energy (KE), which is the energy of a body in motion, is essentially the energy of all moving objects. It is one of the two main types of energy together with potential energy, which is the stored energy existing in objects at rest. 

<h3>To what height will it rise?</h3>

Equating the initial kinetic energy to the final potential energy

At the highest point, the body has only potential energy

initial kinetic energy = .5 * m * v2

velocity = 10.1 km/s

= 10.1 * 10100 m/s

= 10100 m/s

KE = .5 * m * (10100)2 = 51.005 x 106 x m   J

potential energy = m x g x h

max height reached = m x 9.81 x h

m x 9.81 x h =  51.005 x 106 x m   J

h = 5199.28 km

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

#SPJ4

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Answer:

(a) Approximately 10.1\; {\rm V}.

Explanation:

Let C denote the capacitance of a capacitor. Let V be the potential difference (voltage) between the two plates of this capacitor. The energy E stored in this capacitor would be:

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\begin{aligned}V^{2} &= \frac{2\, E}{C} \end{aligned}.

\begin{aligned}V &= \sqrt{\frac{2\, E}{C}} \end{aligned}

The capacitance C of this capacitor is given in nanofarads. Convert that unit to standard unit (farads):

\begin{aligned}C &= 360\; {\rm nF} \\ &= 360\; {\rm nF} \times \frac{1\; {\rm F}}{10^{9}\; {\rm nF}} \\ &= 3.60 \times 10^{-7}\; {\rm F}\end{aligned}.

Given that the energy stored in this capacitor is E = 1.85 \times 10^{-5}\; {\rm J}, the potential difference across the capacitor plates would be:

\begin{aligned}V &= \sqrt{\frac{2 \times 1.85 \times 10^{-5}\; {\rm J}}{3.60 \times 10^{-7}\; {\rm F}}} \\ &\approx 10.1\; {\rm V}\end{aligned}.

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