Question

A pilot of an aeroplane fire a short wing 0.1 kg E with the velocity e 300 per second calculate the total energy of the shot when it is at a height of of 500 m above the ground

Answer 1

Answer:

here you go

final Answer is 4990

Answer 2

Given data : A pilot of an aeroplane fires a shot weighing 0.1 kg with a vertical of 300 m/s. It is at a height of 500 m above the ground.

To find : Calculate the total energy of the shot when it is at a height of 500m above the ground.

Solution : Here, according to the given data;

➜ Mass, m = 0.1 kg

➜ Velocity, v = 300 m/s

➜ Height, h = 500 m

Here, we know that,

➜ Total energy = Kinetic energy + potential energy

Now,

➜ Kinetic energy = ½ * mv²

➜ Kinetic energy = ½ * 0.1 * 300²

➜ Kinetic energy = 0.05 * 9 * 10⁴

➜ Kinetic energy = 0.45 * 10⁴

➜ Kinetic energy = 4500 Joule ----{1}

➜ Potential energy = mgh

Where g is acceleration due to gravity (9.8 m/s²)

➜ Potential energy = 0.1 * 9.8 * 500

➜ Potential energy = 0.98 * 500

➜ Potential energy = 490 Joule ----{2}

➜ Total energy = Kinetic energy + potential energy

Now, from eq. {1} and eq. {2}

➜ Total energy = 4500 + 490

➜ Total energy = 4990 Joule

Answer : Hence, the total energy of the shot when it is at a height of of 500 m above the ground is 4990 Joule.