Question

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy?

Answer 1

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

Answer 2

The plane’s kinetic energy is about 134 J

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Further explanation

Let's recall the Kinetic Energy formula:

$\boxed {E_k = \frac{1}{2}mv^2 }$

Ek = kinetic energy ( J )

m = mass of object ( kg )

v = speed of object ( m/s )

Let us now tackle the problem!

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Given:

mass of model airplane = m = 3.00 kg

velocity component in the east direction = v_x = 5.00 m/s

velocity component in the north direction = v_y = 8.00 m/s

Asked:

kinetic energy of the plane = Ek = ?

Solution:

Firstly , we will find the resultant velocity as follows:

$v^2 = (v_x)^2 + (v_y)^2$

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Next, we could calculate the plane's kinetic energy by using following formula:

$E_k = \frac{1}{2} m v^2$

$E_k = \frac{1}{2} m (v_x^2 + v_y^2)$

$E_k = \frac{1}{2} m v_x^2 + \frac{1}{2} m v_y^2$

$E_k = (\frac{1}{2} \times 3.00 \times 5.00^2) + (\frac{1}{2} \times 3.00 \times 8.00^2)$

$E_k = 133.5 \texttt{ J}$

$E_k \approx 134 \texttt{ J}$

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Learn more

• Velocity of Runner : brainly.com/question/3813437

• Kinetic Energy : brainly.com/question/692781

• Acceleration : brainly.com/question/2283922

• The Speed of Car : brainly.com/question/568302

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Answer details

Grade: High School

Subject: Physics

Chapter: Energy