Respuesta :
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 :
[tex]v=\sqrt{v_1^2+v_2^2}[/tex]
[tex]v=\sqrt{5^2+8^2}=9.43\ m/s[/tex]
Let E is the kinetic energy of the plane. It is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
[tex]E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2[/tex]
E = 133.38 Joules
So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.
The plane’s kinetic energy is about 134 J
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Further explanation
Let's recall the Kinetic Energy formula:
[tex]\boxed {E_k = \frac{1}{2}mv^2 }[/tex]
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:
[tex]v^2 = (v_x)^2 + (v_y)^2[/tex]
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Next, we could calculate the plane's kinetic energy by using following formula:
[tex]E_k = \frac{1}{2} m v^2[/tex]
[tex]E_k = \frac{1}{2} m (v_x^2 + v_y^2)[/tex]
[tex]E_k = \frac{1}{2} m v_x^2 + \frac{1}{2} m v_y^2[/tex]
[tex]E_k = (\frac{1}{2} \times 3.00 \times 5.00^2) + (\frac{1}{2} \times 3.00 \times 8.00^2)[/tex]
[tex]E_k = 133.5 \texttt{ J}[/tex]
[tex]E_k \approx 134 \texttt{ J}[/tex]
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Learn more
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
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Answer details
Grade: High School
Subject: Physics
Chapter: Energy
