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Answer :
Final answer:
The plane decelerates at a rate of 3 m/s².
Explanation:
To calculate the rate at which the plane decelerates, we need to use Newton's second law of motion, which states that force is equal to mass times acceleration. In this case, the force used to bring the plane to a halt is 561,000 Newtons and the mass of the plane is 187,000 kg. Therefore, the acceleration (= force/mass) is 561,000/187,000 = 3 m/s². Therefore, the plane decelerates at a rate of 3 m/s².
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Final answer:
The deceleration rate of the 747 plane is calculated by using the formula Force = Mass x Acceleration. On substituting the values and re-arranging the formula, we find that the plane decelerated at a rate of 3 m/s². Hence, the answer is Option 3.
Explanation:
This question relates to the physical concept of deceleration which explores how the 747 plane slows down on landing. Here we are to calculate the deceleration rate of the plane. Deceleration or negative acceleration can be calculated by the formula:
Force = Mass x Acceleration
In this context, the Force represents the force required to halt the plane, the Mass signifies the mass of the plane and Acceleration points to the deceleration rate of the plane. Given that the force is 561,000 Newtons and the mass being 187,000 kg, we can re-arrange the formula to find the acceleration:
Acceleration = Force / Mass
Therefore, substituting the values, Acceleration is 561,000N / 187,000kg, which calculates to 3 m/s². This answer indicates that the plane decelerated at a rate of 3 m/s², which matches with option 3).
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