Given: Takeoff Speed (u) = 28 m/s
Final velocity (v) = 0.0 m/s
Acceleration (a) = 1.9 m/s²
To calculate: The minimum length of runway (s) =?
Solution: Apply 3rd kinematic equation of motion
v² = u² - 2as
Or, s = ( u² - v²) / 2 a
Or, s = ( 28² - 0²) / 2 × 1.9 m
Or, s = 206.3 m
Hence, the required minimum length of the runway will be 206.3 m
To determine the minimum length of the runway a Cessna 150 airplane would need to take off, you can use a kinematic equation. Plugging in a final velocity of 28 m/s, initial velocity of 0 m/s (since the plane starts from rest), and acceleration of 1.9 m/s/s gives an answer of approximately 207.11 meters.
The subject of your question is related to physics, specifically, kinematics, which studies the motion of objects. In your case, a Cessna 150 airplane needs to accelerate from rest to a speed of 28 m/s for takeoff, with an average acceleration of 1.9 m/s/s, and you're trying to find out the minimum length of the runway required. For this, we can use a kinematic equation.
The equation we can use is the following: v^2 = u^2 + 2as, where v is the final velocity (28 m/s), u is the initial velocity (0 m/s since the plane starts from rest), a is the acceleration (1.9 m/s/s), and s is the distance we want to find out.
When you plug in the known values into the equation, you will get: (28)^2 = (0)^2 + 2 * 1.9 * s. After rearranging the equation, you will have s = [(28)^2 - (0)^2] / 2*1.9 = 207.11 m. Therefore, the minimum length of the runway required for the Cessna 150 airplane to take off is approximately 207.11 meters, assuming constant acceleration and no other factors interfering.
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B. 1.7X10^3 N
C. 8.1X10^3 N
D. 9.0X10^3 N
Answer:
Magnitude of force is
Explanation:
It is given that,
Acceleration of car,
Mass of car,
We have to find magnitude of force. It is given by using Newton's second law of motion which states that force is given by the product of mass and acceleration :
F = m a
or
Hence, the correct option is (c)