Determine the magnitude of the acceleration for the speeding up phase<span>. Express your answer to two significant figures and include the appropriate units. I've tried 2.8 m/s^2, and 1.1 m/s^2, but they are both wrong. </span>Determine the magnitude of the acceleration<span> for the slowing down </span>phase<span>.</span>
Answer:
Correct answer: F = 214.56 N
Explanation:
Given:
V₀ = 0 m/s initial speed
V = 44.7 m/s speed after t = 29 seconds
m = 96 kg
F = ? horizontal force
Taking off an airplane is a uniformly accelerated motion to which the formula applies:
V = V₀ + a t and V₀ = 0 m/s
V = a t ⇒ a = V / t = 44.7 / 20 = 2.235 m/s²
a = 2.235 m/s²
the horizontal force is calculated using the formula:
F = m · a = 96 · 2.235 = 214.56 N
F = 214.56 N
God is with you!!!
4 meters per second. hope this helped
The process is called subduction
The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
<h3>
Magnitude of required force to stop the weight</h3>
The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;
F = ma
F = m(v/t)
F = (mv)/t
F = (5 x 4.5)/0.333
F = 67.6 N
Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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