A 1,492.3-kg airplane travels down the runway. Each of its four engines provides a force of
1 answer:
The acceleration of the air plane is
<u>Explanation:</u>
Given:
The mass of the air plane = 1492.3 kg
Force of each four engine = 1447.5 N
So, the total force of four engines can be calculated as = 4(1447.5) = 5790 N
The force that acts on the object is equal to the product of mass (m) and its acceleration. It can express by the below formula,
The above equation can be written as below to find acceleration,
Now. Substitute the given values, we get,
You might be interested in
Refer to the diagram shown below.
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).