Answer:
-0.8 m/s²
Explanation:
Acceleration is the slope of a velocity vs. time graph.
a = Δv / Δt
a = (0 m/s − 12 m/s) / (15 s − 0 s)
a = -0.8 m/s²
As an airplane moves through the air, its wings cause changes in the
speed and pressure of the air moving past them. These changes result in
the upward force called lift.
The Bernoulli principle states that an increase in the speed of a fluid
occurs simultaneously with a decrease in the pressure exerted by the
fluid.
A wing is shaped and tilted so the air moving over it moves faster than
the air moving under it. As air speeds up, its pressure goes down. So
the faster-moving air above exerts less pressure on the wing than the
slower-moving air below. The result is an upward push on the wing—lift!
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,



So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
40 N/m
Explanation:
F = -kx (This is the Hooke's Law equation)
F is the force the spring exerts = 8 N
-k = spring constant
x = displacement (The distance stretched past it's natural length) = 20cm
x needs to be in meters, and 20 cm is = to 0.2 meters
Finally:
8N = -k (0.2m)
-k = 8N / 0.2 m
k = -40 N/m
Answer:
Maximum acceleration in the simple harmonic motion will be
Explanation:
We have given amplitude of simple harmonic motion is A = 0.43 m
Time period of the oscillation is T = 3.9 sec
We have to find the maximum acceleration
For this we have to find the angular frequency
Angular frequency will be equal to 
Maximum acceleration is given by 
So maximum acceleration in the simple harmonic motion will be