a jet plane flying at 550 m/s makes a curve. The radius of the circle in which the plane is flying is 9.0 X 10^3 m. What centripetal acceleration does the plane experience? Express this acceleration relative to g, the acceleration due to gravity.
centacceleration=v^2/r
divide by 9.81 to get it in "g"
centacceleration=v^2/r
divide by 9.81 to get it in "g"
To find the centripetal acceleration experienced by the plane, we can use the formula:
centripetal acceleration = (velocity^2) / radius
Given:
Velocity of the plane (v) = 550 m/s
Radius of the circle (r) = 9.0 × 10^3 m
Substituting these values into the formula, we get:
centripetal acceleration = (550^2) / (9.0 × 10^3)
First, let's square the velocity:
velocity^2 = 550^2 = 302,500 m^2/s^2
Now, divide that by the radius:
centripetal acceleration = 302,500 / (9.0 × 10^3)
To express the acceleration relative to g, the acceleration due to gravity, we need to divide the centripetal acceleration by the acceleration due to gravity (g). The approximate value of g is 9.8 m/s^2.
centripetal acceleration relative to g = (302,500 / (9.0 × 10^3)) / 9.8
To simplify further, we divide the numerator by 9.8:
centripetal acceleration relative to g = (30,769.23) / (9.8)
centripetal acceleration relative to g ≈ 3,141.54
Therefore, the centripetal acceleration experienced by the plane, relative to g, is approximately 3,141.54.