at the bottom of a loop the speed of an airplane is 400 mph. this causes a normal acceleration of 9 g ft/sec^2, determine the radius of the loop

Question

Visualize a clear blue sky as the backdrop of an exciting aerial scenario. Depict an airplane designed in an industrial style, performing a thrilling loop maneuver at great speed, which is approximately 400 mph. Carefully illustrate this high-speed event, ensuring to capture the plane at the bottom of the loop to symbolize the mentioned normal acceleration of 9 g ft/sec^2. Although you cannot directly depict this sensation, evoke its presence through the plane's swift and powerful motion. Arguably the most crucial detail, visualize the entire loop’s radius by following the plane's traversal, all while excluding any literal representation of mathematical figures or formulas, keeping the image free of textual elements.

at the bottom of a loop the speed of an airplane is 400 mph. this causes a normal acceleration of 9 g ft/sec^2, determine the radius of the loop

Answer 1

a(normal)=(v^2)÷r

r=(v^2)/a(normal)

r(9 ft/sec^2)(32.17) = (400miles/hr x 1610 m/miles x 3.281 ft/m x (1 hr)/3600)^2

r=1190 ft

Answer 2

a(norm) = v^2/R

Answer 3

Why did the airplane go to the doctor?

Because it had a "loop-sy" fever!

Now, to determine the radius of the loop, we'll use the centripetal acceleration formula:

a = v^2 / r

Where:
a = acceleration,
v = velocity,
r = radius.

Given that the speed of the airplane at the bottom of the loop is 400 mph, we need to convert it to feet per second (fps):

400 mph * 1.467 ft/sec = 586.8 ft/sec (approx)

Now we plug the values into the equation:

9 g ft/sec^2 = (586.8 ft/sec)^2 / r

Let's also convert the acceleration from "g" to ft/sec^2:

9 g ft/sec^2 * 32.2 ft/sec^2/g (approx) = 289.8 ft/sec^2

Now we can solve for the radius (r):

289.8 ft/sec^2 = (586.8 ft/sec)^2 / r

Cross-multiplying:

r * 289.8 ft/sec^2 = (586.8 ft/sec)^2

Dividing both sides by 289.8 ft/sec^2:

r = (586.8 ft/sec)^2 / 289.8 ft/sec^2

Calculating:

r ≈ 1190 ft

So the radius of the loop is approximately 1190 feet! Keep on looping!

Answer 4

To determine the radius of the loop, we can use the centripetal acceleration formula:

a = v^2 / r

where:
a = centripetal acceleration
v = speed of the airplane
r = radius of the loop

Given:
v = 400 mph = 400 * 1.46667 ft/sec (1 mph = 1.46667 ft/sec)
a = 9 g ft/sec^2 = 9 * 32.174 ft/sec^2 (1 g = 32.174 ft/sec^2)

Plugging in the values into the formula:

9 * 32.174 = (400 * 1.46667)^2 / r

Simplifying:

288.57 = 400^2 * 1.46667^2 / r

r = 400^2 * 1.46667^2 / 288.57

Calculating:

r ≈ 670.8 feet

Therefore, the radius of the loop is approximately 670.8 feet.

Answer 5

To determine the radius of the loop, we can use the centripetal acceleration formula:

a = v^2 / r

where:
a is the centripetal acceleration,
v is the velocity of the airplane, and
r is the radius of the loop.

In this case, we are given that the speed of the airplane at the bottom of the loop is 400 mph, and the normal acceleration is 9g ft/sec^2.

First, let's convert the speed of the airplane from mph to ft/sec:
400 mph * (5280 ft / 1 mile) * (1 hour / 3600 sec) = 586.67 ft/sec (rounded to two decimal places).

The normal acceleration is given as 9g ft/sec^2, where g is the acceleration due to gravity, approximately 32.2 ft/sec^2.

Substituting the values into the centripetal acceleration formula, we have:
9g ft/sec^2 = (586.67 ft/sec)^2 / r

Now we can solve for r by rearranging the equation:
r = (586.67 ft/sec)^2 / (9 * 32.2 ft/sec^2) ≈ 932.70 ft

Therefore, the radius of the loop is approximately 932.70 ft.