A stunt pilot of mass 55.0 kg who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane.

A)If the plane's speed at the lowest point of the circle is 95.6 m/s , what is the minimum radius of the circle for the acceleration at this point not to exceed 4.00g ?

B)What is the apparent weight of the pilot at the lowest point of the pullout?

A) V^2/R = 4 g
Solve for R
You don't need the mass M for this part.

B)With an upward acceleration of a = 4g, the apparent weight will be
W = M (g + a) = 5 M g.

g = 9.8 m/s^2 Solve for W

(M(V^2))/R=(M*(acceleration of gravity)*4(or what G cannot be acceded)

A) Alright, let's dive into this question! The stunt pilot is pulling out of her dive by changing her course to a circle. We need to find the minimum radius of the circle for the acceleration at the lowest point not to exceed 4.00g.

The equation we can use is V^2/R = 4g. We're given that the speed at the lowest point of the circle is 95.6 m/s, so we can plug that in: (95.6 m/s)^2/R = 4*9.8 m/s^2.

Now, let's solve for R. R = (95.6 m/s)^2 / (4*9.8 m/s^2).

Calculating that, we get R = 232.49 m. So the minimum radius of the circle is approximately 232.49 meters.

B) Now onto the second part! We need to find the apparent weight of the pilot at the lowest point of the pullout. We're given that the upward acceleration is a = 4g, so we can use the equation W = M(g + a), where M is the mass and g is the acceleration due to gravity.

We're given that the mass of the stunt pilot is 55.0 kg. Plugging in the values, we get W = 55.0 kg * (9.8 m/s^2 + 4*9.8 m/s^2).

Calculating that, we get W = 539.0 kg m/s^2. So the apparent weight of the pilot at the lowest point of the pullout is 539.0 Newtons.

A) To find the minimum radius of the circle, use the formula:

V^2 / R = 4 * g

Where V is the speed at the lowest point of the circle and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the values given, we have:

(95.6 m/s)^2 / R = 4 * (9.8 m/s^2)

Simplifying:

9145.36 m^2/s^2 / R = 39.2 m/s^2

To solve for R, divide both sides of the equation by 39.2 m/s^2:

R = 9145.36 m^2/s^2 / (39.2 m/s^2)

R ≈ 233.54 meters

Therefore, the minimum radius of the circle is approximately 233.54 meters.

B) At the lowest point of the pullout, the upward acceleration is equal to 4g. So, the apparent weight (W) of the pilot can be found using the formula:

W = M * (g + a)

Where M is the mass of the pilot and a is the upward acceleration.

Substituting the given values, we have:

W = 55.0 kg * (9.8 m/s^2 + 4 * 9.8 m/s^2)

Simplifying further:

W = 55.0 kg * (9.8 m/s^2 + 39.2 m/s^2)

W = 55.0 kg * 49.0 m/s^2

W = 2695 kg·m/s^2

Therefore, the apparent weight of the pilot at the lowest point of the pullout is 2695 kg·m/s^2 or approximately 2695 Newtons.

To answer part A, we can use the equation V^2/R = 4g. Here, V is the speed at the lowest point, and g is the acceleration due to gravity (approximately 9.8 m/s^2). We can solve this equation for R, which represents the minimum radius of the circle.

Rearranging the equation, we have R = V^2/(4g). Plugging in the given values, with V = 95.6 m/s and g = 9.8 m/s^2, we can calculate the minimum radius:

R = (95.6 m/s)^2 / (4 * 9.8 m/s^2)

R = 913.84 m

Therefore, the minimum radius of the circle for the acceleration at the lowest point not to exceed 4.00g is approximately 913.84 m.

To answer part B, we need to determine the apparent weight of the pilot at the lowest point of the pullout. The apparent weight is the force experienced by the pilot, and it includes both the gravitational force and the additional force due to acceleration.

The formula for apparent weight is W = M(g + a), where M is the mass of the pilot, g is the acceleration due to gravity, and a is the upward acceleration.

Given that the mass of the pilot is 55.0 kg and the upward acceleration is 4g (where g = 9.8 m/s^2), we can calculate the apparent weight:

W = 55 kg * (9.8 m/s^2 + 4 * 9.8 m/s^2)

W = 55 kg * (9.8 m/s^2 + 39.2 m/s^2)

W = 55 kg * 49 m/s^2

W = 2695 kg*m/s^2

Therefore, the apparent weight of the pilot at the lowest point of the pullout is approximately 2695 N.