Question

16.an engineer is designing a turn on a highway. if the speed limit on the turn is 30 m/s, and the engineer wants the maximum centripetal acceleration to be 15 m/s², what should the radius of the turn be? a. 20 meters b. 45 meters c. 60 meters d. 55 meters 17.two objects, a and b, are floating in deep space. object a has a mass that is double that of object b. if the gravitational force of a acting on b is equal to \f\, what is the gravitational force of b acting on a equal to? a. less than f b. greater than f c. equal to f d. zero 18.a 400 kg car drives around a curve on a road. if the radius of the curve is 40 m, and the coefficient of friction between the car and the road is 0.90, what is the fastest speed the car can travel around the curve? a. 19.62 m/s b. 18.78 m/s c. 12.44 m/s d. 13.57 m/s 19.as part of their training, astronauts are placed in machines that spin very quickly, causing them to have a large centripetal acceleration. what are some possible changes that would increase the centripetal acceleration felt by the astronauts? a. increasing the radius of the machines circular path b. decreasing the velocity of the capsule as it rotates c. decreasing the radius of the machines circular path d. increasing the mass of the capsule 20.mars has a mass of approximately 6.39 x 10²³kg and a radius of 3.3895 x 10⁶ m. what is the gravitational field strength experienced by a 75 - kg object sitting on the surface of the red planet? a. 3.71 n/kg b. 2.39 n/kg c. 278 n/kg d. 139 n/kg

Explanation:

Step1: Recall centripetal acceleration formula

The formula for centripetal acceleration is \(a = \frac{v^{2}}{r}\), where \(a\) is centripetal acceleration, \(v\) is velocity, and \(r\) is radius. We need to solve for \(r\), so we can rewrite the formula as \(r=\frac{v^{2}}{a}\).

Step2: Substitute the given values

Given \(v = 30\ m/s\) and \(a=15\ m/s^{2}\). Substitute these values into the formula: \(r=\frac{30^{2}}{15}\).
First, calculate \(30^{2}=900\). Then, \(\frac{900}{15}=60\).

According to Newton's third law of motion (action - reaction law), when object A exerts a gravitational force \(F\) on object B, object B exerts an equal and opposite gravitational force on object A. The gravitational force between two objects is given by \(F = G\frac{m_{A}m_{B}}{r^{2}}\). The magnitude of the force that A exerts on B is the same as the magnitude of the force that B exerts on A, regardless of their masses.

Step1: Determine the centripetal force

The centripetal force \(F_{c}\) is provided by the frictional force \(F_{f}\). The formula for centripetal force is \(F_{c}=m\frac{v^{2}}{r}\), and the formula for frictional force is \(F_{f}=\mu N\). On a level road, \(N = mg\), so \(F_{f}=\mu mg\).
Since \(F_{c}=F_{f}\), we have \(m\frac{v^{2}}{r}=\mu mg\). The mass \(m\) cancels out (because \(m
eq0\)), and we get \(v = \sqrt{\mu gr}\).

Step2: Substitute the given values

Given \(\mu = 0.90\), \(g = 9.8\ m/s^{2}\), and \(r = 40\ m\).
Substitute into the formula: \(v=\sqrt{0.90\times9.8\times40}\).
First, calculate \(0.90\times9.8\times40=(0.90\times40)\times9.8 = 36\times9.8=352.8\). Then, \(\sqrt{352.8}\approx18.78\ m/s\).

The formula for centripetal acceleration is \(a=\frac{v^{2}}{r}\).

• For option a: If we increase the radius \(r\) (while \(v\) is constant), from \(a=\frac{v^{2}}{r}\), \(a\) will decrease.
• For option b: If we decrease the velocity \(v\) (while \(r\) is constant), from \(a = \frac{v^{2}}{r}\), \(a\) will decrease.
• For option c: If we decrease the radius \(r\) (while \(v\) is constant), from \(a=\frac{v^{2}}{r}\), \(a\) will increase.
• For option d: The formula \(a=\frac{v^{2}}{r}\) shows that centripetal acceleration is independent of mass.

Answer:

c. 60 meters