Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

4 years ago

what is the wavelength of radio waves transmitted by an FM station at 90MHz where 1M=10^6, and speed of radiowave is 3*10^8m/s

Igoryamba

Answer:

λ = 3.33 m

Explanation:

Given:

Frequency = f = 9 × 10⁷ Hz

Speed = v = 3 × 10⁸ m/s

Required:

Wavelength = λ = ?

Formula:

v = fλ

Solution:

Putting the givens in the formula

v = fλ

λ = $\frac{v}{f}$

λ = $\frac{3*10^8}{9*10^7}$

λ = 0.33 × 10¹

λ = 3.33 m

5 0

3 years ago