True or false

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

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

Help please explain!!!

zalisa [80]

Newton's first law of motion. An object in motion tends to stay in motion. Since there is no friction in space to slow it down it just keeps going

5 0

3 years ago

The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w

anzhelika [568]

Answer:

594.8 Hz

Explanation:

Parameters given:

Speed of sound, v = 345 m/s

Wavelength = 58 cm = 0.58 m

Speed of a wave is given as:

Speed = wavelength * frequency

Therefore:

Frequency = Speed/Wavelength

Frequency = 345/0.58

Frequency = 594.8 Hz

8 0

3 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

(c) 2.86s

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

The answer for (a) is 1.6m

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }