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gtnhenbr [62]
2 years ago
5

The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c

ross-sectional area 3 air density 4 the falling time 5 terminal velocity depends on all of the 6 given parameters
Physics
1 answer:
ahrayia [7]2 years ago
6 0
<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:  

1. The force of air friction, also called<em> </em><u><em>"drag force"</em></u> D:  

D={C}_{d}\frac{\rho V^{2} }{2}A  (1)

Where:  

C_ {d} is the drag coefficient  

\rho is the density  of the fluid (air for example)

V is the velocity  

A is the transversal area of the object

So, this force is proportional to the transversal area of ​​the falling element and to the square of the velocity.  

2. Its <u>weight </u>due to the gravity force W:  

W=m.g

(2)

Where:  

m is the mass of the object

g is the acceleration due gravity  

So, at the moment <u>when the drag force equals the gravity force, the object will have its terminal velocity:</u>

D=W (3)

{C}_{d}\frac{\rho V^{2} }{2}A=m.g  (4)

V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}  (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

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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
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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

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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.

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