Ask question

Ask question

All categories

3 years ago

7

We will look at an example of a banked curve. In this case the normal force provides the needed radial acceleration. This allows

the vehicle to negotiate the curve without having to rely solely on friction. An engineer proposes to rebuild the curve, banking it so that, at a certain speed v, no friction at all is needed for the car to make the curve. At what angle β should the curve be banked if v = 25 m/s, (56 mi/h) and R=250m?

1 answer:

Lelu [443]3 years ago

7 0

Answer:

The bank angle $\beta\\$ = 14.30 degrees

Explanation:

In engineering, banking a curve reduces the centripetal force on the cars as they turn round the bend. this helps prevent them from sliding off the road.

The bank angle can be calculated using the formula:

$tan \beta =\frac{v^{2}}{gR}$

where v = velocity of the car = 25 m/s

R = radius of the bank = 250 m

g = acceleration due to gravity = $9.81 m/s^{2}$

$tan \beta =\frac{25^{2}}{9.81 \times 250}$

$tan\beta =0.25484$

$\beta = tan^{-1} (0.25484)= 14.30 degrees$

There fore, the bank angle is approximately 14 degrees

You might be interested in

A balanced force has a net force and therefore causes the object to accelerate.

Aleksandr [31]

False it doesn't accelerate

8 0

3 years ago

Read 2 more answers

Hole filling fasteners (for example, MS20470 rivets) should not be used in composite structures primarily because of the?

DIA [1.3K]

Answer:

We can cause delamination.

Explanation:

The reason why is because the probability of causing delamination increase considerably when we use Hole-filling fasteners. If we use a typical rivet, these tends to expands in order to fill the hole.

If we analyze the force applied by the expanded rod will cause that the matrial will be deteriorated and will cause that the material to delaminate around the edges of the hole and we can cause possible control and no protection to the material.

7 0

3 years ago

Which is an example of a social network?

GarryVolchara [31]

Answer:

d

Explanation:

4 0

3 years ago

Read 2 more answers

At which point is the object not moving?

Alenkinab [10]

It’s none of those because it’s moving at a constant rate

8 0

3 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

Elis [28]

Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

$F_v=80062.47\times 10^9$ attractive toward +y axis is the net vertical force

Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

<u>Now the distance between the charge at at origin and the second charge:</u>

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

<u>Now the distance between the charge at at origin and the third charge:</u>

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

<u>Now the force due to second charge:</u>

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

<u>Now the force due to third charge:</u>

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

<u>Now the its horizontal component:</u>

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

<u>Now the its vertical component:</u>

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

3 0

3 years ago