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aev [14]
3 years ago
11

A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if

cars do not slide down the banked road, what is the correct angle of banking for the road
Physics
1 answer:
KengaRu [80]3 years ago
4 0

Answer:

The value of  the correct angle of banking for the road is \theta = 67.76 °

Explanation:

Given data

Velocity (v) = 60 \frac{m}{s}

Radius = 150 m

The velocity of the car in this case is given by

v = \sqrt{r g \tan \theta}

v^{2} = r g \tan \theta

\tan \theta = \frac{v^{2} }{rg}

Put all the values in above formula we get

\tan \theta = \frac{60^{2} }{(150)(9.81)}

\tan \theta = 2.446

\theta = 67.76 °

Therefore the value of  the correct angle of banking for the road is \theta = 67.76 °

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Answer:

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3 years ago
Select the correct answer from each drop-down menu. A device that uses electricity and magnetism to create motion is called a __
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<h3><u>Answer;</u></h3>

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<h3><u>Explanation</u>;</h3>
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4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive
zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

|

|

B1 at 20km/h

|

|

V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0
3 years ago
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