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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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An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform
frosja888 [35]

Answer:

6.16 m/s

0.0105 m

Explanation:

Let the ground 0 for potential reference be at where the spring is compress 0.24 m. The the man would jump from a height h = 2.5 + 0.24 = 2.74 m from it. We can apply the law of energy conservation knowing that as the man jumps, his potential energy converts to kinetic energy, then finally to elastic energy:

E_p = E_e

mgh = kx^2/2

where m = 80 kg is the man mass, g = 9.81 m/s2 is the gravitational acceleration, h = 2.74 m is the potential distance he travels, k N/m is the spring constant and x = 0.24 is the distance it compresses

80*9.81*2.74 = k0.24^2/2

2150.352 = 0.0288k

k = 74665 N/m

Similarly at the position where it compresses by 0.12 m, it's 0.24 - 0.12 = 0.12 m far from ground 0.

E_p = E_{e2} + E_k + E_{p2}

mgh = kx_2^2 + mv^2/2 + mgh_2

2150.352 = 74665*0.12^2/2 + 80v^2/2 + 80*9.81*0.12

2150.352 = 537.588 + 40v^2 + 94.176

40v^2 = 1518.588

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When he steps gently, then his gravity force would equal to his spring force

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x_3 = mg/k = 80*9.81/74665 = 0.0105m

7 0
3 years ago
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