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vovangra [49]
3 years ago
13

A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for

the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g Grade Summary Deductions Potential 12% 88% 789 HOME Submissions Attempts remaining: ] 90 per attempt) detailed view 4% 4% 4% BACKSPACE | DELI CLEAR Submit Feedback I give up! Hints: 0 for a 0% deduction. Hints Feedback: 5% deduction per feedback. ー 50% Part (b) what is the angle of banking of the highway? Give your answer in degrees e-66.84 X Attempts Remain
Physics
1 answer:
Monica [59]3 years ago
3 0

Question

A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g

Part (b) what is the angle of banking of the highway? Give your answer in degrees

Answer:

a. Equation of Tangent

tan(θ) = v²/rg

b. Angle of the banking highway

θ = 0.087°

Explanation:

Given

Radius of the curve = r = 330m

Acceleration of gravity = g = 9.8m/s²

Velocity = v = 8km/h = 8 * 1000/3600

v = 2.22 m/s

a . Write an equation for the tangent of the highway's angle of banking

The Angle is calculated by

tan(θ) = v²/rg

θ = tan-1(v²/rg)

b.

Part (b) what is the angle of banking of the highway? Give your answer in degrees

θ = tan-1(v²/rg)

Substituting the values of v,g and r

θ = tan-1(2.22²/(330 * 9.8)

θ = tan-1(0.001523933209647)

θ = 0.087314873580116°

θ = 0.087°

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Explanation:

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Sum of forces in the vertical direction:

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T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N.  Therefore:

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3 years ago
Consider a boat heading due east at 15 miles/hour. The water's current is moving at 7.1 miles/hour at 45º south of east. Drag ve
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Assuming Earth's gravity, the formula for the flight of the particle is: 

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A uniform electric field of magnitude 7.0 ✕ 104 N/C passes through the plane of a square sheet with sides 5.0 m long. Calculate
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Answer:

1.52*10^6 Nm^2/C

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Given that:

Electrical field E = 7.0 * 10^{-4}N/C

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slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa
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Answer:

Maximum height reached by the rocket is

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total time of the motion of rocket is given as

T = 16.44 s

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a = 2 m/s^2

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

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3 years ago
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