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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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3 years ago
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A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9
Andreyy89

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

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Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin
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Answer

(C).

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Step by step Solution

The work done by a force is defined as the product of the force and the distance traveled in the direction of motion.

The first answer "Only the component of the force perpendicular to the motion is used to calculate the work" is wrong because, the force perpendicular to motion does no work.

The second choice "If the force acts in the same direction as the motion, then no work is done" is wrong because the work in the direction of the force is W=F\times d.

Fourth answer "A force at a right angle to the motion requires the use of the sine of the angle" is wrong because the sin(90)=0 meaning that there is no work done in the direction perpendicular to the motion.

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