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Fittoniya [83]
3 years ago
12

A civil engineer plans to design a curved ramp such that a car may not have to rely on friction to round the curve without skidd

ing. The roadway is tilted toward the inside of the curve. Suppose the designated speed is v and the radius of the curve is R. At what angle should the curve be tilted?

Physics
1 answer:
Delicious77 [7]3 years ago
8 0

Answer:\theta =\tan ^{-1}(\frac{v^2}{gR})

Explanation:

let \theta be the inclination at which curve is tilted

v is the speed of car and R is the radius of curve

m is the mass of car

Suppose R is the reaction offered by road to car

Resolving R in x and y direction we get

R\cos \theta will balance weight and R\sin \theta will provide the necessary centripetal Force

thus R\sin \theta =\frac{mv^2}{R}  ------------1

R\cos \theta =m g  ----------------2

Divide 1 & 2 we get

\frac{R\sin \theta }{R\cos \theta }=\frac{mv^2}{mgR}

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

\theta =\tan ^{-1}(\frac{v^2}{gR})                                      

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The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

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When acceleration is constant, the average velocity is given by

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Take the car's starting position to be at t_0=0\,\mathrm s. Then

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v=v_0+at

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Earth exerts a 100 N gravitational force on a metal box. What is the magnitude of the gravitational force the metal box exerts o
jeka94

We have that the magnitude of the gravitational force  is mathematically given as

f=6.377N

<h3>Force</h3>

Question Parameters:

Earth exerts a 100 N gravitational force on a metal box.

(Mass of the earth is 6e24 kg and radius of the earth is 6.4e6m.)

Generally the equation for the Gravitational mForce  is mathematically given as

F=\frac{GMm}{r^2}\\\\Therefore\\\\F=\frac{100/9.8* 6e116e24}{6.4e6^2}

f=6.377N

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