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1 year ago

The radius of curvature of a highway exit is r = 93.5 m. The surface of the exit road is horizontal, not banked. (See figure.)

Physics

1 answer:

trapecia [35]1 year ago

8 0

Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences

• net vertical force

F [normal] - F [weight] = 0

• net horizontal force

F [friction] = ma = mv²/r

where v is the tangential speed of the car.

It follows that

F [normal] = F [weight] = mg

and when static friction is maximized at the car's maximum speed,

F [friction] = µ F[normal] = 0.402 mg

Solve for v :

0.402 mg = mv²/r ⇒ v = √(0.402 g (93.5 m)) ≈ 19.2 m/s

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Yo sup??

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2 years ago

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2 years ago

a pool ball leaves a 0.60-meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. what is the

inysia [295]

Answer:

0.84 m

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Given in the y direction:

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a = 9.8 m/s²

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2 years ago

You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b

Radda [10]

Answer:

$\theta=34 \textdegree$

Explanation:

From the question we are told that:

Mass $m=55kg$

Angle $\theta =28.0$

Coefficient of static friction $\alpha =0.680$

Generally, the equation for Newtons second Law is mathematically given by

For

$\sum_y=0$

$N=mgcos \theta$

for

$\sum_x=0$

$F_{s}=mgsin\theta$

Where

$F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta$

$F_{s}=0.68*55*9.8*cos 28$

$F_{s}=323.62N$

Therefore

$\alpha mgcos \theta=mg sin \theta$

$\theta=tan^{-1}(0.68)$

$\theta=34 \textdegree$

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1 year ago

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Anon25 [30]

Answer:

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