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2 years 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]2 years 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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The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v

velikii [3]

Answer:

$\theta = n\pi/2, {\rm where~n~is~an~integer.}$

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

$\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k$

Similarly,

$\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k$

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

$\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)$

So,

$\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)$

At time t = 0, this equation becomes

$0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}$

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3 years ago

S_A_V [24]

The distance in meters she would have moved before she begins to slow down is 11.25 m

<h3>LINEAR MOTION</h3>

A straight line movement is known as linear motion

Given that Ann is driving down a street at 15 m/s. Suddenly a child runs into the street. It takes Ann 0.75 seconds to react and apply the brakes.

To know how many meters will she have moved before she begins to slow down, we need to first list all the given parameters.

speed = 15 m/s

time t = 0.75 s

From definition of speed,

speed = distance / time

Make distance the subject of the formula

distance = speed x time

distance = 15 x 0.75

distance = 11.25m

Therefore, the distance in meters she would have moved before she begins to slow down is 11.25 m

Learn more about Linear motion here: brainly.com/question/13665920

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