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2 years ago

11

5 What is the maximum speed at which a car round a curve of 25m radius on a level road if the coefficient of static friction bet

ween the tires and the road is 0.80?

1 answer:

pshichka [43]2 years ago

8 0

Hi there!

On a level road:

∑F = Ff (Force due to friction)

The net force is the centripetal force, so:

mv²/r = Ff

Rewrite the force due to friction:

mv²/r = μmg

Cancel out the mass:

v²/r = μg

Solve for v:

v = √rμg

v = √(25)(9.81)(0.8) = 14.01 m/s

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2 years ago

A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

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2 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

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The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$ ,

$s_1=0+\frac{1}{2}(1.62)(20)^2=324 m$

We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$

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$v_1=32.4 m/s$ is the initial velocity at the beginning of the 2nd phase

$v_2=0$ is the final velocity (the car comes to a stop)

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Substituting,

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So, the total displacement of the car is

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Learn more about accelerated motion:

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brainly.com/question/2562700

#LearnwithBrainly

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