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Makovka662 [10]

2 years ago

3

Wings on race cars push them into the track. The increased normal force makes large friction forces possible. At one Formula One

racetrack, cars turn around a half-circle with diameter 190m at 68m/s .
Part A
For a 610 kg vehicle, the approximate minimum static friction force to complete this turn is. Select best answer
a.15000N
b.6000N
c.30000N
d.18000N
e.24000N

1 answer:

pochemuha2 years ago

7 0

Answer:

a) F = 15000 N

Explanation:

At flat race track the friction force between the tyre and the road will provide the required frictional force in the form of centripetal force

So we will have

$F_f = \frac{mv^2}{R}$

so here we know that

m = 610 kg

v = 68 m/s

R = 190 m

now we have

$F = \frac{610 (68^2)}{190}$

$F = 14845.5 N$

this is nearly

$F = 15000 N$

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You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that

oksano4ka [1.4K]

As we know that reaction time will be

$t = 0.50 s$

so the distance moved by car in reaction time

$d = vt$

$d = 20 \times 0.50$

$d = 10 m$

now the distance remain after that from intersection point is given by

$d = 110 - 10 = 100 m$

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

$v_f = 0$

$v_i = 20 m/s$

now from kinematics equation we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 20^2 = 2(a)100$

$a = \frac{-400}{200} = -2 m/s^2$

so the acceleration required by brakes is -2 m/s/s

Now total time taken to stop the car after applying brakes will be given as

$v_f - v_i = at$

$0 - 20 = -2 (t)$

$t = 10 s$

total time to stop the car is given as

$T = 10 s + 0.5 s = 10.5 s$

3 0

2 years ago

what is the acceleration of a bowling ball that starts at rest and moves 300m down the gutter in 22.4 sec

exis [7]

<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>

7 0

2 years ago

A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass

Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

$L=2\pi R$

where R is the radius of the wheel.

The period of revolution is:

$T=120 s$

Therefore, the tangential speed of the book is:

$v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R$

8 0

2 years ago

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

8 0

2 years ago

A cylinder is sliced in half along its diagonal. Determine the location of the center of mass and the inertia properties relativ

Black_prince [1.1K]

Answer:

See the explanation below

Explanation:

To better understand this problem, a cylinder sketch is attached before and after the cut, we see that after the cut, the shape of this resembles that of a right triangle.

We can find, the centroid in the xy plane, knowing that the centroid for a triangle is located a third of its base.

In the z axis there is no displacement of the centroid.

3 0

2 years ago