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Alona [7]
3 years ago
15

3/50 The standard test to determine the maximum lateral acceleration of a car is to drive it around a 200-ft-diameter circle pai

nted on a level asphalt surface. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. If this maximum speed is 35 mi/hr for a 3000-lb car, determine its lateral acceleration capability an in g's and compute the magnitude F of the total friction force exerted by the pavement on the car tires.
Physics
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

(a): Its lateral acceleration is ac= 8.02 m/s² = 0.81 * g

(b): The magnitude F of the friction force exerted by the paviment is F= 1091.28 N.

Explanation:

r= 100 ft = 30.48 m

Vt= 35 miles/hr = 15.64 m/s

m= 3000 lb = 136.07 kg

w= Vt/r

w= 0.5131 rad/s

ac= w² * r

ac= 8.02 m/s² = 0.818 * g  (a)

F= m * ac

F= 1091.28 N (b)

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5 0
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block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8
dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: \sqrt{800} \,\,\,N

Explanation:

First try to write all forces in  vector component form:

The force F1 acting at 45 degrees would have multiplication factors of \frac{\sqrt{2} }{2} on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is    F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

y-component of F1 is    F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

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x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

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x-component of F3 = 0 N

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x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

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x-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

7 0
3 years ago
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Q = 169 BTU

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4 0
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Help please ASAP !!!
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