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sergejj [24]
3 years ago
7

A mechanic jacks up the front of a car to an angle of 8.0° with the horizontal in order to change the front tires. the car is 3.

05 m long and has a mass of 1130 kg. gravitational force acts at the center of mass, which is located 1.12 m from the front end. the rear wheels are 0.40 m from the back end. calculate the magnitude of the torque exerted by the jack.
Physics
1 answer:
SOVA2 [1]3 years ago
7 0
<span>First draw a free-body diagram. Torque T = Force F x Distance d where force is the component of gravitational force g and d is the lever arm distance to the pivot point. Since the pivot point is at the back tire we subtract that from the length of the car resulting in d = 1.12 - 0.40 = 0.72 meters = d. We are interested in the perpendicular component of the force exerted on the car jack so use sin 8 degrees then T=1130 kg x 9.81 m/s^2 x sin(8 degrees) x0.72 m = 1,110.80 Newton-meters</span>
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<h3>Answer:</h3>

117.6 Joules

<h3>Explanation:</h3>

<u>We are given;</u>

  • Force of the dog is 24 N
  • Distance upward is 4.9 m

We are required to calculate the work done

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In this case;

Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

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viktelen [127]

Answer:

1.7 m

Explanation:

v_x = Velocity of ball in x direction = 4.47 m/s

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t = Time taken

s_y = Vertical displacement = 0.7 m

s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}

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s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}

The passenger should throw the ball 1.7 m in front of the bucket.

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