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Yakvenalex [24]
3 years ago
8

A 7780 ‑kg car is travelling at 30.7 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 404 m al

ong the exit ramp, the car's speed is 13.3 m/s, and it is h = 13.1 m above the freeway. What is the magnitude of the average drag force F drag exerted on the car?
Physics
1 answer:
vredina [299]3 years ago
6 0

Answer:

4899.47 N

Explanation:

mass of the car = 7780 kg, u, initial speed =  30.7 m/s, v, final speed = 13.3 m/s

d = 404 m and h = 13.1 m

initial kinetic energy of  the car = potential energy attained by the car climbing the ramp + the work done by friction against the motion + final kinetic energy

1/2 mu² = mgh + (F × d) + 1/2mv²

3666286.1 J = 998796.4 J + ( 404 F) + 688102.1 J

404 F = 3666286.1 J - 998796.4 J - 688102.1 J

F = 1979387.6 / 404 = 4899.47 N

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A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0
otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

             F_{r} = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0
3 years ago
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Is there a frame of reference one can go into that seems to eliminate gravity as Newton described it?
andriy [413]

Answer:

Yes such a frame exists: a free-fall (free-float frame) frame. This frame of reference is subject only to gravity and no forces such as electromagnetic forces or nuclear forces.

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3 years ago
Hey people! I sorta need help. :))
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Answer:

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A machine can never be 100% efficient because some work is always lost due to .
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Friction
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Eating an orange enables a person to perform about 3.5x10^4 J of work. To what height does eating and orange enable a 55kg woman
crimeas [40]
Joules is a unit for work which may decomposed into N.m. Work is a quantity which is a product of force (in this case, the woman's weight) and the distance she has traveled. 
 
                                    W = F x d      ;    d = W / F

Substituting the given, 
  
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Thus, the woman can climb up to 64.94 meters. 

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