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Phantasy [73]
3 years ago
15

A car has negative acceleration. What information can you infer from this? (4 points) a The car is speeding up and changing its

direction. b The car is speeding up in the same direction. c The car is changing its direction. d The car is slowing down.
Physics
1 answer:
mario62 [17]3 years ago
5 0

Answer:

D. The car is slowing down.

Explanation:

When a car slows down, the acceleration is in the opposite direction as the velocity. With that being said, the car would have negative acceleration.

hope this helps!! p.s. i really need brainliest :)

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4 years ago
A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo
Orlov [11]

Explanation:

(a)   Net force acting on the block is as follows.

           F_{net} = -mg Sin (\theta)

or,           ma = -mg Sin (\theta)[/tex]

                 a = -g Sin (\theta)

                    = -9.8 \times Sin (20^{o})

                    = -3.35 m/s^{2}

According to the kinematic equation of motion,

             v^{2} - v^{2}_{o} = 2as

Distance traveled by the block before stopping is as follows.

     s = \frac{v^{2} - v^{2}_{o}}{2a}

        = \frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}

        = 21.5 m

According to the kinematic equation of motion,

               v = v_{o} + at

      0 = 12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t

   t_{1} = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b)    When the block is moving down the inline then net force acting on the block is as follows.

                 F_{net} = -mg Sin (\theta)

                ma = mg Sin (\theta)

                    a = g Sin (\theta)

                       = 9.8 m/s^{2} \times Sin (20^{o})

                       = 3.35 m/s^{2}

Kinematics equation of the motion is as follows.

                   s = v_{o}t + \frac{1}{2}at^{2}

      21.5 m = 0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}

     t_{2} = \sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}

             = 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

               t = t_{1} + t_{2}

                 = 7.16 sec + 3.58 sec

                 = 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

3 0
4 years ago
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