Question

A car is initially driving at 30 m/s. It hits a large pothole, after which it is traveling in the same direction but at 25 m/s. Draw vectors representing the initial velocity final velocity, and (hange in velocity which describe the equation : + Δ¡. 2. The SI nit for weight is the Newton (not the kilogram!). Express the weight of a 2.8 lbs. watermelon in Newtons. 3. On Mars gravitational acceleration is aMars 3.71 m/s2 or aMars 0.37g, where g is the gravitational acceleration of on object on Earth. If a future astronaut drops a rock from a height of 1.9 m how long will it take to hit the ground on Mars? Compare that number to how long the same rock would take to fall on Earth.

Answer 1

Answer:

The time of Mars is 1.65 times larger on Mars than on Earth

Explanation:

The equation that describes the system is the final speed is equal to the speed minos the speed lost by the collision with the porhole

       Vf = Vo - V pothole

B) let's transform the weight of free groin system and N international system

      1 N = 0.2248 lb

      2.8 lbs (1N / 0.2248lbs) = 12.5 N

c) Kinematic equations are the same in all inertial systems, Mars and Earth, so we can use the height equation, with zero initial velocity

                   

        Y = Vo t - ½ g t²

        Y = - ½ g t²

        t = √ 2Y / g

     

Mars

         gm = 0.37g

         gm = 0.37 9.8

         gm = 3,626 m / s²

         t = √( 2 1.9 / 3.626 )

         t = 1.02 s

Earth

         t = √( 2 1.9 / 9.8)

         t = 0.62 s

To make the comparison of time we are the relationship between the two

         tm / te = 1.02 / 0.62

         tm / te = 1.65

The time of Mars is 1.65 times larger on Mars than on Earth

Answer 2

Answer:

1. The arrows show opposite motion

2 127 N

3. 1 s

Explanation:

1. The arrows will show opposite directions

2. 2.8 lbs = 12.7006 kg

       1 kg = 10 N

Therefore, 12.7006 kg = 12.7006 × 10

                                     = 127 N

3. let g = 3.71 m/s²

         m = 0.37

          h = 1.9

Kinetic energy = mgh

                        = 0.37 × 3.71 × 1.9

                        = 2.6 J

The potential energy will be converted to kinetic energy so, $E_{p} = E_{k}$

Ek = 1/2 mv$^{2}$

velocity  = 3.74 m/s

distance, v$^{2} = u^{2}+ 2gs$

           v = u + at

      3.74 = 0 + 3.71t

          t = 1s