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Elenna [48]
3 years ago
9

A rugby players runs 10.0 m straight down the playing field in 2.00 s. The player is then hit and pushed 5.00 m straight backwar

d in 1.50 s. The player breaks the tackle and runs straight forward another 17.2 m in 4.10 s. Calculate the velocity for the players entire motion.
Physics
1 answer:
s344n2d4d5 [400]3 years ago
3 0

Answer:

2.92 m/s

Explanation:

For the entire motion, the average velocity (V) can be calculated thus;

V = \frac{displacement}{time}

Total displacement of the player = 10.0 + (-5) + 17.2

        = 22.2 m

Total time = 2.00 + 1.50 + 4.10

         = 7.60 s

V = 22.2/7.60

       = 2.92 m/s

Hence, the velocity of the players for the entire motion is 2.92 m/s.

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A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.
kherson [118]

Answer:

a) k = 120 N / m

, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

, e)  a = 5.71 m / s²

, f)   x = 0.200 m

, g)  Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

         F = k x

         k = F / x

          k = 24.0 / 0.200

          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

        w = √ k / m

        w = √ (120 / 4.2)

        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

        f = 0.851 Hz

c) the equation that describes the movement is

        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed ​​is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed ​​is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

5 0
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