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AURORKA [14]
3 years ago
11

An average person can reach a maximum height of about 60 {\rm cm} when jumping straight upfrom a crouched position. During the j

ump itself, the person's bodyfrom the knees up typically rises a distance of around 50 {\rm cm}. To keep the calculationssimple and yet get a reasonable result, assume that the entirebody rises this much during the jump. In terms of this jumper's weight W , what force does the ground exert on him or herduring the jump?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

2.2 W where W is the weight of the jumper.

Explanation:

A) use the formula

V^2 = 2gh= 2*9.81*60

that gives

V = 3.43 m/s

b) W = mg

during the jump , the displacement of the body is d =0.5m

so the acceleration a = (V^2-Vo^2)/2d = 11.76 m/s^2

net force applied by ground = m(g+a) = W/g(g+a)

= 2.2 W

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Which describes an object in projectile motion? Select three options.
Karo-lina-s [1.5K]

Description of an object in projectile motion is;

  • Gravity acts to pull the object down.
  • The object’s inertia carries it forward.
  • The path of the object is curved.

Explanation:

The path of the projectile is usually curved, and NOT straight, due to the influence of gravity on it which is teh only force acting on it-, causing it motion path to fall towards the earth. Most projectiles follow a parabolic path. The projectile, even though it was launched, its motion is then only due to its own inertia – tendency to stay in motion in a straight line, or rest, unless an external force is acting on it - such as drag or friction.  An example of such projectile motion is of ballistic missiles.

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3 years ago
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The motion of a particle is defined by the relation x 5 t 3 2 9t 2 1 24t 2 8, where x and t are expressed in inches and seconds,
shutvik [7]

Answer:

a)  t = 2.0 s,  b)  x_f = - 24.56 m,  Δx = 16.56 m

Explanation:

This is an exercise in kinematics, the relationship of position and time is indicated

          x = 5 t³ - 9t² -24 t - 8

a) ask when the velocity is zero

speed is defined by

         v = \frac{dx}{dt}

let's perform the derivative

        v = 15 t² - 18t - 24

        0 = 15 t² - 18t - 24

let's solve the quadratic equation

      t = \frac{18 \pm \sqrt{18^2 + 4 \ 15 \ 24}  }{2 \ 15}

       t = \frac{18 \pm 42}{30}

       t1 = -0.8 s

      t2 = 2.0 s

the time has to be positive therefore the correct answer is t = 2.0 s

b) the position and distance traveled for a = 0

acceleration is defined by

       a = dv / dt

       a = 30 t - 18

       a = 0

       30 t = 18

       t = 18/30

       t = 0.6 s

we substitute this time in the expression of the position

       

       x = 5 0.6³ - 9 0.6² - 24 0.6 - 8

       x = 1.08 - 3.24 - 14.4 - 8

       x = -24.56 m

we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s

the position for t = 0

       x₀ = -8 m

the position for t = 0.6 s

      x_f = - 24.56 m

the distance

     ΔX = x_f - x₀

     Δx = | -24.56 -(-8) |

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<h3>The flow chart of internal combustion engine</h3>

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The exhaust valve of the engine is the allows for the outflow of used gases from the engine.

From the attached flow chart, Intake valves are opened to allow the flow of an air/fuel mixture into the engine's cylinders prior to compression and ignition, while exhaust valves open to permit the expulsion of exhaust gases from the combustion process after ignition has occurred.

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