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Rom4ik [11]
4 years ago
8

I need help with question 8 and 9

Physics
1 answer:
Wewaii [24]4 years ago
7 0

8) The height of the shelf is 2.04 m

9) The gravitational potential energy is 100 J

Explanation:

8)

The gravitational potential energy of an object is given by:

PE=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object, relative to the ground

For the object in this problem, we have

m = 10 kg

g=9.8 m/s^2

And we know that

PE = 200 J

Therefore we can re-arrange the formula to solve for h, the heigth of the object:

h=\frac{PE}{mg}=\frac{200}{(10)(9.8)}=2.04 m

9)

As before, the gravitational potential energy of the object is given by:

PE=mgh

where in this  case, we have:

m = 10 kg is the mass of the object

h = 2.04 m is the height of the object from the ground

While in this case, the acceleration of gravity on this planet is

g=4.9 m/s^2

Substituting,

PE=(10)(4.9)(2.04)=100 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be x meters.

Horizontal velocity of the ball will always be 26.2\times\cos{34.2\textdegree} until it lands if there's no air resistance.

The ball will arrive at the goal in \displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}} seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0.

For this soccer ball:

  • g = 9.81\;\text{m}\cdot\text{s}^{-2},
  • v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2},
  • h_0 = 0 since the player kicks the ball "from ground level."

\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}

when the ball reaches the goal.

\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x.

Solve this quadratic equation for x, x > 0.

  • x = 65.1 meters when h = 0 meters.
  • x = 6.54 or 58.5 meters when h = 4 meters.

In other words,

  • For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
  • For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

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