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Lemur [1.5K]
3 years ago
10

Read the scenario, then answer the questions. a 2 kg ball is thrown upward with a velocity of 15 m/s. what is the kinetic energy

of the ball as it is being thrown?j what is the potential energy of the ball when it gets to its maximum height just before falling back to the ground?j
Physics
2 answers:
natita [175]3 years ago
4 0

Explanation:

We have,

Mass of the ball is 2 kg and it is thrown upward with a velocity of 15 m/s.

It is required to find the kinetic energy of the ball as it is being thrown. The kinetic energy is possessed due to the motion of motion. It is given by :

E_k=\dfrac{1}{2}mv^2\\\\E_k=\dfrac{1}{2}\times 2\times 15^2\\\\E_k=225\ J

Let potential energy of the ball when it gets to its maximum height just before falling back to the ground will be equal to the kinetic energy i.e. 225 Joules. As the energy of the system remains conserved.

Kaylis [27]3 years ago
3 0
Both answers are 225 J
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Water flows through a horizontal nozzle in steady flow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2
Dvinal [7]

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

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This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

        Q = A v

        v = Q / A

The area of ​​a circle is

       A = π r² = π d² / 4

Let's look at the speeds at each point

       v₁ = Q / A₁ = Q 4 /π d₁²

       v₁ = 10 4 /π 0.5²

       v₁ = 50.93 m / s

       v₂ = Q / A₂

       v₂ = 10 4 /π 0.25²

       v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

       P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

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       P₁ = P2 + ½ rho (v₂² - v₁²)

      P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

      P₁ = 1.013 10⁵ + 2.205 10⁷

      P₁ = 2.215 10⁷ Pa

la definicion de presion es

      P₁ = F₁/A₁

     F₁ = P₁ A₁

     F₁ = 2.215 10⁷ pi d₁²/4

     F₁ = 2.215 10⁷ pi 0.5²/4

     F₁ = 4.3 106 N

     

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Answer:

See explanation

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