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sleet_krkn [62]
3 years ago
6

Isaac drop ball from height og 2.0 m, and it bounces to a height of 1.5 m what is the speed before and after the ball bounce?

Physics
1 answer:
klasskru [66]3 years ago
6 0

Explanation:

It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.

We need to find the speed before and after the ball bounce.

Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,

\dfrac{1}{2}mu^2=mgh\\\\u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2} \\\\u=6.26\ m/s

Let v is the final speed when it bounces to a height of 1.5 m. So,

\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 1.5} \\\\v=5.42\ m/s

So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.

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Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of th
Mashcka [7]

Answer:

   v₂ = 9 v

Explanation:

For this exercise in fluid mechanics, let's use the continuity equation

           v₁ A₁ = v₂ A₂

where v is the velocity of the fluid, A the area of ​​the pipe and the subscripts correspond to two places of interest.

The area of ​​a circle is

           A = π R²

let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint

     

In this case v₁ = v and the area is

            A₁ = π R²

in the second point

           A₂= π (R / 3)²

we substitute in the continuity equation

           v π R² = v₂ π R² / 9

            v = v₂ / 9

           

            v₂ = 9 v

7 0
4 years ago
Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 60 degrees. It reache
Nina [5.8K]

Answer:

R = 0.0503 m

Explanation:

This is a projectile launching exercise, to find the range we can use the equation

       R = v₀² sin 2θ / g

How we know the maximum height

      v_{f}² =v_{oy}² - 2 g y

      v_{f}= 0

      v_{oy} = √ 2 g y

      v_{oy} = √ 2 9.8 / 15

      v_{oy} = 1.14 m / s

Let's use trigonometry to find the speed

    sin θ = v_{oy} / vo

    vo = v_{oy} / sin θ

    vo = 1.14 / sin 60

    vo = 1.32 m / s

We calculate the range with the first equation

     R = 1.32² sin(2 60) / 30

    R = 0.0503 m

3 0
3 years ago
what is the resistance of a 12m long wire having radius( 2*10 to the power - 4) and resistivity (3.14*10 to the power - 8)ohmM
Arisa [49]
R = ρl/A

Where R = Resistance in Ohms, Ω, ρ = Resistivity in Ωm, l = Length in m.

Area in m²

ρ = Resistivity = 3.14 * 10⁻⁸ Ωm,  Length l = 12m, 

Area = πr² = π* (2*10⁻⁴)² m² ≈ 3.14 * (2*10⁻⁴)² m²


R = ρl/A

    ≈ 3.14 * 10⁻⁸ * 12 / (3.14 * (2*10⁻⁴)²)

     ≈ 3

Resistance, R ≈ 3 Ω
3 0
3 years ago
How many atoms of Oxygen (O) are in 3CO2
ikadub [295]

Answer:

2

Explanation:

BRAINLIEST?? AND THANK ME LATER!!! HOPE THIS HELPED!!!

5 0
3 years ago
What speed should a satellite of mass 4,900 kg moving around
Rudiy27

Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.

<u>Given the following data:</u>

  • Gravitational constant = 6.67 × 10⁻¹¹ m/kg²
  • Mass of Moon = 7.36 × 10²² kg
  • Distance, r = 4.2 × 10⁶ m.

<h3>How to determine the speed of this satellite?</h3>

In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.

This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:

Fc = Fg

mv²/r = GmM/r²

<u>Where:</u>

  • m is the mass of the satellite.
  • M is mass of the Moon.

Making v the subject of formula, we have;

v = √(GM/r)

Substituting the given parameters into the formula, we have;

v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)

v = √(1,168,838.095)

v = 1,081.13 m/s.

Speed, v = 1.8 × 10³ m/s.

Read more on speed here: brainly.com/question/20162935

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