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alukav5142 [94]
3 years ago
13

If a ball is 10m high with what velocity will it fall?

Physics
1 answer:
Semmy [17]3 years ago
6 0

14m/s

Explanation:

Given parameters:

Height of the ball = 10m

Unknown:

Velocity of fall or final velocity = ?

Solution:

We are going to use the appropriate equation of motion to solve this problem.

The object is falling with respect to gravity.

  V² = U² + 2gH

where V is the final velocity

            U is the initial velocity

             g is the acceleration due to gravity 9.8m/s²

             H is the height of fall

The initial velocity here is zero and

      V² = 2 x 9.8 x 10 = 196

       V = 14m/s

learn more:

Motion problems brainly.com/question/5248528

#learnwithBrainly

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In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
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the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

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The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

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Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

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