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xxMikexx [17]
2 years ago
11

Doc Brown holds on to the end of the minute hand of the clock atop city hall. The tangential velocity of the minute hand is 0.41

9 m/s. If the minute hand is
4.0 meters long, what is the professor's centripetal acceleration?
Physics
1 answer:
alexdok [17]2 years ago
3 0

The Professor's centripetal acceleration is 0.044 m/s²

Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.

It is given by:

a = v²/r

where v is the velocity and r is the radius.

Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:

a = v²/r = 0.419²/4 =  0.044 m/s²

The Professor's centripetal acceleration is 0.044 m/s²

Find out more at: brainly.com/question/6082363

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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
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Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

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

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        v = Q / A

The area of ​​a circle is

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       v₂ = Q / A₂

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Now we can use Bernoulli's equation in the colon

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

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

       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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