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aev [14]
3 years ago
15

3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000

rpm, completando un total de 12 revoluciones.
d. Determina la aceleración tangencial, centrípeta y la total en m.s-2 cuando el tambor a alcanzado los 60000 rpm

e. Explica lo que ocurre con la magnitud y dirección de los vectores aceleración tangencial, aceleración centrípeta, aceleración total, aceleración angular, velocidad angular cuando la lavadora ha girado desde 3000 rpm hasta 6000 rpm.
Physics
1 answer:
cestrela7 [59]3 years ago
8 0

Answer:

d)    α = 1693.5 rad / s² , a = 392.7 m / s² ,   a_total = α √(R² +1) ,

e)   tan θ = a / α

Explanation:

This is an exercise in linear and angular kinematics.

We initialize reduction of all the magnitudes to the SI system

   w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s

   w = 6000 rev / mi = 628.32 rad / s

   θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad

d) ask for centripetal, tangential and total acceleration.

Let's start by looking for centripetal acceleration, let's use the formula

          w² = w₀² + 2 α θ

          α = (w²- w₀²) / 2θ

we calculate

           α = (628.32²2 - 314.16²) / 2 75.398

           α = 1693.5 rad / s²

the quantity is linear and angular are related

         

the linear or tangential acceleration is

            a =    α  R

where R is the radius of the drum

            a = 1693.5 R

Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m

           a = 1693.5 0.20

           a = 392.7 m / s²

the total acceleration is

           a_total = √(a² + α²)

           a_total = √ (α² R² + α²)

           a_total = α √(R² +1)

e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant

Tangential acceleration is tangency to radius and its value varies proportionally radius

the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry

            tan θ = a / α

the angular velocity increases linearly when with centripetal acceleration

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Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m
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Answer:

U₂ = 20 J

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Explanation:  

Given that

H= 12 m

m = 0.5 kg

h= 4 m

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U₁ = m g H

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U₁ = 60 J

The potential energy at position 2

U₂ = m g h

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The kinetic energy at position 1

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From energy conservation

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By putting the values

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By putting the values

40 = 1/2 x 0.5 x v²

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3 0
2 years ago
Read 2 more answers
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