Physics, 06.05.2020 06:02, nautiannaharper59901

Spin cycles of washing machines remove water from clothes by producing a large radial acceleration at the rim of the cylindrical tub that holds the water and clothes. Suppose that the diameter of the tub in a typical home washing machine is 50 cm.

1.What is the rotation rate, in rev/min, of the tub during the spin cycle if the radial acceleration of points on the tub wall is 3g?
2.At this rotation rate, what is the tangential speed in m/s of a point on the tub wall?

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answered: Hiadamcom

1. To determine the rotation rate, in rev/min, of the tub during the spin cycle is 103.5 rpm.

2. At this rotation rate, the tangential speed in m/s of a point on the tub wall is 25.88 m/s.

Given the following data:

Diameter = 50 cm to m = 0.5 mCentripetal acceleration = 3g

Radius = $\frac{Diameter}{2} = \frac{0.5}{2} =0.25 \;meter$

Acceleration due to gravity = 9.8 m/s^2

1. To determine the rotation rate, in rev/min, of the tub during the spin cycle:

Mathematically, the centripetal acceleration in rotational motion is given by the formula:

$\alpha =r\omega^2\\\\3g = r \omega^2\\\\\omega^2 = \frac{3g}{r}\\\\\omega = \sqrt{\frac{3g}{r}}$

Substituting the given parameters into the formula, we have;

$\omega = \sqrt{\frac{3\times 9.8}{0.25}} \\\\\omega = \sqrt{\frac{29.4}{0.25}} \\\\\omega = \sqrt{117.6} \\\\\omega = 10.84\;rad/s$

In rev/min:

1 rad/s = 9.5493 rpm

10.84 rad/s = $9.5493 \times 10.84 = 103.5\;rpm$

Rotation rate = 103.5 rpm.

2. To find the tangential speed in m/s of a point on the tub wall, at the above rotation rate:

Mathematically, tangential speed is given by the formula:

$V = r\omega\\\\V = 0.25 \times 103.5$

Tangential speed, V = 25.88 m/s

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answered: mairealexander87

Given Information:

Diameter of the cylindrical tub = d = 50 cm = 0.50 m

Acceleration = α = 3g

Required Information:

1. Rotation rate in rev/min = ω = ?

2. Tangential speed in m/s = v = ?

1. ω = 103.5 rev/min

2. v = 2.71 m/s

Explanation:

We know that centripetal acceleration is given by

α = ω²r

Where ω is the angular speed or rotation rate and r is the radius.

The relation between diameter and radius is given by

r = D/2

r = 0.50/2

r = 0.25 m

Since it is given that the acceleration is equal to 3g where g is the gravitational acceleration 9.81 m/s².

α = ω²r

3g = ω²r

ω² = 3g/r

ω = √(3g/r)

ω = √(3*9.81/0.25)

ω = 10.84 rad/s

To convert rad/s into rev/s divide it by 2π

ω = 10.84/2π

ω = 1.752 rev/s

To convert rev/s into rev/min multiple it by 60

ω = 1.752*60

ω = 103.5 rev/min

Therefore, the rotation rate is 103.5 rev/min

2. The tangential speed can be found using

v = ωr

Where ω is the rotation rate in rad/s and r is the radius.

v = 10.84*0.25

v = 2.71 m/s

Therefore, the tangential speed is 2.71 m/s

answered: Guest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

answered: Guest

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