Ask question

Ask question

All categories

3 years ago

8

Operating speed of an automatic washing machine is 5.5 rad s-1. After loading dirty clothes and pressing a start button, the tub

of the washer can reach its operating speed with an average angular acceleration of 4.15 rad s-2. Later at the spin-dry mode, it is starting from rest and reaching an angular speed of 6.0 revolution per second in 7.0 s.
a) How long does it take for the clothes to come up to the speed during the washing mode?
b) Now the tub of the washer goes into it spin-dry cycle. At this point, the person doing the laundry opens the lid and a safety switch turns off the washer. The tub slows to rest in 13.0 s. How many revolutions does the tub turn during this 20.0 s interval?

1 answer:

AlekseyPX3 years ago

4 0

The equations for rotary motion is based on motion in a circular path

a) Time taken by clothes to come up to speed during washing is approximately 1.325 s

b) The tub turns approximately 24.016 revolutions in the 20.0 s interval

Reasons:

Known parameter are;

Operating speed of an automatic washing machine = 5.5 rad·s⁻¹

Average angular acceleration of the washing machine, α = 4.15 rad·s⁻²

Angular speed in the spin-dry mode, ω = 6.0 revolution per second

Time it takes to reach angular speed in spin-dry mode = 7.0 s

a) The relationship between angular acceleration and time are;

$\alpha = \dfrac{\Delta \omega}{\Delta t}$ , $\Delta t = \dfrac{\Delta \omega}{\alpha}$

Where;

Δω = ω₂ - ω₁ = 5.5 rad·s⁻¹

Δt = The time it takes to accelerate

Therefore;

$\Delta t = \dfrac{5.5 \ rad\cdot s^{-1}}{4.15 \ rad \cdot s^{-2}} \approx 1.325 \ s$

The time it takes for the clothes to come up to speed during the washing mode, Δt ≈ 1.325 s

b) The angular acceleration of the spin-dry mode is given as follows;

6 revolutions per second = 2·π ×6 rad per second = 12·π rad/s

$\alpha = \dfrac{\Delta \omega}{\Delta t}$

The acceleration during spin up

$\therefore \alpha = \dfrac{12 \cdot \pi }{7.0} \ rad \cdot s^{-2} \approx 5.39 \ rad \cdot s^{-2}$

The acceleration during slowing down;

$\therefore \alpha = \dfrac{12 \cdot \pi }{13.0} \ rad \cdot s^{-2} \approx 2.9 \ rad \cdot s^{-2}$

The angle turned in the 20.0 second interval is therefore;

$\dfrac{1}{2} \times 5.39 \times 7^2 + 12 \cdot \pi \times 7-\dfrac{1}{2} \times 2.9 \times 13^2 \approx 150.9$

The angle turned in the 20.0 s is approximately 150.9 radians

$Number \ of \ revolutions \ =\dfrac{\theta}{2 \cdot \pi}$

Therefore;

$Number \ of \ revolutions \ in \ 20.0 \ s =\dfrac{150.9 }{2 \cdot \pi} \approx 24.016$

The number of revolutions of the tub in the 20.0 s interval is approximately 24.016 revolutions

Learn more about constant angular acceleration here:

brainly.com/question/14057630

You might be interested in

A proton is traveling horizontally to the right at 1.8 × 106 m/s. (a) Find the magnitude and direction of the weakest electric f

Mars2501 [29]

Answer:

528398.4375 N/C opposite to the direction of the proton

$3.56\times 10^{-8}\ s$

288.24609375 N/C in the same direction of the motion of the electron

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity = $1.8\times 10^{6}\ m/s$

s = Displacement = 3.2 cm

a = Acceleration

Mass of electron = $9.11\times 10^{-31}\ kg$

Mass of electron = $1.67\times 10^{-27}\ kg$

q = Charge of particle = $1.6\times 10^{-19}\ C$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(1.8\times 10^6)^2}{2\times 0.032}\\\Rightarrow a=-5.0625\times 10^{13}\ m/s^2$

Electric field is given by

$E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{1.67\times 10^{-27}\times -5.0625\times 10^{13}}{1.6\times 10^{-19}}\\\Rightarrow E=−528398.4375\ N/C$

The electric field is 528398.4375 N/C opposite to the direction of the proton

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-1.8\times 10^6}{-5.0625\times 10^{13}}\\\Rightarrow t=3.56\times 10^{-8}\ s$

The time taken is $3.56\times 10^{-8}\ s$

$E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{9.11\times 10^{-31}\times -5.0625\times 10^{13}}{-1.6\times 10^{-19}}\\\Rightarrow E=288.24609375\ N/C$

The electric field is 288.24609375 N/C in the same direction of the motion of the electron

7 0

4 years ago

7. An engineer is using a wire that has a resistance of 1.5 . This resistance is too high for the application he is designing. T

Veronika [31]

Answer and Explanation:

We know that resistance $R=\frac{\rho l}{A}$ from the given equation of resistance it is clear that resistance depends on resistivity length and area of the material but we can not change the length because it is given that the length must be 2.5 cm long.

So we can do two two things to reduce the resistance

increase the cross sectional area
decrease the resistivity of the material

7 0

3 years ago

Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

$\int dW = \int_0^{0.1} k.x dx$

$W = \int_0^{0.1} k.x dx$

$W = k[\dfrac{x^2}{2}]_0^{0.1}$

$10 = k\dfrac{0.1^2}{2}$

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

$W = \int_{0.1}^{0.2} 2000 x dx$

$W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx$

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0

3 years ago

A 10 m long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 30 m/s when a 68

timama [110]

Answer:

v = 30.39 m/s

Explanation:

given,

mass of glider,M = 680 Kg

mass of the skydiver, m = 68 Kg

horizontal velocity,V = 30 m/s

when skydiver releases, velocity,v = 30 m/s

velocity if the glider,v' = ?

use the conservation of momentum

M V = m' v' + m v

m' = 680-68 = 612 Kg

680 x 30 = 612 x v + 60 x 30

612 v = 18600

v = 30.39 m/s

since the skydiver's speed will be the same as before release

The glider should continue to travel at 30.39 m/s since there are no external forces acting on it

5 0

3 years ago

The heat flux for a given wall is in the x-direction and given as q^n = 11 W/m^2, the walls thermal conductivity is 1.7 W/mK and

MrMuchimi

Answer:

$\frac{dT}{dx} = 6.47 ^oC/m$

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

Explanation:

As we know that heat flux is given by the formula

$q^n = K\frac{dT}{dx}$

here we know that

K = thermal conductivity

$\frac{dT}{dx}$ = temperature gradient

now we know that

$q^n = 11 W/m^2$

also we know that

K = 1.7 W/mK

now we have

$11 = 1.7 \frac{dT}{dx}$

so temperature gradient is given as

$\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m$

also in other unit it will be same

$\frac{dT}{dx} = 6.47 ^oC/m$

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

5 0

3 years ago