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natali 33 [55]
3 years ago
15

During the spin cycle the time dependent angular speed of a washing machine drum is given by the equation ω(t) = at + bt2 - ct4

where a = 2.6 rad/s2, b = 0.85 rad/s3 and c = 0.035 rad/s5. At time t = 0 s, a point P on the washer drum is located at θ0 = 0.75 rad. Write an equation for the angular acceleration of the drum, as a function of time, in terms of the given parameters.
Physics
2 answers:
Darya [45]3 years ago
6 0

Answer:

a(t)=a+2bt-4ct^{3}

Explanation:

We have the following angular speed equation :

w(t)=at+bt^{2}-ct^{4}

We also know the parameters :

a=2.6\frac{rad}{s^{2}}

b=0.85\frac{rad}{s^{3}}

c=0.035\frac{rad}{s^{5}}

We need the equation of the angular acceleration in terms of a,b and c.

We know that the angular acceleration is the rate of change of angular speed respect time.

Therefore, to obtain the equation a(t) we need to derivate w(t) in respect of  the variable t.

w(t)=at+bt^{2}-ct^{4} ⇒ derivating in respect of t ⇒

a(t)=a+2bt-4ct^{3}

And that is the angular acceleration equation in terms of a,b and c.

anastassius [24]3 years ago
5 0

Answer:

α = 2.6 +1.7 t - 0.14 t³

Explanation:

Given that

ω(t) = at + bt² - ct⁴

where a = 2.6 rad/s², b = 0.85 rad/s³ and c = 0.035 rad/s⁵

We  know that angular acceleration is the rate of change of angular velocity

α = dω/dt

ω(t) = at + bt² - ct⁴

dω/dt= a + 2 b t - 4 ct³

So

α = a + 2 b t - 4 ct³

By putting the values of a b and c

α = a + 2 b t - 4 ct³

α = 2.6 + 2 x 0.85 t - 4 x 0.035 t³

α = 2.6 +1.7 t - 0.14 t³

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kolbaska11 [484]

Answer:

Final volumen first process V_{2} = 98,44 cm^{3}

Final Pressure second process P_{3} = 1,317 * 10^{10} Pa

Explanation:

Using the Ideal Gases Law yoy have for pressure:

P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

n_{1} = n_{2} = n

In an isobaric process the pressure is constant so:

P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

Replacing : T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}

V_{2} = 98,44 cm^{3}

Replacing on the ideal gases formula the pressure at this piont is:

P_{2} = 3,92 * 10^{9} Pa

For Temperature the ideal gases formula is:

T = \frac{P V }{n R }

For the second process you have that T_{2} = T_{3}  So:

\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }

P_{2} V_{2}  = P_{3} V_{3}

P_{3} = \frac{P_{2} V_{2}}{V_{3}}

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Answer:

True

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Momentum of an object can be defined as the product of its mass and velocity at which it is travelling. With that in mind, momentum = 3*100=300(kg⋅m/s).

One thing to note is the units mentioned. The SI unit of momentum is kg * m/s as it is the product of mass(kilograms) and velocity(meter per second) and not Newton.

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3 years ago
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af
vekshin1

Answer:

T=51.64^\circ F

t=180.10s

Explanation:

The Newton's law in this case is:

T(t)=T_m+Ce^{kt}

Here, T_m is the air temperture, C and k are constants.

We have

70^\circ F in t=0

So:

T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F

And we have 60^\circ F in t=30 s, So:

T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061

Now, we have:

T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)

Applying (1) for t=1 min=60s:

T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F

Applying (1) for T=30^\circ F:

30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s

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