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rjkz [21]
3 years ago
10

Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g

t; 0 is a constant. Suppose that A = A0 cos(ωt) for some constants A0 and ω. That is, the ambient temperature oscillates (for example night and day temperatures). a) Find the general solution. b) In the long term, will the initial conditions make much of a difference? Why or why not?
Physics
1 answer:
lys-0071 [83]3 years ago
5 0

Answer:

a)X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b)Does not affect the long term.

Explanation:

Given that

\dfrac{dx}{dt}=-k(x-A)

A = A0 cos(ωt)

\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))

\dfrac{dx}{dt}+kx=kA_o cos(\omega t)

This is linear equation so integration factor ,I

I=e^{\int kdt}

I=e^{kt}

Now by using linear equation property

e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C

e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C

X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b)

at t= 0

X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C

X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+e^{-kt}\times \left ( X(0)-\dfrac{k^2A_o}{\omega^2+k^2} \right )

So the initial condition does not affect the long term.

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A pogo stick has a spring with a force constant of 2.50 × 104 N/m , which can be compressed 11.2 cm. what maximum height, in met
AleksandrR [38]

Answer:

h = 36.4 cm

Explanation:

given,

spring constant = 2.5 x 10⁴ N/m

compressed distance = 11.2 cm = 0.112 m

mass of the child = 44 kg

maximum height = ?

by energy of conservation

K.E_i +P.E_i= K.E_f + P.E_f

\dfrac{1}{2}kx^2 = mgh

\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h

\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h

156.8 = 44 \times 9.8 \times h

h = \dfrac{156.8}{44 \times 9.8}

h = 0.364 m

h = 36.4 cm

7 0
3 years ago
Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees
Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt)  ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1  

70 = Tamb + (100 - Tamb)*e^(-10k)   ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k)   ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)  

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k)   .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0
3 years ago
The speed of water flowing through the 4in diameter section of the piping system is 3ft/s. What is the volume flow rate of water
olga_2 [115]
1 ft =12 in 
4 in = 0.333 ft 
volume = (п/4)*(0.333)² = 0.087 ft²
vol. flow = spead *volume
              =3 ft/s * 0.087 ft² 

vol flow = 0.261 ft³/s
3 0
3 years ago
What type of radioactive decay is shown in this equation?
svp [43]
There is no <span>radioactive decay</span>
6 0
3 years ago
Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an
Artyom0805 [142]

a) For the motion of car with uniform velocity we have , s = ut+\frac{1}{2}at^2, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 m/s^2

Substituting

       520 = u*223\\ \\u = 2.33 m/s

 The constant velocity of car a = 2.33 m/s

b) We have s = ut+\frac{1}{2} at^2

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

      520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2

Now we have v = u+at, where v is the final velocity

Substituting

        v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 m/s^2

7 0
3 years ago
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