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Natasha_Volkova [10]
3 years ago
11

In the heat equation, what does c represent

Physics
2 answers:
Bingel [31]3 years ago
5 0
Heat equation, Q = m.c.Δt
Here, c represents " the specific heat of the substance "

Hope this helps!
sveticcg [70]3 years ago
5 0

Answer : 'c' represent the 'Specific heat of substance'.

Explanation :

Specific heat : It is defined as the amount of heat required to raise the temperature of 1 kilogram of substance by one degree Celsius. The S.I unit of specific heat is, Joule per kelvin (J/K).

Formula for specific heat,

c=\frac{Q}{m\Delta T}

where,

c = specific heat capacity of substance

Q = heat required

m = mass of substance

\Delta T = change in temperature

Hence, 'c' represent the 'Specific heat of substance'.

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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
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Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet x=\frac{4}{12} =\frac{1}{3}feet

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

W=kx ⇒ k=\frac{W}{x}

The spring constant , k=\frac{24}{1/3}

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    m\frac{d^2x}{dt} +kx=0

    \frac{3}{4} \frac{d^2x}{dt} +72x=0

  \frac{d^2x}{dt} +96x=0

Auxiliary equation is, m^2+96=0

                                 m=\sqrt{-96}

                               =\frac{+}{} i4\sqrt{6}

Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

                                 x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2  4\sqrt{6} cos4\sqrt{6t}

The mass is released from the rest x'(0) = 0

                    =-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2 4\sqrt{6} cos4\sqrt{6(0)} =0

                                                    c_2 4\sqrt{6} =0

                                     c_2=0

Therefore , x(t)=c_1 cos 4\sqrt{6t}

Since , the mass is released from the rest from 4 inches

                    x(0)= -4 inches

c_1 cos 4\sqrt{6(0)} =-\frac{4}{12} feet

   c_1=-\frac{1}{3} feet

Therefore , the equation of motion is  -\frac{1}{3} cos4\sqrt{6t}

7 0
2 years ago
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