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mina [271]
3 years ago
10

A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125

F. (a) What does the thermometer read after 4 minutes
Physics
1 answer:
frez [133]3 years ago
7 0

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

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The volume of the balloon increases in the upper atmosphere.

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swat32

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A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the
svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength  is given by

           \lambda = \dfrac{2L}{n}

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

            f = n\dfrac{v}{2L}

now, wavelength calculation

      n = 1

           \lambda_1 = \dfrac{2\times 2}{1}

                    λ₁ = 4 m

      n = 2

           \lambda_2 = \dfrac{2\times 2}{2}

                   λ₂ = 2 m

      n =3

           \lambda_3 = \dfrac{2\times 2}{3}

                    λ₃ = 1.333 m

now, frequency calculation

      n = 1

            f = n\dfrac{v}{2L}

            f_1 =1\times \dfrac{50}{2\times 2}

                    f₁ = 12.5 Hz

      n = 2

            f = n\dfrac{v}{2L}

            f_2 =2\times \dfrac{50}{2\times 2}

                    f₂= 25 Hz

      n = 3

            f = n\dfrac{v}{2L}

            f_3 =3\times \dfrac{50}{2\times 2}

                    f₃ = 37.5 Hz

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