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velikii [3]
3 years ago
11

what was the initial temperature is 250 calories reply .1 kg of gold the final temperature of the gold was 175°c ? the specific

heat cold is 0.175 °c​
Physics
1 answer:
weqwewe [10]3 years ago
3 0

Answer:

115.2^{\circ}C

Explanation:

When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by \Delta T, according to the equation

Q=mC_s \Delta T

where C_s is the specific heat capacity of the substance.

In this problem, we have:

Q=250 \cdot 4.184 =1046 J is the amount of heat supplied to the sample of gold

m = 0.1 kg = 100 g is the mass of the sample

C_s = 0.175 J/gC is the specific heat capacity of gold

Solving for \Delta T, we find the change in temperature

\Delta T = \frac{Q}{m C_s}=\frac{1046}{(100)(0.175)}=59.8^{\circ}

And since the final temperature was

T_f = 175^{\circ}

The initial temperature was

T_i = T_f - \Delta T= 175 -59.8=115.2^{\circ}C

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You are playing a violin, where the fundamental frequency of one of the strings is 440 Hz, as you are standing in front of the o
Natalka [10]

Answer:

a)   L = 440 cm

Explanation:

In the open tube on one side and cowbell on the other, we have a maximum in the open part and a node in the closed part, therefore the resonance frequencies are

             λ₁ = 4L             fundamental

             λ₃ = 4L / 3       third harmonic

             λ₅ = 4L / 5       five harmonic

             

The violin string is a fixed cure in its two extracts, so both are nodes, their length from resonance wave are

              λ₁ = 2L                    fundamental

             λ₂ = 2L / 2              second harmonic

             λ₃ = 2L / 3              third harmonic

             λ₄= 2L / 4               fourth harmonic

They indicate that resonance occurs in the fourth harmonic, let's look for the frequency

              v =λ f

for the fundamental

              v = λ₀ f₀

              V = 2L f₀

for the fourth harmonica

              v = λ₄  f ’

              v = L / 2  f'

             2L f₀ = L / 2 f ’

             f ’= 4 f₀

             f ’= 4 440

             f ’= 1760 Hz

for this frequency it has the resonance with the tube

           f ’= 4L

           L = f ’/ 4

           L = 1760/4

           L = 440 cm

b) let's find the frequency of the next harmonic in the tube

             λ₃ = 4L / 3

             λ₃ = 4 400/3

             λ₃ = 586.6 cm

            v = λf

            f = v / λlam₃

            f₃3 = 340 / 586.6

            f3 = 0.579

as the minimum frequency on the violin is 440 Beam there is no way to reach this value, therefore there are no higher resonances

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