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Mars2501 [29]
3 years ago
14

Brass is an alloy made from copper and zinc a 0.59 kg brass sample at 98.0 is dropped into 2.80 kg of water at 5.0 c if the equi

librium temperature is 6.8 what is the specific heat capacity
Physics
2 answers:
MariettaO [177]3 years ago
6 0

Answer:

393.399 J/kg.°C

Explanation:

Specific heat capacity: This is the quantity of heat required to raise the temperature of a unit mass of a substance through a degree rise in temperature.

Heat lost by the brass = heat gained by water

CM(t₁-t₃) = cm(t₃-t₂)........................ Equation 1

Where C = specific heat capacity of the brass, M = mass of the brass, t₁ = initial temperature of the brass, t₂ = initial temperature of water, t₃ = temperature of the mixture.

Making C the subject of the equation

C = cm(t₃-t₂)/M(t₁-t₃)............................... Equation 2

Given: M = 0.59 kg, m = 2.8 kg, t₁ = 98 °C, t₂ = 5.0 °C, t₃ = 6.8 °C

Constant: c = 4200 J/kg.°C

Substitute into equation 2,

C = 2.8×4200(6.8-5.0)/0.59(98-6.8)

C = 21168/53.808

C = 393.399 J/kg.°C

Thus the specific heat capacity of the brass = 393.399 J/kg.°C

zhannawk [14.2K]3 years ago
3 0
The specific heat capacity of brass would be ranked between 0 and infinity
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3 years ago
As one moves farther and farther from the Sun, the distance between adjacent planets is _____.
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3 years ago
In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn
kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

s=ut+\dfrac{1}{2}at_{A}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times1.6\times t_{A}^2

t_{A}=\sqrt{\dfrac{30\times2}{1.6}}

t_{A}=6.12\ sec

We need to calculate the time for B

Using equation of motion

s=ut+\dfrac{1}{2}at_{B}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times2.0\times t_{B}^2

t_{B}=\sqrt{\dfrac{30\times2}{2.0}}

t_{B}=5.48\ sec

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t'=t_{A}-t_{B}

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t'=0.64\ s

Hence, B can take 0.64 sec for the longest nap .

4 0
4 years ago
A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H
Aleks04 [339]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

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Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

Dont report my answer please

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3 years ago
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