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Ratling [72]
3 years ago
10

1kg slab of concrete loses 12,000j of heat when it cools from 30 Celsius to 26 Celsius. Determine the specific heat capacity of

concrete.
Physics
1 answer:
OleMash [197]3 years ago
3 0

The specific heat capacity of concrete is 3.0 J/(g^{\circ}C)

Explanation:

When a certain amount of energy Q is supplied/given off to/from a sample of substance with mass m, the temperature of the substance increases/decreases by an amount \Delta T, according to the equation

Q=mC_s \Delta T

where

m is the mass of the substance

C_s is the specific heat capacity of the substance

\Delta T is the change in temperature of the substance

In this problem, we have:

m = 1 kg = 1000 g is the mass of the concrete slab

Q = -12,000 J is the amount of energy lost by the slab

\Delta T = 30-26= -4^{\circ}C is the change in temperature of the slab

Solving the equation for C_s, we find the specific heat capacity of concrete:

C_s = \frac{Q}{m \Delta T}=\frac{-12,000}{(1000)(-4)}=3.0 J/(g^{\circ}C)

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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Answer:

D. Asthenosphere

Explanation:

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4 years ago
A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre
Alisiya [41]

Answer:

0.546 \hat k

Explanation:

From the given information:

The force on a given current-carrying conductor is:

F = I ( \L  \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})

where the length usually in negative (x) direction can be computed as

\L ^ {\to }  = -x\hat i \\dL\limits ^ {\to }- dx\hat i

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})

F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)

F = I \int^3_1  - 9.0x^2 \ dx \hat k

F = I  (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k

F = I  (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ]  \hat k

where;

current I = 7.0 A

F = (7.0 \ A)  (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ]  \hat k

F = (7.0 \ A)  (9.0) \bigg [\dfrac{26}{3} \bigg ]  \hat k

F = 546 × 10⁻³ T/mT \hat k

F = 0.546 \hat k

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From the image shown in the question, there are two systems that contain only the fundamental frequency and they are the systems C and D.

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