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Rashid [163]
3 years ago
15

Does the reference point selected for the properties of a substance have any effect on thermodynamic analysis? Why?

Mathematics
1 answer:
MrMuchimi3 years ago
4 0

Answer:

No, the reference point selected for the properties of a substance does't have any effect on thermodynamic analysis.

Step-by-step explanation:

Consider the provided information.

We need to determine that the reference point selected for the properties of a substance have any effect on thermodynamic analysis.

The answer for this question is No.

The reference point selected for the properties of a substance does't have any effect on thermodynamic analysis.

Because in thermodynamic analysis we are deal with the change in the properties.

If we are doing calculation, the reference state chosen is irrelevant as long as we use values in a single consistent array of tables and charts.

Hence, The reference point selected for the properties of a substance does't have any effect on thermodynamic analysis.

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Out of a salary of $4500, I kept 1/3 as savings. Out of the remaining money, I spend 50% on food and 20% on house rent. How much
jeka94

Answer:

$2100

Step-by-step explanation:

1/3 of 4500 is 1500

to find 1/3 of a number is to multiply by 1/3

1/3 * 4500 = 1500

$1500 is for savings

subtract 1500 from 4500

4500 - 1500 = 3000

the remaining money is $3000

to find 50% of something, divide by 2

3000/2 = 1500

$1500 is spent on food.

20% of 3000 is 600

to find 20% of something, multiply the decimal value (0.2)

0.2 * 3000 = 600

$600 are spent on rent.

$450 leftover

1500 + 600 = 2100

Food and rent cost $2100

hope this helps:)

3 0
2 years ago
Please answer, will choose brainliest
Vikentia [17]
B) would be tha answer :)
6 0
3 years ago
Read 2 more answers
What are the roots of the polynomial equation? x² + 7x + 12 = 0 Enter your answers in the boxes. x1= x2=
oksian1 [2.3K]
The equation factors as
.. (x +3)(x +4) = 0
By the zero-product rule, the roots are
.. x1 = -4
.. x2 = -3

3 0
4 years ago
Read 2 more answers
The circumference of the equator of a sphere was measured to be 82 82 cm with a possible error of 0.5 0.5 cm. Use linear approxi
True [87]

Answer:

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.

Step-by-step explanation:

The circumference (s), in centimeters, and the surface area (A_{s}), in square centimeters, of a sphere are represented by following formulas:

A_{s} = 4\pi\cdot r^{2} (1)

s = 2\pi\cdot r (2)

Where r is the radius of the sphere, in centimeters.

By applying (2) in (1), we derive this expression:

A_{s} = 4\pi\cdot \left(\frac{s}{2\pi} \right)^{2}

A_{s} = \frac{s^{2}}{\pi^{2}} (3)

By definition of Total Differential, which is equivalent to definition of Linear Approximation in this case, we determine an expression for the maximum error in the calculated surface area (\Delta A_{s}), in square centimeters:

\Delta A_{s} = \frac{\partial A_{s}}{\partial s} \cdot \Delta s

\Delta A_{s} = \frac{2\cdot s\cdot \Delta s}{\pi^{2}} (4)

Where:

s - Measure circumference, in centimeters.

\Delta s - Possible error in circumference, in centimeters.

If we know that s = 82\,cm and \Delta s = 0.5\,cm, then the maximum error is:

\Delta A_{s} \approx 8.3083\,cm^{2}

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.

6 0
3 years ago
1. Define a variable and write each phrase as an algebraic expression.
katen-ka-za [31]

Answer:

6x=6c?

30-=ml

Step-by-step explanation:

I'm sorry if i got this wrong

7 0
3 years ago
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