Is 2 3/4 <, >, or = to 2.75%

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

MrMuchimi

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.

4 0

3 years ago

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

$V = \frac{1}{3}\pi r^{2}h$

Differentiate the equation with respect to time t, such that

$\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)$

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)$

To differentiate the product,

Let r² = u, so that

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

$V = \frac{1}{3}\pi r^{2}h$

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

3 0

3 years ago

Look at the coordinate plane.

Taya2010 [7]

Answer:

y = -3

Step-by-step explanation:

Desmos

7 0

2 years ago

Fiona wrote the linear equation y = x – 5. when henry wrote his equation, they discovered that his equation had all the same sol

djverab [1.8K]

I would go with A but I am not 100% sure so you may want to check however you can.

6 0

3 years ago

Read 2 more answers

There are 35 students in a drama club two out of every five boys in the club are boys how much diamonds in the club are boys

andrew11 [14]

Given:

Number of students = 35

Two out of every five boys in the club are boys.

To find:

The number of students in the club that are boys.

Solution:

Two out of every five boys in the club are boys. It means, the ratio of boys to the total number of student is 2:5.

Let the number of boys be 2x and number of students be 5x.

According to the question,

$5x=35$

$x=7$

Now, the number of boys is

$2x=2(7)$

$2x=14$

Therefore, 14 students in the club are boys.

8 0

3 years ago