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Fantom [35]
3 years ago
13

The density of a fluid is given by the empirical equation rho=70.5 exp(8.27 X 10^-7 P) where rho is density (lbm/ft^3) and P is

pressure (lbf/in.^2) a) What are the units of 70.5 and 8.27 x 106-7? b) Calculate the density in g/cm3 for a pressure of 9.00 x10^6 N/m^2. e) Derive a formula for rho (g/cm^3) as a function of P(N/m^2).
Mathematics
1 answer:
jasenka [17]3 years ago
5 0

Answer:

(a) The unit of 70.5 is lbm/ft^3 and the unit of 8.27×10^-7 is in^2/lbf

(b) density = 0.1206g/cm^3

(c) rho = 0.1206exp(8.27×10^-7P)

Step-by-step explanation:

(a) The unit of 70.5 is the same as the unit of rho which is lbm/ft^3. The unit of 8.27×10^-7 is the inverse of the unit of P (lbf/in^2) because exp is found of a constant. Therefore, the unit of 8.27×10^-7 is in^2/lbf

(b) P = 9×10^6N/m^2

rho = 70.5exp(8.27×10^-7× 9×10^6) = 70.5exp7.443 = 70.5×1.71 = 120.6kg/m^3

rho = 120.6kg/m^3 × 1000g/1kg × 1m^3/10^6cm^3 = 0.1206g/cm^3

(c) Formula for rho (g/cm^3) as a function of P (N/m^2) is

rho = 0.1206exp(8.27×10^-7P) (the unit of 0.1206 is g/cm^3)

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Michael and Imani go out to eat for lunch. If their food and beverages cost $25.30 and there is an 8% meals tax, how much is the
blondinia [14]

Answer:

The bill total is $27.32

Step-by-step explanation:

how to figure out tax amount: 25.30 x .08 = $2.024

add tax to cost of meal : 25.30 + 2.024 = $27.32

5 0
3 years ago
A certain drug dosage calls for 17 mg per kg per day and is divided into two doses (1 every 12 hours). If a person weighs 255 po
vekshin1

Answer:

985.15mg

Step-by-step explanation:

Calculation for how many milligrams of the drug should he receive every 12 hours

First step is to calculate the weight in kg

255 lbs = 255/2.2 = 115.9 kg

Second step is multiply mg/kg in order to get total medicine per day

Dose per day=115.9kg × 17 mg

Dose per day= 1,970.3

Third step is to calculate how many milligrams of the drug should he receive by dividing the day dose by 2

Dose milligrams=1,970.3/2

Dose milligrams=985.15mg

Therefore how many milligrams of the drug should he receive every 12 hours will be 985.15mg

3 0
3 years ago
Quadrilateral ABCD is dilated by a scale factor of 1 over 2 centered around (2, 2). Which statement is true about the dilation?
Musya8 [376]

Answer:

Option (1)

1) Segment B'D' will run through (2, 2) and will be shorter than segment BD.

Step-by-step explanation:

The rest of the question is the attached figure (1).

The statement options are:

1) Segment B'D' will run through (2, 2) and will be shorter than segment BD.

2) Segment B'D' will run through (2, 2) and will be longer than segment BD.

3) Segment B'D' will parallel to segment BD and will be shorter than segment BD.

4) Segment B'D' will be parallel to segment BD and will be longer than segment BD.

==============================================================

See the attched figure (2), which represents Quadrilateral ABCD and the image A'B'C'D'

As shown:

The quadrilateral ABCD with vertices A(1,2), B(2,3), C(4,2) and D(2,1).

Quadrilateral ABCD is dilated by a scale factor of 1 over 2 centered around (2, 2).

So,

The quadrilateral A'B'C'D' will be with vertices:

A'= (1.5,2), B'= (2,2.5), C'= (3,2), D'= (2,1.5)

Comparing the options with the figure (2):

So, the answer is option (1)

1) Segment B'D' will run through (2, 2) and will be shorter than segment BD.

6 0
3 years ago
Inverse laplace of [(1/s^2)-(48/s^5)]
Katen [24]
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
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3 years ago
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adelina 88 [10]

Answer:

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Step-by-step explanation:

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3 years ago
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