DIRECTIONS:
Use the food diary chart below to record everything that you eat in a 24-hour period. Be specific. For example, if you eat a hot dog with ketchup and mustard, don’t forget to include the ketchup and mustard. Be sure to include the quantity of food eaten as well. Under “Amount or Quantity,” do your best to estimate in cups, ounces, or other measurement. You may add additional rows if needed.

Answer:

The formulas are down below

Explanation:

[tex]w = \frac{o}{t} [/tex]

From V=rw

[tex]w = \frac{v}{r} [/tex]

(a) The average pressure exerted by the person is 40,000 N/m².

(b) The average pressure exerted by the elephant has 295,000 N/m².

The average pressure exerted by the elephant is about 7 times greater than the average pressure exerted by the person.

The average pressure exerted by each is defined as the force per unit area.

Mathematically, the formula for average pressure is given as;

P = F/A

where;

F is the applied force

A is the area of each object

P = W / A

where;

W is weight of the person

A is the area of surface in contact

The average pressure exerted by the person is calculated as follows;

P = (640 N) / (0.016)

P = 40,000 N/m²

The average pressure exerted by the elephant is calculated as follows;

P = (4.13 x 10⁴ N) / (0.14)

P = 295,000 N/m²

Thus, the average pressure exerted by the elephant and the person depends on their weight and area of their shoes.

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The luminosity of this star in units of the solar luminosity would be: 483.7L.

To calculate the luminosity, we would use the different values given and the formula for luminosity.

Temperature = 9305K

Star's radius = [tex]5.90 * 10^{9} m\\[/tex]

Luminoisty of the star

Luminosity of the sun

= [tex]\frac{4π * (5.90 * 10^9)^2 * 5.67 * 10^-8 * 9305^4 W}{3.846 * 10^26 W}[/tex]

= 483.7L

This is the unit for luminosity.

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The molecules of O2 that are  present in 3.90 L flask at  a temperature of 273 K and a pressure of 1.00 atm is 1.047 x 10^23 molecules  of O2

Step  1:  used the ideal gas equation to calculate the moles of O2

that is Pv=n RT  where;

P(pressure)= 1.00 atm

V(volume) =3.90 L

n(number of moles)=?

R(gas constant) = 0.0821 L.atm/mol.K

T(temperature) = 273 k

by  making n the subject of the formula by  dividing  both side by RT

n= Pv/RT

n=[( 1.00 atm x 3.90 L)  /(0.0821 L.atm/mol.k  x273)]=0.174  moles

Step 2: use the Avogadro's  law constant  to calculate  the number of molecules

that  is  according to Avogadro's law

                          1  mole =  6.02 x10^23  molecules

                            0.174 moles=? molecules

by  cross  multiplication

the number of  molecules

= (0.174  moles x  6.02 x10^23  molecules)/ 1 mole  =1.047 x 10^23 molecules of O2

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The final volume of the gas, when the pressure changes from 80.8 kPa to 101.3 kPa at constant temperature, is approximately 652.9 cm³.

To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

Boyle's Law can be represented by the equation: P₁V₁ = P₂V₂

Where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.

Given:

Initial volume, V₁ = 817 cm³

Initial pressure, P₁ = 80.8 kPa

Final pressure, P₂ = 101.3 kPa

We need to find the final volume, V₂.

Using Boyle's Law equation, we can rearrange it to solve for V₂:

V₂ = (P₁V₁) / P₂

Plugging in the given values:

V₂ = (80.8 kPa * 817 cm³) / 101.3 kPa

Simplifying the expression:

V₂ ≈ 652.9 cm³

Therefore, the final volume of the gas, when the pressure changes from 80.8 kPa to 101.3 kPa at constant temperature, is approximately 652.9 cm³.

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Answer:

[tex]{\Delta L=0.012 \ m[/tex]

Explanation:

Given:

[tex]L_0=10.0 \ m\\\Delta T_0=25.0 \ \textdegree C\\\Delta T_f=75.0 \ \textdegree C\\\alpha_{Al}=2.40 \times 10^{-5} \ \textdegree C^{-1}[/tex]

Find:

[tex]\Delta L= \ ?? \ m[/tex]

Using the formula for linear expansion.

[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Formula for Linear Expansion:}}\\\\ \Delta L=\alpha L_0 \Delta T\end{array}\right}[/tex]

Where...

[tex]\hrulefill[/tex]

Plug the known values into the formula for linear expansion.  

[tex]\Delta L=\alpha L_0 \Delta T\\\\\Longrightarrow \Delta L=(2.40 \times 10^{-5})(10.0)(75.0-25.0)\\\\\therefore \boxed{\boxed{\Delta L=0.012 \ m}}[/tex]

Thus, the change in length is found.

The change in length of the aluminum beam due to the thermal linear expansion is 0.012 m.

When a material's temperature increases, a phenomenon known as linear expansion occurs, which results in an increase in the material's length.

The length of a material that is one-unit long changes as the temperature rises by ten degrees Celsius, which is how the coefficient of linear expansion is stated.

Length of the aluminum beam, L = 10 m

Initial temperature of the beam, T₁ = 25°C

Final temperature of the beam, T₂ = 75°C

The coefficient of linear expansion of aluminum, α = 2.4 x 10⁻⁵⁻⁵⁻⁻C⁻

The equation for the change in length of the aluminum beam is given by,

ΔL = αLΔT

ΔL = 2.4 x 10⁻⁵x 10 x(75 - 25)

ΔL = 2.4 x 10⁻⁴x 50

ΔL = 0.012 m

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An annotated bibliography is a list of citations to books, articles, and documents. Each citation is followed by a brief descriptive and evaluative paragraph, the annotation. The purpose of the annotation is to inform the reader of the relevance, accuracy, and quality of the sources cited.

The cue column is typically located on the left-hand side of the page and is used to jot down keywords or questions that serve as cues for recalling the main points of the lecture or reading. The note-taking area is located on the right-hand side of the page and is used to write down detailed notes about the lecture or reading.

The summary section is located at the bottom of the page and is used to summarize the key points of the notes. Overall, the Cornell method is an effective way to organize and retain information during lectures and readings.

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In the given pith ball experiment with charges of 6.8 x 10^-6 C and 10.5 x 10^-6 C, and a mass of 45 grams each, positioned 0.50 m apart at an angle of 27.72°, the magnitudes of the tension in each string, gravitational force, and electrostatic force are approximately 0.456 N, 0.441 N, and 4.704 N, respectively.

To solve this problem, we need to analyze the forces acting on each pith ball: the tension in the strings, the gravitational force, and the electrostatic force.

1. Tension in each string, T:

Using the given values:

T = (0.045 kg × 9.8 m/s^2) / cos(27.72°)

T ≈ 0.456 N (rounded to three decimal places)

2. Gravitational force, Fg:

Using the given values:

Fg = 0.045 kg × 9.8 m/s^2

Fg ≈ 0.441 N (rounded to three decimal places)

3. Electrostatic force, Fq:

Using the given values and Coulomb's law:

Fq = (8.99 × 10^9 N m^2/C^2) × (6.8 × 10^(-6) C) × (10.5 × 10^(-6) C) / (0.50 m)^2

Fq ≈ 4.704 N (rounded to three decimal places)

Therefore, the values for the magnitudes of the tension in each string (T), the gravitational force (Fg), and the electrostatic force (Fq) are approximate:

T ≈ 0.456 N

Fg ≈ 0.441 N

Fq ≈ 4.704 N

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