Every month, Frank pays an $80 membership fee at a fitness
center so he can avail himself of the unlimited use
of its facilities. On average, he goes to the center 10 times
a month. What is the average cost of each trip he makes
to the center? What is the marginal cost of an additional
work-out session?

Answer:

The answer is "[tex]3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}[/tex]"

Explanation:

Given data:  

Initial temperature of tank [tex]T_1 = 300^{\circ}\ C= 573 K[/tex]

Initial pressure of tank [tex]P_1= 400 \ kPa[/tex]

Diameter of throat [tex]d* = 2 \ cm[/tex]

Mach number at exit [tex]M = 2.8[/tex]

In point a:

calculating the throat area:

[tex]A*=\frac{\pi}{4} \times d^2[/tex]

      [tex]=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2[/tex]

Since, the Mach number at throat is approximately half the Mach number at exit.  

Calculate the Mach number at throat.  

[tex]M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4[/tex]

Calculate the exit area using isentropic flow equation.

[tex]\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}[/tex]

Here: [tex]\gamma[/tex] is the specific heat ratio. Substitute the values in above equation.

[tex]\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm[/tex]

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

[tex]\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K[/tex]

Calculate the velocity at exit.  

[tex]V*=M*\sqrt{ \gamma R T*}[/tex]

Here: R is the gas constant.  

[tex]V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}[/tex]

Calculate the density of air at inlet

[tex]\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}[/tex]

Calculate the density of air at throat using isentropic flow equation.  

[tex]\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}[/tex]

Calculate the mass flow rate.  

[tex]m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}[/tex]

Beginning with model year 2008,  Controller Area Network (CAN) is required for diagnostics.

Diagnostics are defined as the process of identifying an injury, ailment, or disease based on the symptoms and signs a person is exhibiting. In order to confirm or rule out illnesses and diseases, diagnostic tests are utilized. Before creating a treatment plan, your doctor needs particular information from a diagnostic test to make an accurate diagnosis.

The robust automotive bus standard known as a Controller Area Network (CAN) enables microcontrollers and other devices to interact with each other's applications without the need for a host computer. Devices and CAN busses are frequently found in industrial and automotive systems. Bosch created the Controller Area Network (CAN) bus, which saw its debut in manufacturing in 1991.

Thus, beginning with model year 2008,  Controller Area Network (CAN) is required for diagnostics.

To learn more about diagnostics, refer to the link below:

https://brainly.com/question/12814536

#SPJ2

Answer:

d

Explanation:

Answer:fan blades, rotating , abrasive wheel machinery

Explanation:

Answer:

The first one.

Explanation:

The first one without a doubt

this is perfect.........

Answer:

Statement 1: All balls hit the ground at the same time

Explanation:

When there is no resistance of air, the acceleration due to gravity experienced by all the bodies are same. So for falling bodies, neglecting the air resistance, the falling object will be weightless and therefore all the objects will hit the ground at the same time when there is nor air resistance and the objects are considered to be falling in vacuum.

Answer:

[tex]250\ \text{lbm/min}[/tex]

[tex]625\ \text{ft/min}[/tex]

Explanation:

[tex]A_1[/tex] = Area of section 1 = [tex]10\ \text{ft}^2[/tex]

[tex]V_1[/tex] = Velocity of water at section 1 = 100 ft/min

[tex]v_1[/tex] = Specific volume at section 1 = [tex]4\ \text{ft}^3/\text{lbm}[/tex]

[tex]\rho[/tex] = Density of fluid = [tex]0.2\ \text{lb/ft}^3[/tex]

[tex]A_2[/tex] = Area of section 2 = [tex]2\ \text{ft}^2[/tex]

Mass flow rate is given by

[tex]m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}[/tex]

The mass flow rate through the pipe is [tex]250\ \text{lbm/min}[/tex]

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

[tex]m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}[/tex]

The speed at section 2 is [tex]625\ \text{ft/min}[/tex].

(a) The mass flow rate will be "250 lbm/min".

(b) At section 2, the speed will be "625 ft/min".

According to the question,

Section 1 area, A₁ = 10 ft²

Section 2 area, A₂ = 2 ft²

Water's velocity, V₁ = 100 ft/min

Volume at section 1, v₁ = 4 ft³/lbm

(a) We know the formula,

Mass flow rate, m = ρA₁V₁

                              = [tex]\frac{A_1 V_1}{v_1}[/tex]

By substituting the values,

                              = [tex]\frac{10\times 100}{4}[/tex]

                              = [tex]\frac{1000}{4}[/tex]

                              = 250 lbm/min

(2) The speed will be:

→ m = ρA₂V₂

or,

  V₂ = [tex]\frac{m}{\rho A_2}[/tex]

By substituting the values,

       = [tex]\frac{250}{0.2\times 2}[/tex]

       = [tex]\frac{200}{0.4}[/tex]

       = 625 ft/min

Thus the responses above are correct.  

Find out more information about speed here:

https://brainly.com/question/4931057

Answer:

suspended load; bed load

Explanation:

Suspended load is those particles in a river that are transported and float between and around the middle and top of the river, they often change the color of the river through their suspension such as clay.

Bed load is those particles in a river that are transported at the river bed or floor. This is as a result of their weight. Examples are relatively big stones.

Hence, in this case, Clay sized particles in a river are likely to be transported as SUSPENDED LOAD whereas gravel-sized particles are likely to be transported as BED LOAD

Based on the information provided, Technician B is correct because the drivetrain on a vehicle includes the axles and wheels.

An automotive engineer can be defined as a professional who has completed and acquired his or her bachelor's degree in mechanical engineering and has been trained on how to create designs for new automobile vehicles (cars), test prototypes, troubleshoots problems that arise, and oversees the manufacturing process.

A drivetrain can be defined as the group of components or mechanical system that are fitted in automobile vehicles, which is designed and developed to connect the transmission to the drive axles such as:

In this context, we can reasonably infer and logically deduce that Technician B is correct because the drivetrain on a vehicle includes the axles and wheels.

Read more on drivetrain here: https://brainly.com/question/24083630

#SPJ1

Answer: drafting blueprints

Explanation:

Answer:

(a) Electrons

Explanation:

When there is a current flow in a transistor of NMOS so it is because of electrons as in NMOS the current is flowing from drain to source. Since the flow of electrons is from source to drain. Also the current would be inverse direction for electrons flow

Therefore the option A is correct

hence, the other options are incorrect

Answer:

Following are the proving to this question:

Explanation:

[tex]\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}[/tex]

using the energy equation for entry and exit value :

[tex]\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}[/tex]

where

[tex]\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}[/tex]

         [tex]= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\[/tex]

         [tex]= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\[/tex]

[tex]\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\[/tex]

[tex]\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})[/tex]

L.H.S = R.H.S

Answer:

6 amps

Explanation:

current =v÷r

12v÷2 ohms= 6amps

Answer:

The answer is "Neither".

Explanation:

Engine cranking implies that certain energy is applied to both the starter motor to motivate the engine only with one function: crank the wind-shaft of the engine and create an appropriate compressor in at most one piston for a good activation.

In this question, both Technician was wrong because the pressure motor oil is being used, a motor can crank very slowly or can't start throughout the cold weather.

Answer:

13.335 CM (1 ft, 1.335 cm)

I am 80% sure this is the answer, but i am not too keen on math so if i am wrong let me know and i will try my best to fix it!

I hope this helped! Have a good day :]

Answer:

D would be correct because it maximizes it.

Answer:

Fourier coefficient, Ck:

[tex]{ \rm{ c_{k}} = \int ^{ \infin} _{ - \infin } \{x(t) {e}^{jk \omega _{0} t} \} }[/tex]

Fourier transform:

[tex]{ \rm{x( \omega) = \int ^{ \infin} _{ - \infin} } \{x(t) {e}^{j \omega} t} \}[/tex]

Answer:

15psig

Explanation:

The probability that the wavelengths of the next photosynthetically active radiations is at most 82 nanometer is 0.35.

Photosynthetically active radiation (PAR) is light with wavelengths ranging from 400 to 700 nm that is used by plants for photosynthesis.

The photon flux density of photosynthetic photon flux density (PPFD) is defined as the photon flux density of PAR.

Photosynthesis and plant growth require Photosynthetically Active Radiation. Higher PAR promotes plant growth, and monitoring PAR is critical to ensure plants receive enough light for this process.

Given that,

a = 75

b =95

Using uniform distribution

P(x c) = (c - a) / (b - a)

P(x 82) = (82 - 75) / (95 - 75)=0.35

Probability=0.35

Thus, the probability of the given scenario is 0.35.

For more details regarding photosynthetically active radiations, visit:

https://brainly.com/question/10582848

#SPJ1

Answer:

[tex]2.9\times 10^{-6}[/tex]

Explanation:

[tex]Q_s[/tex] = Energy for defect formation = 1.86 eV

T = Temperature = [tex]573^{\circ}\text{C}=573+273.15=846.15\ \text{K}[/tex]

k = Boltzmann constant = [tex]8.62\times 10^{-5}\ \text{eV/K}[/tex]

The fraction of lattice sites that are Schottky defects is given by

[tex]\dfrac{N_s}{N}=e^{-\dfrac{Q_s}{2kt}}\\\Rightarrow \dfrac{N_s}{N}=e^{-\dfrac{1.86}{2\times 8.62\times 10^{-5}\times 846.15}}\\\Rightarrow \dfrac{N_s}{N}=2.9\times 10^{-6}[/tex]

The required ratio is [tex]2.9\times 10^{-6}[/tex].

A percentage increase from 80 to 100 is equal to 25%.

A percentage can be defined as any number that is expressed as a fraction of hundred (100). This ultimately implies that, a percentage indicates the hundredth parts of any given number.

In Mathematics, there are two main types of percentage and these include the following:

Mathematically, percentage increase can be calculated by using this formula:

Percentage increase = [Final value - Initial value]/Initial value × 100

Substituting the given parameters into the formula, we have;

Percentage increase = [100 - 80]/80 × 100

Percentage increase = [20]/80 × 100

Percentage increase = 1/4 × 100

Percentage increase = 25%.

Read more on percentage increase here: brainly.com/question/11360390

#SPJ1

Answer: capacitance decreases as distance increases

Explanation:  when the plates on the capacitor are being pulled what happens is it will decrease in capacitance so the charge will decrease in the capacitor and because the battery is pulled away and if the plate is fully charged the capacitance as the distance increases between the plates the capacitance drops and heres something else i found on it and the capacitance  

Capacitance C∝d1 when plates of a capacitor are moved farther, the capacitance decreases. After disconnecting the battery, the charge on capacitor remains constant, therefore the energy stored by capacitor U (=2Cq), increases

Answer:

You can see there are four different types of measurement scales (nominal, ordinal, interval and ratio).

Some Important Formulas:

Hope this helps!

Don't forget to mark me as Brainliest.