Define a named tuple Player that describes an athlete on a sports team. Include the fields name, number, position, and team.

Answer:

[tex]F=6200\ \text{N}\\[/tex]

Explanation:

In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:

In this problem the flexural strength is defined with the following formula:

[tex]\sigma=\cfrac{3FL}{2bd^2}[/tex]

where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.

The force is then defined as:

[tex]F=\cfrac{2\sigma bd^2}{3L}=6200\ \text{N}[/tex]

In this exercise we have to use electronic knowledge to calculate the voltage value from the resistance. In this way we can conclude that:

So from the data given in the text, we have that:

So with the formula given below, we can calculate what is being asked by it:

[tex]Peak \ to \ peak \ ripple = I (load)/(f)(c)[/tex]

They are using the data previously informed and putting it in the given formula, we would be with:

[tex]I (load) = load \ current = 30/600 = 0.05 A\\Peak \ to \ peak \ ripple = 0.05/6\\= 8.33mV[/tex]

The average Dc something produced power for a complete wave exist double :

[tex]Vd_c= 0.637 * 30\\Vd_c =19.11V[/tex]

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Answer:

The question continues ; Determine the expressions for the velocity and acceleration of a point P as a function of time t, and determine the maximum velocity of point P during one cycle. Use the values r = 75mm and w = pie-rads/s

Explanation:

The diagram and the detailed step by step explanation is as shown in the attachment

In the Scotch-yoke mechanism, if the displacement is given by x = r sin(t), the velocity v = r cos(t) and acceleration a = -r sin(t). The negative sign in acceleration indicates that it is in the opposite direction to displacement.

The Scotch-yoke mechanism which converts rotary motion into reciprocating motion can be analyzed using principles of kinematics. If we have the displacement given by x = r sin(t), the velocity and acceleration can be derived from this displacement equation.

The velocity (v) is the time derivative of the displacement function, i.e., v = dx/dt = r cos(t).

The acceleration (a) is the time derivative of the velocity function, so a = dv/dt = -r sin(t).

The negative sign signifies that acceleration is in the opposite direction to displacement.

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Answer:

(a) Current density at P is [tex]J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\[/tex].

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

[tex]J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\[/tex]

where [tex]\textbf{a}_{\rho}[/tex] and [tex]\textbf{a}_{\phi}[/tex] unit vectors.

(a) In order to find the current density at a specific point (P), we can simply replace the coordinates in the current density equation.  Therefore

[tex]J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\[/tex]

(b) Total current flowing outward can be calculated by using the relation,

[tex]I=\int {\textbf{J} \, \textbf{ds}[/tex]

where integral is calculated through the circular band given in the question. We can write the integral as below,

[tex]I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\[/tex]

due to unit vector multiplication. Then,

[tex]I=10\int\(\rho^3z.dz.d\phi[/tex]

where [tex]\rho=3,\ 0<\phi<2\pi, \ 2<z<2.8[/tex]. Therefore

[tex]I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A[/tex]

Following are the solution to the given points:

Calculating the current density:

 [tex]\bold{J = 10 \rho^2 z a_{\rho} − 4 \rho \cos^2 \phi a_{\phi} \frac{mA}{m^2}}{}[/tex]

For point a:

Calculating the current density at point P:  

[tex]\to p=2 \\\\ \to \phi =60^{\circ}\\\\ \to Z=3[/tex]

[tex]\to J= 10(2)^2 3 (3) a_{\rho} - (4\times 2 \times ( \cos 60^{\circ})^2) a_{\phi}\\\\[/tex]

       [tex]=120 \ a_{\rho} - 4 \times 2 \times \frac{1}{2^2} a_{\phi} \\\\= 120 a_{\phi} - 2 a_{\phi} \frac{mA}{m^2}\\\\[/tex]

For point b:

Calculating the total current:  

[tex]\to I=\int \int \vec{J} \cdot \vec{ds}\\\\[/tex]

       [tex]= \int \int (10 \rho^2 z a_{\rho} − 4 \rho \cos^2 \phi a_{\phi}) \cdot (\rho d \phi \cdot dz) a_{\rho}[/tex]

Note:  

[tex]\text{p=constont then} \vec{ds} = \rho d \phi dz \hat{a}_{\rho} \\\\[/tex]

[tex]I= \int \int (10 \rho^2 z) \cdot ( d \phi dz)[/tex]

  [tex]= (10 \rho^3 \int^{8}_{3} z dz \int^{2 \pi}_{\pi} 1 \cdot d \phi\\\\= (10 \times 4^3 [\frac{z^2}{2}]^{8}_{3} \times [\phi]^{2 \pi}_{\pi}\\\\= 10 \times 64 \times \frac{(64-9)}{2} \times (2\pi-\pi) MA \\\\= 55292 \ MA \\\\= 55.292\ A[/tex]

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Question:

Current density is given in cylindrical coordinates as J = −10^6z^1.5az A/m² in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0.

(a) Find the total current crossing the surface z = 0.1 m in the az direction.

(b) If the charge velocity is 2 × 10^6 m/s at z = 0.1 m, find ρν there.

(c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.

Answer:

a. -39.8μA

b. -15.81mC/m³

c. 29.05m/s

Explanation:

Given

Density = J = −10^6z^1.5az A/m²

Region: 0 ≤ ρ ≤ 20 µm

ρ ≥ 20 µm

J = 0.

a. Total current is calculated by.

J * ½((ρ1)² - (ρ0)²) * 2 π * φdza.

Where J = Density = -10^6 * z^1.5

ρ1 = Upper bound of ρ = 20

ρ0 = Lower bound of ρ = 0

π = 22/7

φdza = 10^-6

z = 0.1

Total current

= -10^6 * z^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= 10^6 * 0.1^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= −39.7543477278310

= -39.8μA

b. Calculating velocity charge density at (ρv)

Density (J) = ρv * V

Where J = Density = -10^6 * z^1.5

V = 2 * 10^6

z = 0.1

Substitute the above values

-10^6 * 0.1 ^1.5 = ρv * 2 * 10^6

ρv = (-10^6 * 0.1^1.5)/(2 * 10^6)

ρv = -0.1^1.5/(2)

ρv = -0.015811388300841

ρv = -0.01581 --------- Approximated

ρv = -15.81mC/m³

c. Calculating Velocity

Velocity = J/V

Where Velocity Charge Density = -2000 C/m3

Where J = -10^6 * z^1.5

z = 0.15

J = -10^6 * 0.15^1.5

J = -58094.75019311125

Velocity = -58094.75019311125/-2000

Velocity = 29.047375096555625m/s

Velocity = 29.05m/s

A) The total current crossing the surface z = 0.1 m in the z^ direction is; I_tot ≈ -39.8μA

B) If the charge velocity is 2 × 10⁶ m/s, then ρv is; -15.81 mC/m³

C) If the volume charge density at z = 0.15 m is −2000 C/m³, the Charge velocity is; 29.05m/s

We are given;

Current Density; J = −10⁶z^(1.5) (z^) A/m²

Region: 0 ≤ ρ ≤ 20 µm

At ρ ≥ 20µm,  J = 0.

I_tot = J × ½((ρ1)² - (ρ0)²) × 2π × φdza.

Where;

J is current Density = −10⁶z^(1.5) A/m²

ρ1 is Upper bound of ρ = 20 µm

ρ0 is Lower bound of ρ = 0 µm

 φdza = 10⁻⁶

z = 0.1

Thus plugging in the relevant values, we have;

Total current;

I_tot = -10⁶ × 0.1^(1.5) * ½(20² - 0²) × 2 × π × 10⁻⁶

I_tot ≈ -39.8μA

ρv = J/V

where;

J is current density = −10⁶z^(1.5) A/m²

 V is charge velocity = 2 × 10⁶ m/s

z = 0.1

Thus;

 ρv = ( −10⁶ × 0.1^(1.5))/(2 × 10^6)

ρv = -¹/₂(0.1^(1.5))

ρv ≈  -0.01581 C/m³

Thus;

ρv = -15.81 mC/m³

Charge Velocity = J/V

Where;

J is current density = −10⁶z^(1.5) A/m²

V is volume charge density = -2000 C/m³

At z = 0.15;

Charge velocity = (−10⁶ × 0.15^(1.5))/(-2000)

Charge velocity ≈ 29.05m/s

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Answer:

The engine's thermal efficiency is 0.32

Explanation:

Thermal efficiency = work done ÷ quantity of heat supplied

Work done = 210 J

Quantity of heat supplied = work done + waste heat = 210 + 440 = 650 J

Thermal efficiency = 210 ÷ 650 = 0.32

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

                           mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

                           BE / BC = 45 / 60 = 0.75

Hence,                E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

                           mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

                           vec(EF) = < -15 , -110 , 30 >

                           unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension            T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

                           mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

                           vec(AD) = < 48 , -12 , 36 >

                           unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

                           unit (AD) = <0.7845 , -0.19612 , 0.58835 >

Next:

                           M_AD = unit(AD) . ( E x T_EF)

                           [tex]M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right][/tex]

                            M_AD = 1835.73 + 116.49528 - 593.0568

                            M_AD = 1359.17 lb-in

The moment of the force about the line joining points A and D is; 617.949 lb.in

We are given;

Force exerted by cable EF at E; T_EF = 46 lb.

From the diagram of the guy wire, we can draw a triangle and we will have the following coordinates;

A(0, 0, 0)

D(48, -12, 36)

E(E_x, 96, E_z)

Also, we can get that;

BC² = 48² + 36²

BC = √(48² + 36²)

BC = 60 in

Also, from similar triangles, we will have the coordinate of E as;

E(36, 96, 27)

Position of Vector of EF is;

EF = {(21 - 36)i + (-14 - 96)j + (57 - 27)k} in

EF = {-15i - 110j + 30k} in

Magnitude of EF from online calculation = 115 in

Force along cable EF is;

F_EF = 46{(-15i - 110j + 30k)/115}

F_EF = {-6i - 44j + 12k} lb

Position vector of AE is {36i + 96j + 27k} in

Position vector of AD is {48i - 12j + 36k} in

Magnitude of AD = 61.188 N

Unit vector of AD; λ_ad = {48i - 12j + 36k}/61.188

λ_ad = 0.7845i - 0.1961j + 0.5883k

M_ad = λ_ad × r_ea × T_EF

M_ad = [tex]\left[\begin{array}{ccc}0.7845&-0.1961&0.5883\\36&96&27\\6&-44&12\end{array}\right][/tex]

Solving this gives;

M_ad = 617.949 lb.in

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The expression for the bonding energy E₀ in terms of the equilibrium interionic separation r₀ and the given constants is;

E₀ = rD[(exp(-r₀/ρ) + exp(-r/ρ)] - EN/r₀

We are given the expression;

EN = -C/r + Dexp(-r/ρ)

where;

EN is net potential energy

r is the interionic separation

C, D, and rho(ρ) are constants whose values depend on the specific material.

The formula for the bonding energy is usually;

E₀ = -C/r₀ + D exp(-r₀/ρ)

Step 1; We are to differentiate EN with respect to r. Thus, we have;

dEN/dr = C/r² - D exp(-r/p)

Step 2; We are to solve for C in terms of D, rho(ρ) and r₀. Thus;

E₀ + (C/r₀) = -D*exp(-r₀/ρ)

⇒ C/r₀ = -Dexp(-r₀/ρ) - E₀

Thus, multiplying both sides by r₀ gives;

C = -r₀(Dexp(-r₀/ρ) + E₀)

Step 3; We are to determine the expression for E₀ by substitution for C in the equation given. This gives us;

EN = -r₀(Dexp(-r₀/ρ) + E₀)/r + Dexp(-r/ρ)

EN = r₀*D*exp(-r₀/ρ) - (r₀E₀/r) + D*exp(-r/ρ)

EN + (r₀E₀/r) = r₀*D*exp(-r₀/ρ) + D*exp(-r/ρ)

r₀E₀/r = D[(exp(-r₀/ρ) + exp(-r/ρ)] - EN

Thus;  E₀ = rD[(exp(-r₀/ρ) + exp(-r/ρ)] - EN/r₀

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The electronics can be operated for approximately 25.19 hours from the fully charged battery, which stores 1.26 kWh of energy. The cost of the energy per kilowatt hour for the battery, given its lifecycle, is around $0.20.

The electronics aboard a sailboat consume 50 watts from a 12.6-V source, and the storage battery used is rated for 12.6 V and 100 ampere hours (Ah). To calculate the number of hours the electronics can be operated without recharging, we use the power consumption and the voltage to find the current drawn by the electronics:

I = P/V = 50W / 12.6V = 3.968 Ah

Thus, the electronics draw approximately 3.97 Ah from the battery. Since the battery has a capacity of 100 Ah, the operating time before a recharge is needed can be calculated as:

operating time = battery capacity / current draw = 100 Ah / 3.97 Ah ≈ 25.19 hours

To find how much energy in kilowatt-hours is stored in the battery:

Energy = Voltage  imes Capacity = 12.6V  imes 100Ah = 1260 Wh = 1.26 kWh

Considering the battery's life of 300 charge-discharge cycles and cost of $75, the cost per kilowatt hour can be calculated as follows:

cost per kWh = total cost / (energy storage  imes number of cycles) = $75 / (1.26 kWh  imes 300 cycles) ≈ $0.20 per kWh

Answer:

[tex]W=-52 800\ \text{J}=-52.8\ \text{kJ}[/tex]

Explanation:

First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.

To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:

[tex]W=\int_{0.42}^{0.12}-1200V+500dV=-52.8\ \text{kJ}[/tex]

Answer:

Advantage of satellite communication are as follow

- it help in mobile and wireless communications

- it can covers the wide area in single time

- installation of satellite communication is easy

- it can be used for any form of communication i.e. video, audio, etc.

- service can be available in uniformly

Explanation:

Advantage of satellite communication are as follow

- it helps in mobile and wireless communications

- it can cover the wide-area in a single time

- installation of satellite communication is easy

- it can be used for any form of communication i.e. video, audio, etc.

- service can be available in uniformly

Distance insensitive communication system is describe in terms of limitation      discovered area. Example of this is mobile and satellite. The mobile discover area is less than satellite.  

a) The water content is 85.3%.

b) The void ratio is approximately 2.31.

c) The degree of saturation is 99.6%.

How do we determine the water content?

a) We shall calculate the water content (w) using the formula:

[tex]\frac{M_{w} }{Ms} * 100[/tex]

where:

[tex]M_{w}[/tex] = mass of water

[tex]M_{s}[/tex] = mass of solids (oven-dried mass)

Given:

[tex]M_{w}[/tex] = [tex]417g -225g = 192g[/tex]

[tex]M_{s}[/tex] = [tex]225g[/tex]

We shall now calculate water content (w):

[tex]w = (192/225) * 100[/tex]

[tex]w = 0.853 * 100[/tex]

[tex]w = 85.3[/tex]%

Thus,  the water content is 85.3%.

b) We shall compute the Void ratio (e) by using the formula:

[tex]e = \frac{V_{v} }{V_{s} }[/tex]

where:

[tex]V_{v} =[/tex] volume of voids

[tex]V_{s} =[/tex] volume of solids

Let us find [tex]V_{s}[/tex] first:

[tex]V_{s} =[/tex] [tex]M_{s} /[/tex]Specific gravity * Density of water

[tex]V_{s} = \frac{225g}{2.70 * 1g/cm^{3} }[/tex]

[tex]V_{s} = \frac{225g}{2.70g/cm^{3} }[/tex]

[tex]V_{s} = 83.33cm^{3}[/tex]

[tex]V_{v} = V_{t} - V_{s}[/tex]

[tex]V_{v} = 276cm^{3} - 83.33cm^{3}[/tex]

= 192.67cm³

We calculate e (Void ratio):

e = [tex]192.67/83.33[/tex]

[tex]e = 2.31.[/tex]

Hence, the void ratio is ≈ 2.31.

c) We can find the degree of saturation (S) using the formula:

[tex]S = \frac{V_{w} }{V_{v} }[/tex]

Given:

[tex]V_{w}[/tex] = 192cm³

And we have [tex]V_{v}[/tex] = 192.67cm³

[tex]S = (192/192.67) * 100[/tex]

[tex]S = 0.996 * 100[/tex]

S = 99.6%

So, the degree of saturation is 99.6%.

To determine the minimum cross-sectional area of the steel cables, calculate the allowable stress and use it along with the provided load. The result is a minimum area of 50 mm² for each cable.

The question involves calculating the minimum cross-sectional area of each steel cable designed to support a load with a safety factor, given the yield stress of the material. First, determine the allowable stress by dividing the yield stress by the factor of safety. In this case, the allowable stress is 200 MPa (400 MPa / 2.0). To find the minimum cross-sectional area (A), use the formula A = P / σ, where P = 10 kN = 10,000 N (the load) and σ (sigma) is the allowable stress in N/m². Convert 200 MPa to N/m² to get 200,000 N/m². Therefore, the minimum cross-sectional area required for each cable is 50 mm² (10,000 N / 200,000 N/m²).

To determine the minimum required diameter of an anchor rod, we must calculate the allowable tensile stress using the factor of safety and the ultimate failure stress, then solve the area formula for a circular cross-section for the rod's diameter.

To calculate the minimum required diameter of an anchor rod designed using Allowable Stress Design and considering a factor of safety (F.S.), we must ensure that the designed stress ( au_{allow}) does not exceed the allowable stress. The allowable stress is derived from the material's ultimate (failure) stress ( au_{fail}) divided by the factor of safety (F.S.). Given the allowable load (F_{allow}) the rod must support and the factor of safety (F.S.), the failure load (F_{fail}) is F_{fail}= F_{allow} \times F.S.

Since  au_{allow} = F_{allow} / A, rearranging to solve for the cross-sectional area (A) needed gives us A = F_{allow} /  au_{allow}. And for a circular cross-section rod, A = \pi (d/2)^{2}, where d is the diameter of the rod.

With the failure stress given as  au_{fail} = 60 MPa and the factor of safety F.S. = 1.6, the allowable tensile stress ( au_{allow}) is  au_{allow} =  au_{fail} / F.S. = 60 MPa / 1.6.

Using the formula A = \pi (d/2)^{2} and the given load F_{allow} = 11 kN, we can solve for d. First, we convert the load to Newtons (since 1 kN = 1000 N), then substitute the values of F_{allow} and  au_{allow} into the area equation and solve for the diameter (d).

Upon calculating d, we obtain the minimum required diameter of the rod that will ensure it supports the given load with the required factor of safety.

Factor of safety: The factor of safety is defined as the ratio of the load that causes failure to the load the member can safely support. In this case, the factor of safety for tension failure is given as 1.6. A factor of safety larger than 1 ensures structural stability.

Calculating required diameter: Using the allowable stress design approach, the minimum required diameter of the rod can be calculated to ensure it can support the load safely without failing in tension. Given the material failure stress and the factor of safety, the required diameter can be determined based on uniform stress distribution.

Engineering analysis: Engineers use factors of safety and stress calculations to design structural elements that can safely support loads. Understanding stress, strain, and material properties are essential in ensuring the structural integrity and safety of a design.

Answer: This is EASY!

Explanation:

To make it easy, you would convert those binary numbers and to denary. And this gives:

84 104 105 115 32 105 115 32 69 65 83 89 33

Then, the denary numbers generated can be converted to ASCII code using this ASCII Table in the attachment below. And the result is: This is EASY!

Answer:

G = 0.424

Explanation:

Ds = ( 0.278tr * V ) + (0.278 * V²)/ ( 19.6* ( f ± G))

Where Ds = stopping sight distance = 415miles = 126.5m

G = absolute grade road

V = velocity of vehicle = 52miles/hr

f = friction = 0 because the road is wet

tr = standard perception / reaction time = 2.5s

So therefore:

Substituting to get G

We have

2479.4G = 705.6G + 751.72

1773.8G = 751.72

G = 751.72/1773.8

G = 0.424

Based on the calculations, the grade of the road is equal to 2.43 %.

Given the following data:

Assumed data:

Mathematically, the stopping distance for an inclined surface with a coefficient of friction is given by this formula:

[tex]SD = 0.278Vt + \frac{V^2}{254} (f \pm 0.01G)[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]135 = 0.278 \times 83.6859 \times 2.5 + [\frac{83.69^2}{254} \times (\frac{3.5}{9.8} + 0.01G)]\\\\135 = 58.14 + [27.58 \times (0.36 + 0.01G)]\\\\135 = 58.14 +9.93+0.2758G\\\\0.2758G=135 - 58.14 -9.93\\\\0.2757G=66.9923\\\\G=\frac{66.93}{0.2757}[/tex]

G = 242.76 ≈ 243

G = 2.43 %

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Answer:

59.9%

Explanation:

R^2 =(-0.774)^2  = 0.599

59.9% of fuel is accounted for

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

In this exercise we have to use the parallel plate and capacitor knowledge to find the values, so:

a) 0.948

b) 117.5µf

Capacitor is a component that stores electrical charges in an electrical field, accumulating an internal electrical charge imbalance.

Given the information that:

Knowing that the formula is;

[tex]P= S sin \theta - (p/cos \theta) sin \theta\\= P tan \theta (cos^{-1} (0.8))\\=2.4 tan(36.87)= 1.8KVAR\\S= 2.4 + j1. 8KVA[/tex]

Continues the calculus we have:

[tex]S1= 1.5 +1.5j KVA\\S1 + S2= S\\2.4+j1.8= 1.5+1.5j + S2\\S2= 0.9 + 0.3j KVA\\S2= 0.949= 18.43 \\Pf= cos(18.43) = 0.948[/tex]

b.) pf to 0.9, a capacitor is needed.

[tex]Pf = 0.9\\Cos \theta= 0.9\\\theta = 25.84\\(WC) V^2= P (tan \theta_1 - tan \theta_2)\\2400 ( tan (36. 87) - tan (25.84)) /2 \pi f * 120^2\\[/tex]

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Answer:

The pipe is open ended and the length of pipe is 3.4 m.

Explanation:

For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below

Equation for nth frequency for open end pipe is given as

[tex]f_n=\frac{nv}{2L}[/tex]

For (n+1)th value the frequency is

[tex]f_{n+1}=\frac{(n+1)v}{2L}[/tex]

Taking a ratio of both equation and solving for n such that the value of n is a whole number

[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\[/tex]

So n is a whole number this means that the pipe is open ended.

For confirmation the  nth frequency for a closed ended pipe is given as

[tex]f_n=\frac{(2n+1)v}{4L}[/tex]

For (n+1)th value the frequency is

[tex]f_{n+1}=\frac{(2n+3)v}{4L}[/tex]

Taking a ratio of both equation and solving for n such that the value of n is a whole number

[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\[/tex]

As n is not a whole number so this is further confirmed that the pipe is open ended.

Now from the equation of, with n=6, v=340 m/s and f=300 Hz

[tex]f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m[/tex]

The value of length is 3.4m.

Answer:

P= 541.5 kW.

Explanation:

Given that

velocity of water after leaving the nozzle ,v= 95 m/s

The mass flow rate of the water , m= 120 kg/s

The power generated P is given as

[tex]P=\dfrac{1}{2}mv^2[/tex]

Now by putting the values in the above equation we get

[tex]P=\dfrac{1}{2}\times 120\times 95^2\ W[/tex]

P=541500  W

The  power in kW will be 541.5 kW.

Therefore the answer will be 541.5 kW

P= 541.5 kW.

The power generation potential of the water jet is 542.25 kW.

To determine the power generation potential of the water jet, we need to calculate the kinetic energy of the water jet and then convert it to power. The kinetic energy of the water jet can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass flow rate of the water and v is the velocity of the water jet. Given that the flow rate is 120 kg/s and the velocity is 95 m/s, we can calculate the kinetic energy to be KE = 0.5 * 120 * 95^2 = 0.5 * 120 * 9025 = 542,250 J/s.

To convert the kinetic energy to power, we divide by the time taken to deliver the energy. Since the flow rate is given in kg/s, we can assume the time taken is 1 second. Therefore, the power generation potential of the water jet is 542,250 J/s, or 542.25 kW.

Answer:

V2 = final volume = 8.3m^3

Explanation:

Given P1 = 445 kPa, V1 = 2.6 m^3, P2 = 140 kPa

From PV = constant; P1V1 =P2V2 , where V2 = final volume

V2 = P1V1/P2

Substituting in the equation ;

V2 = 445 x 2.6 / 140

V2 = final volume = 8.3m^3

Answer: For non-precision approaches, the maximum acceptable descent rate acceptable should be one that ensures the aircraft reaches the minimum descent altitude at a distance from the threshold that allows landing in the touch down zone. Otherwise, a decent rate greater than 1000fpm is unacceptable.

Explanation: For non-precision approaches, a descent rate should be used that ensures the aircraft reaches the minimum decent altitude at a distance from the threshold that allows landing in the touchdown zone (TDZ) . On many instrument approach procedures, this distance is annotated by a visual descent point (VDP) If no VDP is annotated, calculate a normal descent point to the TDZ. To determine the required rate of descent, subtract the TDZ elevation (TDZE) from the final approach fix (FAF) altitude and divide this by the time inbound. For illustration, if the FAF altitude is 1,000 feet mean sea level (MSL), the TDZE is 200 feet MSL and the time inbound is two minutes, an 400 fpm rate of descent should be applied.

A descent rate greater than approximately 1,000 fpm is unacceptable during the final stages of an approach (below 1,000 feet AGL). Operational experience and research shows that this is largely due to a human perceptual limitation that is independent of the airplane or helicopter type. As a result, operational practices and techniques must ensure that descent rates greater than 1,000 fpm are not permitted in either the instrument or visual portions of an approach and landing operation.

Answer:

As a heating engineer and considering a house as a dynamic system , and that without a heater, the average temperature in the house would vary over a 24-h period.

What might you consider for inputs, outputs, and state variables for a simple dynamic model?

State variables: according to the weather conditions of the area where the house was built:

State variable # 1: minimum temperture during a day in an specific season (*4);

State variable # 2: maximum temperature during the day, in an specific season (*4) as well;

State variable # 3: average temperature during the day in an specific season (*4).

That makes 16 state variables all of them in Centigrade degrees.

Input variables:

# 1: one degree over each of the state variables given.

# 2: one degree below each of the state variables, all of them in Centigrade           degrees.

Output variables:

# 1 are the temperatures reached after adding one degree to each of the input variables.

# 2 are the temperatures reached after decreasing one degree, all of them in Centigrade degrees.

How would you expand your model so that it would predict temperatures in several rooms of the house?

I would add output variables in a "Y" system to predict temperatures in several rooms of the house.

How does the installation of a thermostatically controlled heater change your model?

It would change on the "Y" variables as they will get a control system  designed for sensors to produce from some input variables to make the system respond.

Explanation:

State-determined system models using well defined physical systems is of highly interest to engineers.

For flow over a plate, the variation of velocity with vertical distance y from the plate, the correct relation for the wall shear stress is  [tex]\( \tau = \mu (a - 2by) \)[/tex]. The correct option is B.

The wall shear stress ([tex]\( \tau \)[/tex]) for flow over a plate is given by the following relation, assuming the fluid has constant viscosity ([tex]\( \mu \)[/tex]):

[tex]\[ \tau = \mu \frac{du}{dy} \][/tex]

Where:

[tex]\( \mu \)[/tex] = dynamic viscosity of the fluid,

[tex]\( \dfrac{du}{dy} \)[/tex] = rate of change of velocity with respect to the vertical distance (y) from the plate.

In your case, the velocity profile [tex]\( u(y) = ay - by^2 \)[/tex] is given. To find the rate of change of velocity with respect to y, we differentiate [tex]\( u(y) \)[/tex] with respect to y:

[tex]\[ \frac{du}{dy} = a - 2by \][/tex]

Now, substitute this into the formula for wall shear stress:

[tex]\[ \tau = \mu \left( a - 2by \right) \][/tex]

So, the correct relation for the wall shear stress ([tex]\( \tau \)[/tex]) in terms of a, b, and [tex]\( \mu \)[/tex] is:

[tex]\[ \tau = \mu \left( a - 2by \right) \][/tex]

Thus, the correct option is B.

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Your question seems incomplete, the probable complete question is:

Question:

For flow over a flat plate, the velocity variation with vertical distance y from the plate is given by [tex]\( u(y) = ay - by^2 \)[/tex], where a and b are constants. What is the correct expression for the wall shear stress ([tex]\( \tau \)[/tex]) in terms of a b, and dynamic viscosity ([tex]\( \mu \)[/tex])?

A) [tex]\( \tau = \mu (a - by) \)[/tex]

B) [tex]\( \tau = \mu (a - 2by) \)[/tex]

C) [tex]\( \tau = \mu (a + by) \)[/tex]

D) [tex]\( \tau = \mu (a + 2by) \)[/tex]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, [tex]e_{min}[/tex] = 0.51

Void ratio in the loosest state, [tex]e_{max}[/tex] = 0.87

Now,

Dry density, [tex]\gamma_d=\frac{\gamma_t}{1+w}[/tex]

[tex]=\frac{18}{1+0.05}[/tex]

= 17.14 kN/m³

Also,

[tex]\gamma_d=\frac{G\gamma_w}{1+e}[/tex]

here, G = Specific gravity = 2.7 for sand

[tex]17.14=\frac{2.7\times9.81}{1+e}[/tex]

or

e = 0.545

Relative density = [tex]\frac{e_{max}-e}{e_{max}-e_{min}}[/tex]

= [tex]\frac{0.87-0.545}{0.87-0.51}[/tex]

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

or

S = 0.2477

or

S = 0.2477 × 100% = 24.77%

Answer:

γ[tex]_{xy}[/tex] =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²;    1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ[tex]_{xy}[/tex] = displacement / total height

     = 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also,  τ = Gγ[tex]_{xy}[/tex]

By comapring both equations, we get

P/A = Gγ[tex]_{xy}[/tex]   ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

Answer:

a) Power = Av⁴

b) Energy = Av⁴/2 or (A²v⁶)/2m

Explanation:

Given Y = Av³

A = a constant, v = Velocity, Y = system variable.

a) If Y = Force, F, Find Power in the element.

Power is the dot product of Force and velocity and it's a scalar quantity.

i.e. P = F.v = F.v (cos θ) where θ is the angle between the force and velocity vector.

But in this case, average power is simply given by Fv.

P(avg) = Fv = Yv = (Av³) × v = Av⁴

b) If Y = linear momentum, p, Find the energy stored in the element.

Energy is related to linear momentum by the relationship between kinetic energy and linear momentum.

p = mv and E = mv²/2 = (mv)(v)/2, so,

E = pv/2

For this question, p = Y = Av³

E = Yv/2 = (Av³)v/2 = Av⁴/2

Kinetic energy is often related to momentum through this expression too,

p = mv; p² = m²v²

E = mv²/2; E = (mv²/2) × (m/m) = m²v²/2m = p²/2m

Therefore, E = Y²/2m = (Av³)²/2m = (A²v⁶)/2m

Hope this helps!

Defining the radius of area throat to inlet we would have that the proportional relationship would be [tex]\frac{A_2}{A_1}.[/tex]

This equation or relationship is obtained from continuity where

[tex]A_1V_1 = A_2V_2[/tex]

[tex]\frac{V_2}{V_1} = \frac{A_2}{A_1}= 0.8[/tex]

Now applying the Bernoulli equation between inlet and throat section we have,

[tex]\frac{p_1}{\rho g}+ \frac{v_1^2}{2g}+z_1=\frac{p_2}{\rho g}+ \frac{v_2^2}{2g}+z_2[/tex]

Here,

[tex]z_1 = z_2[/tex]

Then for a Venturi duct, the velocity of the airplane [tex]V_1[/tex] will be

[tex]V = \sqrt{\frac{2(p_1-p_2)}{\rho[(\frac{A_1}{A_2})^2-1]}}[/tex]

Our values are,

[tex]\frac{A_2}{A_1} = 0.8[/tex]

[tex]\rho = 0.002377slug/ft^3[/tex]

[tex]p_1 = 2116lb/ft^2[/tex]

[tex]p_2 = 2100lb/ft^2[/tex]

Replacing,

[tex]V= \sqrt{\frac{2(2116-2100)}{(0.002377)[(\frac{1}{0.8})^2-1]}}[/tex]

[tex]V = 154.7ft/s[/tex]

Therefore the velocity of the airplane is 154.7ft/s

Answer:

Your stratified squamous epithelium is difficult to penetrate!

Explanation:

The epithelium is the tissue formed by one or several layers of cells attached to each other that cover all the free surfaces of the organism, and constitute the internal lining of the cavities, organs, hollows, ducts of the body and skin and also form the Mucous and glands. The epithelia also forms the parenchyma of many organs, such as the liver.

Flat or squamous epithelia: Formed by flat cells, with much less height than width and a flattened nucleus. It is one of the most external being part of the epidermis and generating some inconveniences when penetrating with needles to perform blood extractions when you have certain characteristics of hardness.

Answer:

The fluid level difference in the manometer arm = 22.56 ft.

Explanation:

Assumption: The fluid in the manometer is incompressible, that is, its density is constant.

The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

And P(gage) = ρgh

ρ = density of the manometer fluid = 60 lbm/ft³

g = acceleration due to gravity = 32.2 ft/s²

ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

h = fluid level difference between the two arms of the manometer = ?

P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²

1353.6 = ρg × h = 60 lbf/ft³ × h

h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!