The real power delivered by a source to two impedances, ????1=4+????5⁡Ω and ????2=10⁡Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current.

The time that the maximum queue length occurs would be c. 49.4 min

To find the time at which the maximum queue length occurs, we need to determine when the arrival rate equals the departure rate.

The queue length increases when the arrival rate exceeds the departure rate and decreases when the departure rate exceeds the arrival rate. The maximum queue length occurs when the arrival rate equals the departure rate.

Substituting the given functions into the equation, we get:

[tex]\[ 1.8 + 0.25t - 0.0030t^2 = 1.4 + 0.11t \][/tex]

Rearranging the terms, we get a quadratic equation:

[tex]\[ -0.0030t^2 + 0.25t - 0.11t + 1.8 - 1.4 = 0 \][/tex]

[tex]\[ -0.0030t^2 + 0.14t + 0.4 = 0 \][/tex]

Now, we can solve this quadratic equation to find the value(s) of t at which the maximum queue length occurs. We can use the quadratic formula:

[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Where:

a = -0.0030

b = 0.14

c = 0.4

Calculate the values of t using this formula. Then, we'll choose the appropriate value based on the physical meaning of the problem.

Using the quadratic formula:

[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

[tex]\[ t = \frac{-0.14 \pm \sqrt{(0.14)^2 - 4(-0.0030)(0.4)}}{2(-0.0030)} \][/tex]

[tex]\[ t = \frac{-0.14 \pm \sqrt{0.0196 + 0.0048}}{-0.0060} \][/tex]

[tex]\[ t = \frac{-0.14 \pm 0.156}{-0.0060} \][/tex]

Now, we have two possible values for t :

1. [tex]\( t_1 = \frac{-0.14 + 0.156}{-0.0060} \)[/tex]

2. [tex]\( t_2 = \frac{-0.14 - 0.156}{-0.0060} \)[/tex]

Calculate these values:

1. [tex]\( t_1 = \frac{0.016}{-0.0060} = -2.67 \)[/tex]

2. [tex]\( t_2 = \frac{-0.296}{-0.0060} = 49.4[/tex]

Since t represents time, it cannot be negative. The maximum queue length occurs at approximately t = 49.4 minutes after the beginning of the observation.

Answer:

Super insulation are obtained by using layers of highly reflective sheets separated by glass fibers in an vacuumed space. Radiation heat transfer between any of the surfaces is inversely proportional to the number of sheets used and thus heat lost by radiation will be very low by using these highly reflective sheets which will an effective way of heat transfer.

Explanation:

Answer

given,

oil barrel weight  = 1.5 k N = 1500 N

weight of the barrel = 110 N

Assuming volume of barrel = 0.159 m³

weight of oil = 1500-110

                     = 1390 N

[tex]specific\ weight = \dfrac{weight}{volume}[/tex]

[tex]specific\ weight = \dfrac{1390}{0.159}[/tex]

            = 8742.14 N/m³

[tex]mass = \dfrac{weight}{g}[/tex]

[tex]mass = \dfrac{1390}{9.8}[/tex]

              = 141.84 kg

[tex]density = \dfrac{mass}{volume}[/tex]

[tex]density = \dfrac{141.84}{0.159}[/tex]

                    = 892.05 kg/m³

[tex]Specific\ gravity = \dfrac{density\ of\ oil}{density\ of\ water}[/tex]

[tex]Specific\ gravity = \dfrac{892.05}{1000}[/tex]

                    = 0.892

Answer:

(a) Aluminum Indium Gallium Phosphide (AlInGaP). Band gap = 1.81eV  ≈ 2eV

(b) Gallium Nitride (GaN). Band Gap = 3.4eV

(c) Aluminium Gallium Arsenide (AlGaA). Band Gap = 1.42eV ≈ 2.16eV

(d) Zinc Selenide (ZnSe). Band Gap = 2.82eV

(e) Gallium Phosphide (GaP). Band Gap = 2.24eV

Explanation:

LED's are semi-conducting materials that convert electrical energy to light energy. The light color emitted from the LED depends on the semi-conducting material and other compositions.

The band gap of the semi-conductor determines its wavelength. High band gap semi-conductors emit lower wavelengths which means greater power(UV semi-conducting macterials fall under this category).

Answer:

Dendrogram

Explanation:

Dendrogram is referred to as the tree structure that represents the hierarchy between the object in a cluster. it is also referred to as an output that is drawn from clustering.

The dendrogram is interpreted by observing the object situated at the higher side in a scatter plot. By joining the object at the same level it represents the order of cluster. another purpose of modifying form of Dendrogram is to help in calculating the distance between the objects in clusters.

To calculate the theoretical density of Niobium with a BCC structure, we use its atomic radius and atomic weight, converting these into the cube's edge length using the BCC relation, then apply the density formula.

To calculate the theoretical density of Niobium (Nb), which has a body-centered cubic (BCC) crystal structure, we first use the known values: atomic radius = 0.143 nm (or 0.143 × 10-9 m) and atomic weight = 92.91 g/mol. The formula for the density (ρ) in a BCC structure is ρ = (2 × M) / (a3 × NA), where M is the atomic mass, a is the edge length of the cube, and NA is Avogadro's number (6.022 × 1023 atoms/mol).

Since it's a BCC structure, the atomic radius relates to the cube's edge length (a) as a = 4r / √3. Substituting the given atomic radius, we find a = 4 * 0.143 × 10-9 m / √3. Then, to find the density, we substitute M (92.91 g/mol), a, and NA into the density formula. This calculation will give us the theoretical density of Niobium in g/cm3.

To determine the rate of heat supply by the fan-motor assembly, the electrical input power is calculated based on the 0.25-hp shaft output and 60% efficiency of the motor. The resulting heat supply to the room is the same as the electrical power input, which is 310.83 watts.

The question asks to determine the rate of heat supply by a fan-motor assembly used to circulate chilled water through a heat exchanger for cooling a room. Given that the fan has a shaft output of 0.25 horsepower (hp) and that small electric motors driving such equipment typically have an efficiency of 60 percent, we can calculate the electrical power input needed to run the fan.

The electrical power input (Pinput) can be calculated as:

Pinput = Poutput / Efficiency

Where Poutput is the shaft output power (0.25 hp) and 'Efficiency' is the efficiency of the electric motor (60%, or 0.60 in decimal form).

Pinput = (0.25 hp) / 0.60

To convert horsepower to watts, we use the conversion factor 1 hp = 746 watts.

Pinput = (0.25 hp × 746 watts/hp) / 0.60

Pinput = 310.83 watts (rounded to two decimal places)

The rate of heat supply to the room will be equal to the electrical power input to the fan, which is 310.83 watts. This accounts for both the useful work done and all inefficiencies in the system that convert electrical energy into heat.

Answer:

T = 5416.67 N

T = -2083.5 N

T = 0

Explanation:

Forward thrust has positive values and reverse thrust has negative values.

part a

Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

[tex]T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (150 - 41.67)\\\\T = 5416.67 N[/tex]

Answer: The thrust force T = 5416.67 N

part b

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:

[tex]T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (0 - 41.67)\\\\T = -2083.5 N[/tex]

Answer: The thrust force T = -2083.5 N reverse direction

part c

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.

Answer: T = 0 N

Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = [tex]\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}[/tex]

a)(100)

a=5.28 Å

Distance = [tex]\frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}[/tex]=5.28 Å

b)(110)

Distance = [tex]\frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}}[/tex] = 3.73 Å

c)(111)

Distance= [tex]\frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}[/tex]= 3.048 Å

Answer:

a. Self declared by the manufacturer

Explanation:

Recycled content refers to the portion of materials used in a product that have been diverted from the solid waste stream. If those materials are diverted during the manufacturing process, they are be referred to as pre-consumer recycled content (sometimes referred to as post-industrial). If they are diverted after consumer use, they are

post-consumer .

Answer:

bypassed fraction B will be B= 0.105 (10.5%)

Explanation:

doing a mass balance of SO₂ at the exit

total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂

F*(1-er) =  Fs*(1-es) + Fb

where

er= required efficiency

es= scrubber efficiency

Fs and Fb = total mass inflow of  SO₂ to the scrubber and to the bypass respectively

F= total mass inflow of  SO₂

and from a mass balance at the inlet

F= Fs+ Fb

therefore the bypassed fraction B=Fb/F is

F*(1-er) =  Fs*(1-es) + Fb

1-er= (1-B)*(1-es) +B

1-er = 1-es - (1-es)*B + B

(es-er) = es*B

B= (es-er)/es = 1- er/es

replacing values

B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)

Answer:

V = 56.8 mV

Explanation:

When a current I flows across a circuit element, if we assume that the dimensions of the circuit are much less than the wavelength of the power source creating this current, there exists a fixed relationship between the power dissipated in the circuit element, the current I and the voltage V across it, as follows:

P = V*I

By definition, power is the rate of change of energy, and current, the rate of change of the charge Q, so we can replace P and I, as follows:

E/t = V*q/t ⇒ E = V*Q

Solving for V:

V = E/Q = 94.2 mJ /1.66 C = 56.8 mV

Answer: 62 cos(50t - 70°)

Explanation:

First we need to convert all sines into cosines because phasor forms are represented through cosine. For that we will use the fact that sin(wt + ∅) = cos(wt + ∅ - 90°)

Therefore, 37sin50t = 37cos(50t - 90°)

Now we have 37cos(50t - 90°)+ 30 cos(50t – 45°). We need to convert them into phasor form to add the terms. For that we will use the fact Acos(wt+∅)=A∠∅ which can be represented using real and imaginary parts as A [cos(∅)+jsin(∅)].

So,

37cos(50t - 90°)

= 37∠-90°

= 37[cos(-90°)+jsin(-90°)

=37[0+j(-1)]

= -j37

Similarly,

30 cos(50t – 45°)

=30∠-45

=30[cos(-45)+jsin(-45)

=30[0.707-j0.707]

=21.21 - j21.21

37cos(50t - 90)+ 30 cos(50t – 45°) = -j37+21.21 - j21.21 = 21.21 - j58.21

Now we need to convert the real and imaginary parts back to cosine form. We will first calculate the magnitude by the formula √a²+b² where a and b are the real and imaginary parts respectively.

Here a=21.21 and b=58.21

magnitude = √(21.21)²+(58.21)²=61.95≅62

For calculating phase ∅ the formula is ∅=inversetan (b/a) where a and b are the real and imaginary parts respectively.

∅=inversetan(-58.21/21.21)

= -69.9°≅-70°

So the final answer is 62cos(50t-70°)

The value of [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] is [tex]\(61.97 \cos(50t - 70^\circ)\)[/tex].

To solve [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] using phasors, we follow these steps:

Step-by-Step Calculation:

1. Express each term as a phasor:

- The term [tex]\(37 \sin 50t\)[/tex] :

    - Convert to cosine form: [tex]\(37 \sin 50t = 37 \cos(50t - 90^\circ)\)[/tex]

    - Phasor form: [tex]\(37 \angle -90^\circ\)[/tex]

  - The term [tex]\(30 \cos(50t - 45^\circ)\)[/tex] :

    - Phasor form: [tex]\(30 \angle -45^\circ\)[/tex]

2. Convert the phasors to rectangular form:

  - [tex]\(37 \angle -90^\circ\)[/tex] :

    [tex]\[ 37 \cos(-90^\circ) + j 37 \sin(-90^\circ) = 0 - j 37 = -j 37 \][/tex]

- [tex]\(30 \angle -45^\circ\)[/tex]:

    [tex]\[ 30 \cos(-45^\circ) + j 30 \sin(-45^\circ) = 30 \left(\frac{\sqrt{2}}{2}\right) - j 30 \left(\frac{\sqrt{2}}{2}\right) = 21.2132 - j 21.2132 \][/tex]

3. Add the rectangular components:

 [tex]\[ -j 37 + (21.2132 - j 21.2132) \][/tex][tex]\[ = 21.2132 - j (37 + 21.2132) \][/tex]

  [tex]\[ = 21.2132 - j 58.2132 \][/tex]

4. Convert the result back to polar form:

  - Magnitude:

   [tex]\[ R = \sqrt{21.2132^2 + 58.2132^2} = \sqrt{449.54 + 3388.78} = \sqrt{3838.32} = 61.97 \][/tex]

  - Angle:

   [tex]\[ \theta = \tan^{-1}\left(\frac{-58.2132}{21.2132}\right) = \tan^{-1}(-2.743) \approx -70^\circ \][/tex]

5. Express the final result:

  Using the positive magnitude, the expression in cosine form is:

  [tex]\[ 37 \sin 50t + 30 \cos(50t - 45^\circ) = 61.97 \cos(50t - 70^\circ) \][/tex]

Explanation:

Please kindly share your problem two with us as to know the actual problem we are dealing with, the question looks incomplete

Answer:

The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²

Explanation:

The electric field intensity due to a thin conducting sheet is given by the following formula:

Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)

From this formula:

Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)

We have the following data:

Electric Field Intensity = 1.5 N/C

Permittivity of free space = 8.85 x 10^-12 C²/N.m²

Therefore,

Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)

Surface Charge Density = 26.55 x 10^-12 C/m²

Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².

The surface charge density on the conductor is; σ = 13.275 × 10⁻¹² C/m²

The formula for surface charge density on a conductor in an electric field just outside the surface of conductor is;

σ = E * ϵ₀

​where;

E is electric field = 1.5 N/C

ϵ₀ is permittivity of space = 8.85 × 10⁻¹² C²/N.m²

Thus;

σ = 1.5 * 8.85 × 10⁻¹²

σ = 13.275 × 10⁻¹² C/m²

Read more about Surface Charge Density at; https://brainly.com/question/14306160

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

 Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

 Eq 3

Finally, solving for unknown pressure as a function of hoop strain:

 Eq 4

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

 Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

 Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

 Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

 Eq 8

For Wheat stone bridge:

 Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

To determine the final volume and work done during the heating of water in a piston-cylinder assembly at constant pressure, tabulated data like steam tables are required since water is not an ideal gas. The work done is calculated using the formula W = PΔV.

The student has been asked to find the final volume and the work done during a constant pressure process in which 5 kg of water is heated in a piston-cylinder assembly from an initial state of 5 bar and 300°C. To solve for the final volume and work done, one would typically use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat input minus the work output. However, since water at this state is not an ideal gas, tabulated data from steam tables or software would be used to determine the specific volume at the final state and then multiplied by the mass to find the total volume. The work done in a constant pressure process is equal to the pressure times the change in volume (W = PΔV). Without the final specific volume from the tables, we cannot compute the final volume or work directly.

Answer:

a) 69,630KW

b) 203 KW

Explanation:

The data obtained from Tables A-4, A-5 and A-6 is as follows:

[tex]h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2} = 255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\[/tex]

[tex]x_{4} = \frac{s_{4} - s_{f} }{s_{fg} } \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\[/tex]

The power produced and consumed by turbine and pump respectively are:

[tex]W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW[/tex]

Answer: (a). W = 4KJ and Q = 4.25KJ

(b). W = -2KJ and ΔPE = 2KJ

Explanation:

(a).

i. We are asked to calculate the work done during the expansion process considering gas as system.

from W = [tex]\int\limits^a_b {p} \, dV[/tex] where a = V₂ and b = V₁

so W = P(V₂-V₁)

W = (2 × 10²) (0.12 - 0.10)

W = 4 KJ

ii.  We apply the energy balance to gas as system

given Q - W = ΔE

Where ΔE = ΔU + ΔKE + ΔPE

since motion of the system is constrained, there is no change in both the potential and kinetic energy i.e. ΔPE = ΔKE = 0

∴ Q - W = ΔU

Q = ΔU + W

Q = 0.25 + 4

Q = 4.25 KJ

(b).

i. to calculate the work done during the expansion process considering piston as system;

W = [tex]\int\limits^a_b {(Patm - Pgas)} \, dV[/tex]where a and b represent V₂ and V₁ respectively.

W = (Patm - Pgas)(V₂ - V₁)

W = (1-2) ×10² × (0.12-0.1)

W = -2KJ

ii. We apply the energy balance to gas as system

given Q - W = ΔE

Where ΔE = ΔU + ΔKE + ΔPE

Q = 0 since the piston and cylinder walls are perfectly insulated.

for piston, we neglect the change in internal energy and kinetic energy

ΔU = ΔKE = 0

from Q - W = ΔU + ΔKE + ΔPE

0 - (-2) = 0 + 0 + ΔPE

∴ ΔPE = 2KJ

Answer:

The element that is oxidized is carbon.

Its oxidation state increased. It increased from -4 to +4

Explanation:

Oxidation is a process that involves increase in oxidation number.

The oxidation number of carbon in CH4 is -4

C + (1×4) = 0

C + 4 = 0

C = 0 - 4 = -4

The oxidation number of carbon in CO2 is +4

C + (2×-2) = 0

C - 4 = 0

C = 0+4 = 4

Increase in the oxidation number of carbon from -4 to +4 means carbon is oxidized

Answer:

5.994 V

Explanation:

The pressure as a function of hoop strain is given:

[tex]P = \frac{4*E*t}{D}*\frac{e_{h} }{2-v}[/tex]

[tex]e_{h} = \frac{D*P*(2-v)}{4*E*t} .... Eq1[/tex]

For wheat-stone bridge with equal nominal resistance of resistors:

[tex]V_{out} = \frac{GF*e*V_{in} }{4} .... Eq2[/tex]

Hence, input Eq1 into Eq2

 [tex]V_{out} = \frac{GF*e*V_{in}*D*P*(2-v) }{16*E*t} .....Eq3\\[/tex]

Given data:

P = 253313 Pa

D = d + 2t = 0.09013 m

t = 65 um

GF = 2

E = 75 GPa

v = 0.33

Use the data above and compute Vout using Eq3

[tex]V_{out} = \frac{2*6*0.09013*253313*(2-0.33) }{16*75*10^9*65*10^-6} \\\\V_{out} = 0.006285 V\\\\change in V = 6 - 0.006285 = 5.994 V[/tex]

Answer:

The score for both exams 88 is above the 90 percentile, so then Brynne qualified for honors. See the explanation below.

Explanation:

Assuming the following question:"At Westtown High School, the mean score on the French final examination was 81 with a standard deviation of 5, while the mean score on the Spanish final examination was 72 with a standard deviation of 12.

To earn a language honor at graduation, students must score in the 90th percentile on all their language final exams. Brynne scored 88 on both the French exam and the Spanish exam. Is Brynne qualified for honors?"

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

French case

Let X the random variable that represent the scores for the French exam of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(81,5)[/tex]  

Where [tex]\mu=81[/tex] and [tex]\sigma=5[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.1[/tex]   (a)

[tex]P(X<a)=0.90[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=1.28<\frac{a-81}{5}[/tex]

And if we solve for a we got

[tex]a=81 +1.28*5=87.4[/tex]

So the value for the scores that separates the bottom 90% of data from the top 10% is 87.4 (90th percentile).

And since the score of Brynne is 88 is above the 90 percentile  

Spanish case

Let X the random variable that represent the scores for the Spanish exam of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(72,12)[/tex]  

Where [tex]\mu=72[/tex] and [tex]\sigma=12[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.1[/tex]   (a)

[tex]P(X<a)=0.90[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=1.28<\frac{a-72}{12}[/tex]

And if we solve for a we got

[tex]a=72 +1.28*12=87.36[/tex]

So the value for the scores that separates the bottom 90% of data from the top 10% is 87.36 (90th percentile).

And since the score of Brynne is 88 is above the 90 percentile  

Answer with explanation:

As is the question the answer would be 19 hours, and the key to solving it is in the phrase in 15 more hours, basically what they are saying is that the small pipe takes 15 hours more than both the big and the small to fill the tank. Since both pipes working together can fill the tank in 4 hours we need to add 4 and 15 to solve the problem.

If the question is how many hours would it take to fill the tank using only the big pipe? Then we could solve t for the following equation:

[tex]\frac{1}{4+15} + \frac{1}{t} = \frac{1}{4}[/tex]

Getting as a result: 5.06

Note that the equation is the result of taking the rate of the small pipe (what we solved before), plus the unknown rate of the big one equals the rate of both.

Answer:

a)29.9 b) 9.81

Explanation:

Wt% = mass of solute / mass of solvent × 100

0.749 = mass of solute / mass of solvent

a) Mass of perchloric acid = 0.749 × 39.1 = 29.29

b) Mass of water = 39.1 - 29.29 = 9.81

The calculation of the exit temperature of water in the heated pipe involves using the energy balance equation, considering the heat lost through the pipe walls by convection, and then finding the change in thermal energy of the water via the heat transfer equation to solve for the exit temperature.

Exit Temperature of Water in a Heated Pipe

To determine the exit temperature of water at the end of an over-pressurized heated pipe, we must consider the energy balance for the water flowing through the pipe. Based on the first law of thermodynamics, the change in thermal energy of the water will be equal to the heat lost through the pipe walls by convection:

Q = mc_p
(Exit Temperature - Inlet Temperature)

In this case, Q will be negative, since the water is losing heat to the surrounding air. The heat transfer from the pipe to the air is given by:

Q = hA(T_surface - T_air)

The area A for heat transfer is the external surface area of the pipe (
2
classes Math.PI
* radius * length of the pipe). Since we have the heat transfer coefficient h, the surface temperature of the pipe T_surface, and the air temperature T_air, we can calculate Q. Then we can use the mass flow rate m and the constant pressure specific heat c_p to find the exit temperature of the water.

To solve, we first calculate Q, then rearrange the first equation to solve for the Exit Temperature.

Answer:

Answer: No of anti-nodes increases

Answer: wavelength remains same.

Answer: fundamental frequency decreases

Explanation:

a)

The number of nodes (n) would have (n-1) anti-nodes.

The relation of Length of string with n is given below:

[tex]L = \frac{n*lambda}{2}[/tex]

Hence, n and L are directly proportional so as string length increases number of nodes and anti-nodes also increases.

Answer: No of anti-nodes increases

b)

Wavelength is dependent on the frequency:

[tex]lambda = \frac{v}{f}[/tex]

The speed v of the string remains same through-out and frequency of generator is unchanged!

Hence according to above relationship lambda is unchanged.

Answer: wavelength remains same.

C

Fundamental frequency equates to 1st harmonic that 2 nodes and 1 anti-node. Wavelength is = 2*L

Hence, if L increases wavelength increases

Using relation in part b

As wavelength increases fundamental frequency decreases

Answer: fundamental frequency decreases

Answer:

00111111101000000000000000000000

Explanation:

View Image

0   01111111   01000000000000000000000

The first bit is the sign bit. It's 0 for positive numbers and 1 for negative numbers.

The next 8-bits are for the exponents.

The first 0-126₁₀ (0-2⁷⁻¹) are for the negative exponent 2⁻¹-2⁻¹²⁶.

And the last 127-256₁₀ (2⁷-2⁸) are for the positive exponents 2⁰-2¹²⁶.

You have 1.25₁₀ which is 1.010₂ in binary. But IEEE wants it in scientific notation form. So its actually 1.010₂*2⁰

The exponent bit value is 127+0=127 which is 01111111 in binary.

The last 23-bits are for the mantissa, which is the fraction part of your number. 0.25₁₀ in binary is 010₂... so your mantissa will be:

010...00000000000000000000

Answer:

W = 1.7593 * 10 ^ (-7) KJ

Explanation:

The work done by the bubble is given:

[tex]W = sigma*\int\limits^2_1 {} \, dA \\\\W = sigma*( {A_{2} - A_{1} } ) \\\\A = pi*D^2\\\\W = sigma*pi*(D^2_{2} - D^2_{1})\\\\W = 0.07 * pi * (0.03^2 - 0.01^2)*10^(-3)\\\\W = 1.7593 *10^(-7) KJ[/tex]

Answer: W = 1.7593 * 10 ^ (-7) KJ

Answer:

C = 132.5 MPa

R = 86.20 MPa

Explanation:

Given

σx = 175 MPa

σy = 90 MPa

τxy = 75 MPa

For the given state of stress at a point in a structural member, determine the center C and the radius R of Mohr’s circle.

We apply the following equation for the center C

C = (σx + σy) / 2

C = (175 MPa + 90 MPa) / 2

C = 132.5 MPa

The Radius can be obtained as follows

R = √(((σx - σy) / 2)² + (τxy)²)

R = √(((175 MPa - 90 MPa) / 2)² + (75 MPa)²)

R = 86.20 MPa

Answer:

The resistance (R) of the resistor is 2.4 ohm

The maximum voltage (V) is 4.33V

The maximum current (I) is 1.80A

Explanation:

Resistance (R) = resistivity×length/area

Resistivity = 0.003 ohm meter, length = 2mm = 2/1000 = 0.002m, width = 1.25mm = 1.25/1000 = 0.00125m, height = 0.5mm = 0.0005m, area = width × height = 0.00125m × 0.0005m = 6.25×10^-7m^2

R = 0.003×0.002/6.25×10^-7 = 3.2 ohm

Power (P) = V^2/R

V^2 = P × R = 7.81 × 3.2= 24.992

V = √24.992 = 4.99V

P = IV

I = P/V = 7.81/4.99 = 1.57A