Urgent!!!
List the assumptions that can be taken into account in torsion analysis.​

Answer:

The answer is "[tex]\$29.7072[/tex]".

Explanation:

Please find the complete question in the attachment file.

As the original prices are 24 euros, [tex]\$ 1.2378[/tex] must be multiplied by 24.  

[tex]\to 1.2378 \times 24 = \$29.7072[/tex]

Therefore the new price value will be [tex]\$ 29.7072[/tex]

This question is incomplete, the complete question is;

Plexiglas, polymethylmethacrylate (PMMA), is often used as display windows and cases in art galleries and museums. If its flame imparts 32 kW/m^2 to the surface when it burns estimate the energy release rate of a 3 m^2 square sheet (one side). Assume its vaporization temperature is 350°C. Calculate with units, provide work.

a ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10⁻¹¹ kW/m².K⁴

b ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] -  [tex]q_{rad[/tex]

c ⇒ [tex]m"[/tex] =  [tex]q"[/tex]/L, where L = 1.6 kJ/g

d ⇒ Q =  [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g

Answer:

a) [tex]q_{rad[/tex] = 8.54 kW/m²

b) [tex]q"[/tex] = 23.46 kW/m²

c) [tex]m"[/tex] = 14.6625 g/m²

d) Q = 1095.2888 kJ

Explanation:

Given the data in the question;

[tex]q_{flame[/tex] = 32 kW/m²

Area; A = 3m²

vaporization temperature; T = 350°C = ( 350 + 273 )K = 623 K

Now,

a) ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10¹¹ kW/m².K⁴

we substitute

[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × ( 623 K)⁴

[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × 150644120641 K⁴

[tex]q_{rad[/tex] = 8.54 kW/m²

b) ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] - [tex]q_{rad[/tex]

we substitute

[tex]q"[/tex] = 32 kW/m² - 8.54 kW/m²

[tex]q"[/tex] = 23.46 kW/m²

c) ⇒ [tex]m"[/tex] =  [tex]q"[/tex]/L, where L = 1.6 kJ/g

we substitute

[tex]m"[/tex] =  23.46 / 1.6

[tex]m"[/tex] = 14.6625 g/m²

d) ⇒ Q =  [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g

we substitute

Q =  14.6625 × 3 × 24.9

Q = 1095.2888 kJ

Answer:

Tradicionalmente, se han distinguido 5 tipos de mantenimiento, que se diferencian entre sí por el carácter de las tareas que incluyen:

Explanation:

Mantenimiento Correctivo: Es el conjunto de tareas destinadas a corregir los defectos que se van presentando en los distintos equipos y que son comunicados al departamento de mantenimiento por los usuarios de los mismos.

Mantenimiento Preventivo: Es el mantenimiento que tiene por misión mantener un nivel de servicio determinado en los equipos, programando las intervencions de sus puntos vulnerables en el momento más oportuno. Suele tener un carácter sistemático, es decir, se interviene aunque el equipo no haya dado ningún síntoma de tener un problema.

Mantenimiento Predictivo: Es el que persigue conocer e informar permanentemente del estado y operatividad de las instalaciones mediante el conocimiento de los valores de determinadas variables, representativas de tal estado y operatividad. Para aplicar este mantenimiento, es necesario identificar variables físicas (temperatura, vibración, consumo de energía, etc.) cuya variación sea indicativa de problemas que puedan estar apareciendo en el equipo. Es el tipo de mantenimiento más tecnológico, pues requiere de medios técnicos avanzados, y en ocasiones, de fuertes conocimientos matemáticos, físicos y/o técnicos.

Mantenimiento Cero Horas (Overhaul): Es el conjunto de tareas cuyo objetivo es revisar los equipos a intervalos programados bien antes de que aparezca ningún fallo, bien cuando la fiabilidad del equipo ha disminuido apreciablemente de manera que resulta arriesgado hacer previsiones sobre su capacidad productiva. Dicha revisión consiste en dejar el equipo a Cero horas de funcionamiento, es decir, como si el equipo fuera nuevo. En estas revisiones se sustituyen o se reparan todos los elementos sometidos a desgaste. Se pretende asegurar, con gran probabilidad un tiempo de buen funcionamiento fijado de antemano.

Mantenimiento En Uso: es el mantenimiento básico de un equipo realizado por los usuarios del mismo. Consiste en una serie de tareas elementales (tomas de datos, inspecciones visuales, limpieza, lubricación, reapriete de tornillos) para las que no es necesario una gran formación, sino tal solo un entrenamiento breve. Este tipo de mantenimiento es la base del TPM (Total Productive Maintenance, Mantenimiento Productivo Total).

Answer:

[tex]C_v= 24.942[/tex] J / mol K

Explanation:

Molar Heat Capacity

using the equipartition, the equation :

[tex]$C_v=\frac{d}{2} R$[/tex]

Here, [tex]C_v[/tex] = molar heat capacity

          d = degree of freedom

          R = Universal gas constant

Degree of freedom, d is 6          

Universal gas constant, R = 8.317 J/ mol K

Therefore, the molar heat capacity is :

[tex]$C_v=\frac{d}{2} R$[/tex]

[tex]$C_v=\frac{6}{2} \times 8.314$[/tex]

[tex]C_v= 24.942[/tex] J / mol K

Complete Question

Nitrogen (N2) flows through a pipe, entering at at 4 m/sec at 1000 kPa, 2270C. For a pipe inside diameter of 3 cm, find the volumetric flow rate (m3/sec) and the mass flow rate of the gas (kg/sec) assuming you have an ideal gas Then using your ideal gas mass flow rate find the rate at which enthalpy enters the pipe (kJ/sec) NO Cp, Cv, k permitted

Answer:

[tex]H=9.91kJ/sec[/tex]

Explanation:

From the question we are told that:

Velocity [tex]v=4 m/sec[/tex]

Pressure [tex]P=1000kPa[/tex]

Temperature [tex]T=227 \textdegree C[/tex]

Diameter [tex]d=3cm=>0.03m[/tex]

Generally the equation for volumetric Flow Rate is mathematically given by

[tex]V_r=(\frac{\pi*d^2}{4}v)[/tex]

[tex]V_r=(\frac{\pi*(0.03)^2}{4} *4)[/tex]

[tex]V_r=0.002827m^3/s[/tex]

Generally the equation for mass Flow Rate is mathematically given by

[tex]m_r=\frac{PV_r}{RT}[/tex]

[tex]m_r=\frac{1000*0.002827}{0.297*(227+273)}[/tex]

[tex]m_r=0.019kg/sec[/tex]

Generally the equation for mass Flow Rate is mathematically given by

Using gas Table for enthalpy Value

[tex]T=500K=>h=520.75kg[/tex]

Therefore

[tex]H=mh[/tex]

[tex]H=0.019*520.75[/tex]

[tex]H=9.91kJ/sec[/tex]

Answer:

Explanation:

Two that come to mind:

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

A) Classifying the soil according to USCS system

 ( using 2nd image attached below )

description of sand :

The soil is a coarse sand since  ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Answer:

Explanation:

Be bop

Answer:

  parallel

Explanation:

All components in this circuit are tied in parallel. Each component experiences the same voltage from one terminal to the other. It is a parallel circuit.

Answer:

On my console displays for the ISS visiting vehicles, three units are used. The Americans use pounds per square inch (psi). The Russians use kilopascals (kPa). The Japanese use Torr - millimeters of mercury (mmHg). A fourth unit is simply the atmosphere, or multiples of it. So, for example, sea level air pressure (which is what we use onboard ISS) is defined as 1 atmosphere. That is equivalent to 14.7 psi, 101.3 kPa, or 760 mmHg.Here N represents newton which is SI unit of Force which is same as Kg.m/s2." role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">2.2.

m represent metre which is SI unit of length.

Kg represent Kilogram which is SI unit of Mass.

m2SI" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">2SI2SI unit of Area.

Hope it helps.

Thanks.

Answer:

Pounds per square inch (psi)

Kilopascals (kPa)

Millimeters of mercury (mmHg)

Pascal (Pa)

Megapascal (MPa)

Atmospheric pressure (atm)

Hope this helps!

Answer:

Inlet

Explanation:

Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the INLET of the pipe.

This is because the friction factor is experienced at the highest level when a laminar forced flow is at the tube inlet where the thickness of the boundary layer is zero. Also, the friction factor decreases step by step at a lower rate to the fully augmented value.

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Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N

If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

Answer:

Only Technician A

Explanation:

Continuously variable transmission (CVT) does not shift gears because it simply doesn't have gears. It instead uses 2 pulleys which change constantly and continuously in size, and they are linked by a belt. Continuously variable transmission (CVT) is automatic transmission as mentioned in the question above, largely due to the power the belt continuously transmits, making the vehicle's engine work in the optimum power range.

While Technician B saying, that a continuously variable transmission has a maximum of six distinct gear ranges, which is not correct because continuously variable transmission (CVT) works with an unlimited number of gear ratios within a fixed range.

Answer:

I don't understand the question

Answer:

Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);

Types of Land Reform

Abolition of Intermediaries

The first step taken by the Indian government under land reforms post-independence was passing the Zamindari Abolition Act. The primary reason of a backward agrarian economy was the presence of intermediate entities like, jagirdars and zamindar who primarily focussed on collecting sky-rocketing rents catering to their personal benefits, without paying attention to the disposition of farms and farmers. Abolition of such intermediaries not only improved conditions of farmers by establishing their direct connection with the government but also improved agricultural production.

Regulation of Rents

This was in direct response to the unimaginably high rents which were charged by intermediaries during British rule, which resulted in a never-ending cycle of poverty and misery for tenants. Indian government implemented these regulations to protect farmers and labourers from exploitation by placing a maximum limit on the rent that could be charged for land.

Tenure Security

Legislations were passed in all states of the country to grant tenants with permanent ownership of lands and protection from unlawful evictions on expiry of the lease. This law protects tenants from having to vacate a property immediately after their tenure is over unless ordered by law. Even in that case, ownership can be regained by tenants with the excuse of personal cultivation.

Answer: Biometrics

Explanation:

Dual factor authentication refers to an electronic authentication method whereby a user will only be granted an access to an application or a website after the user has successfully been able to present two pieces of evidence which then grants access to the application or website.

Since the technician wants to implement a dual factor authentication system, the biometrics should be implemented during the authorization stage.

Biometrics refers to the body measurements and the calculations that are related to the characteristics of humans. Biometric authentication is used as a form of identification.

complete Question

A motor driving a 1000-W water pump has a power factor of 0.80 lagging; a second motor driving a 600-W water pump has a power factor of 0.60 lagging assuming the motors are working under 120-Vrms, 60-HZAC. When both motors working together what is the combined power factor? If a 200-μF capacitor is connected to the above system (two motors) what is the new combined power factor?

Answer:

[tex]p.f'=0.960[/tex]

Explanation:

First motor Power [tex]P=1000W[/tex]

First motor Power factor [tex]P.f=0.80[/tex]

Second motor Power [tex]P=600W[/tex]

Second motor Power factor p.f=0.60

Voltage [tex]V=120Vrms[/tex]

Frequency [tex]F=60Hz[/tex]

Capacitor [tex]C=200\mu F[/tex]

Generally power in Var is given as

For First Motor

[tex]Q=\frac{1000}{0.8}\sqrt{1-0.8^2}[/tex]

[tex]Q=750Var[/tex]

For Second Motor

[tex]Q=\frac{600}{0.6}\sqrt{1-0.6^2}[/tex]

[tex]Q=800Var[/tex]

Generally the equation for The Reactive Power is mathematically given by

[tex]Q_c=\frac{V^2}{X_c}[/tex]

Where

[tex]X_c=\frac{1}{2 \pi fc}[/tex]

[tex]X_c=\frac{1}{2 \pi 60*200*10^{-6} }[/tex]

[tex]X_c=13.3[/tex]

Therefore

[tex]Q_c=-\frac{120^2}{13.3}\\\\Q_c=-1085.97j[/tex]

Giving

Total Power Drawn by Supply

[tex]P_t=(1000+j750)+(600+800)-j1085.97[/tex]

[tex]P_t=1600+464.03j[/tex]

Therefore

[tex]p.f'=\frac{1600}{\sqrt{1600^2+464.03^2}}[/tex]

[tex]p.f'=0.960[/tex]

Answer:

a. Covalent modification = Seconds to minutes

b. Allosteric control = Milliseconds

c. Gene expression = Hours

Explanation:

Covalent modifications refer to the addition and/or removal of chemical groups by the action of particular enzymes such as methylases, acetylases, phosphorylases, phosphatases, etc. For example, histones are chromatin-associated proteins covalently modified by enzymes that add methyl groups (histone methylation), acetyl groups (histone acetylation), phosphate groups (histone phosphorylation), etc. Moreover, allosteric control, also known as allosteric regulation, is a type of regulation of the enzyme activity by binding an effector molecule (allosteric modulator) at a different site than the enzyme's active site, thereby triggering a conformational change on the enzyme upon binding of an effector. Finally, gene expression encompasses the cellular processes by which genetic information flows from genes to proteins (i.e., transcription >> translation). In metabolic pathways, enzymes that are able to catalyze irreversible reactions represent sites of control (for example, during glycolysis, pyruvate kinase is an enzyme that catalyzes an irreversible reaction, thereby serving as a control site). In turn, enzymatic activity is modulated by covalent modifications or reversible binding of allosteric effectors. Finally, metabolic pathways are also modulated by gene regulatory mechanisms that control the transcription of specific enzymes required for such pathways. During these processes, the times required for allosteric regulation, covalent modification (e.g., phosphorylation) and transcriptional control can be counted in milliseconds, seconds, and hours, respectively.

The given parameters include;

From the image uploaded;

The resolution of the vectors along the sides of the triangle is as follows;

forces -------angle (θ) --------y-component-------x-component

3p -------------  60⁰ --------- ----2.598 p ------------------- 1.5 p

5p -------------- 60⁰ ------------ (-4.33p) ------------------- 2.5 p

7p -------------- 60⁰ -------------  0   -------------------------- (-7p)

sum:                                   ( -1.732 p) y                  (-3P)x

The resultant of the vectors:

R² = (-1.732 p)²  +  (-3P)²

R² = 11.998 p²

R = √(11.998 p²)

R = 3.46 p

The direction of the vector:

[tex]\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{-1.732}{-3} )\\\\\theta = tan^{-1} (0.577)\\\\\theta = 29.98 \ ^0[/tex]

The position of the vector = 180 - θ = 180 - 29.98 = 150⁰ (second quadrant)

To learn more about resultant and direction of forces, please visit: https://brainly.in/question/21981854

Answer:

a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.

Explanation:

Maximum theoretical efficiency for a power cycle ([tex]\eta_{r}[/tex]), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:

[tex]\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\%[/tex] (1)

Where:

[tex]T_{C}[/tex] - Temperature of the cold reservoir, in Kelvin.

[tex]T_{H}[/tex] - Temperature of the hot reservoir, in Kelvin.

The maximum theoretical efficiency associated with this power cycle is: ([tex]T_{C} = 400\,K[/tex], [tex]T_{H} = 1200\,K[/tex])

[tex]\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%[/tex]

[tex]\eta_{r} = 66.667\,\%[/tex]

In exchange, real efficiency for a power cycle ([tex]\eta[/tex]), no unit, is defined by this expression:

[tex]\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\%[/tex] (2)

Where:

[tex]Q_{C}[/tex] - Heat released to cold reservoir, in kilojoules.

[tex]Q_{H}[/tex] - Heat gained from hot reservoir, in kilojoules.

[tex]W_{C}[/tex] - Power generated within power cycle, in kilojoules.

A power cycle operates irreversibly for [tex]\eta < \eta_{r}[/tex], reversibily for [tex]\eta = \eta_{r}[/tex] and it is impossible for [tex]\eta > \eta_{r}[/tex].

Now we proceed to solve for each case:

a) [tex]Q_{H} = 900\,kJ[/tex], [tex]W_{C} = 450\,kJ[/tex]

[tex]\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta = 50\,\%[/tex]

Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.

b) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{C} = 300\,kJ[/tex]

[tex]\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta = 66.667\,\%[/tex]

Since [tex]\eta = \eta_{r}[/tex], the power cycle operates reversibly.

c) [tex]W_{C} = 600\,kJ[/tex], [tex]Q_{C} = 400\,kJ[/tex]

[tex]\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%[/tex]

[tex]\eta = 60\,\%[/tex]

Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.

d) Since [tex]\eta > \eta_{r}[/tex], the power cycle is impossible.

Answer:

not understand language

Answer:

Explanation:

Increased by the constant. Take a very simple case.

4  +  5 +  6 = 15

The mean is 5  (obtained by dividing the total (15) by the number of terms (3).

Now add a constant say 6

4 + 6 = 10

5 + 6 = 11

6 + 6 = 12

Total = 33/3 = 11

So the mean 5 is increased by the constant 6.

Now do the same thing more symbolically.

4 + c

5 + c

6 + c

Total = 15 + 3c

Divide by 3 you get 5 + c

If you want a more formal proof involving n terms, leave a note.

Answer:

[tex]D=0.016m[/tex]

Explanation:

From the question we are told that:

Discharge Rate [tex]F_r=0.5kgls[/tex]

Pressure [tex]P=15Kpa[/tex]

Temperature [tex]T=25=>298K[/tex]

Ambient pressure is 1 atm.

Generally the equation for Density is mathematically given by

[tex]\rho=\frac{PM}{RT}[/tex]

[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]

[tex]\rho=16.958kg/m^2[/tex]

Generally the equation for Flow rate is mathematically given by

[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]

Where

[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]

[tex]Q=1.4[/tex]

[tex]\mu= Discharge\ coefficient[/tex]

[tex]\mu=0.68[/tex]

Therefore

[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]

[tex]A=2.129*10^{-4}[/tex]

Where

[tex]A=\frac{\pi}{4}D^2[/tex]

[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]

[tex]D=0.016m[/tex]

Answer:

Explanation:

The main function of the fins that are to be added is to ensure the speedy transfer of heat from the Hot air.

The fins should be attached outside the tube because the convection coefficient of air is higher inside the tube than the convection coefficient of the outside air ( atmospheric air ),  BUT

When convection coefficient of air inside the tube is equal to the atmospheric air outside the tube, it is recommended that the fins can be added on both sides of the tube ( i.e. in and outside the tube )

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Answer:

  see attached

Explanation:

Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.

  (5 cm³/s)(1.06 g/cm³) = 5.3 g/s

If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:

  (5.3 g/s)/2 = 2.65 g/s

__

The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...

  (5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s

Downstream, we have half the volumetric flow and a smaller area.

  (2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s