three forces equal to 3p,5p,7p act simultaneously along the three side AB, BC,and CA of equilateral triangle ABC of side a. find the magnitude, direction and position of the resultant.​

Answer:

The governments receiving aid were generally experienced in industrial development. ... During the 1950s, little attention was given to differences in the Third World's conditions and needs, until these appeared to create obstacles to achieving high levels of industrial output

I hope it is helpful

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:

[tex]Q_2 = 32[/tex] mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, [tex]Q_1[/tex] = 1 mL/s

Initial diameter, [tex]D_1= D_0[/tex]

Initial length, [tex]L_1=L_0[/tex]

The initial pressure difference to maintain the flow, [tex]P_1=P_0[/tex]

We know for a viscous flow,

[tex]$\Delta P = \frac{32 \mu V L}{D^2}$[/tex]

[tex]$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$[/tex]

[tex]$Q \propto \Delta P \times D^4$[/tex]

[tex]$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{32}$[/tex]

∴ [tex]Q_2 = 32[/tex] mL/s

The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s

We are given;

Initial flow rate; Q1 = 1 mL/s

Initial uniform diameter; D0

Initial Length; L0

Initial Pressure difference; P0

Relationship between pressure, flow rate and diameter for vicious flow is given by;

Q1/Q2 = (P1/P2) × (D1/D2)⁴

Where;

Q1 is initial flow rate

Q2 is final flow rate

P1 is initial pressure difference

P2 is final pressure difference

D1 is initial diameter

D2 is final diameter

We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;

P2 = 2P0

D2 = 2D0

Thus;

1/Q2 = (P0/2P0) × (D0/2D0)⁴

>> 1/Q2 = ½ × (½)⁴

1/Q2 = 1/32

Q2 = 32 mL/s

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Answer:

a)  [tex]m=0.17kg/s[/tex]

b)  [tex]Ma=0.89[/tex]

Explanation:

From the question we are told that:

Pressure [tex]P=60kPa[/tex]

Diameter [tex]d=3cm[/tex]

Generally at sea level

[tex]T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4[/tex]

Generally the Power series equation for Mach number is mathematically given by

[tex]\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}[/tex]

[tex]\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}[/tex]

[tex]Ma=0.89[/tex]

Therefore

Mass flow rate

[tex]\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]

[tex]\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]

[tex]\rho=0.848kg/m^3[/tex]

Generally the equation for Velocity at throat is mathematically given by

[tex]V=Ma(r*T_0\sqrt{T_e}[/tex])

Where

[tex]T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}[/tex]

[tex]T_e=248[/tex]

Therefore

[tex]V=0.89(1.4*288\sqrt{248})\\\\V=284[/tex]

Generally the equation for Mass flow rate is mathematically given by

[tex]m=\rho*A*V[/tex]

[tex]m=0.84*\frac{\pi}{4}*3*10^{-2}*284[/tex]

[tex]m=0.17kg/s[/tex]

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]

Solution :

Let [tex]$x(t) = \frac{\sin (20 \pi t)}{\pi t}$[/tex]

[tex]$T_s = 20$[/tex] ms, so [tex]$f_s=\frac{1}{T_s}[/tex]

                           [tex]$=\frac{1}{20}$[/tex]

                           = 0.05 kHz

[tex]$f_s=50 $[/tex] Hz , ws = [tex]$2 \pi f_s = 100 \pi$[/tex]  rad/s

We know that,

FT → [tex]$\frac{\sin (20 \pi \omega)}{\pi \omega}$[/tex]

The sampled signal is :

[tex]$XS(\omega) = \frac{1}{T_s} \sum_{k=- \infty}^{\infty}X (\omega-k\omega S)[/tex]

So, [tex]$XS(\omega) = \frac{1}{20 \times 10^{-3}} \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]

     [tex]$XS(\omega) = 50 \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]

Answer:

Answer to the following question is as follows;

Explanation:

It is critical to communicate well during negotiations and conversation in order to attain your objectives. Communication also becomes essential in business and corporation . Effective communication may help you as well as your employees develop a strong professional relationship, which can boost morale and efficiency.

Answer:

The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.

Answer:

B

Explanation:

Anxiety is very common especially nowadays but it's especially common in psychological disorders

Answer:

80⁰C

Explanation:

80°C.

Normal flat plate collectors can deliver heat at temperatures up to 80°C. Deficiency rates for normal flat plate collectors can be classified as visual losses, which produce with cumulative angles of the incident sunshine, and thermal losses, which upsurge fast with the working temperature intensities

Answer:

I didn't understand your question or is it a fun fact

Answer:

i really font onow why tbh eot you

Explanation:

Concentration overpotential, ηc,

I hope it helps you

Answer:

Explanation:

From the given information:

weight of fiber [tex]w_f[/tex] = 3.0 g

weight of composite specimen [tex]w_c[/tex] = 4.0 g

specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g

specific gravity of fiber [tex]S_f[/tex] = 2.4

specific gravity of matrix [tex]S_m[/tex] = 1.3

The weight of the matrix = weight of the composite - the weight of fiber

⇒ (4.0 - 3.0) g

= 1.0 g

The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]

[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]

[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]

[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]

The experimental density [tex]\rho _{ce}[/tex] is determined  by using the equation:

[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]

[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]

The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]

[tex]= \dfrac{1.980-1.620}{1.980}[/tex]

= 0.1818

Answer:

  v = 8.95 rad / s,    a = 22.3 rad / s²

Explanation:

This is an exercise in kinematics, where we must use the definitions of velocity and acceleration

          v = dr / dt

we perform the derivative

          v = 0+ 2 (-sin 5t) 5

          v = -10 sin 5t

we calculate for t = 0.85 t,

remember angles are in radians

           v = -10 sin (5 0.85)

           v = 8.95 rad / s

acceleration is defined by

           a = dv / dt

we perform the derivatives

          a = -10 (cos 5t) 5

          a = - 50 cos 5t

we calculate for t = 0.85 s

          a = -50 are (5 0.85)

          a = 22.3 rad / s²

Answer:

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Explanation:

jwhwhwhwhwhwwhhahwhahahwh

Answer:

f = 0.04042

Explanation:

temperature = 0°C = 273k

p = 600 Kpa

d = 40 millemeter

e = 10 m

change in  P = 235 N/m²

μ = 2m/s

R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042

Answer:

10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .

Explanation:

Answer:

True

Explanation:

All big companies are pretty much required in today's day and ages to complete these reports whether they truly believe it.

Answer:

p q output ¬(p ∧ q)

0 0 1

0 1 1

1 0 1

0 0 0

p q output ¬p ∨ ¬q

0 0 1

0 1 1

1 0 1

0 0 0

Explanation:

We'll create two separate truth tables for both sides of the equation, and see if they match.

The expressions in the question use AND, OR and NOT operators.

Let's start with ¬(p ∧ q)

Now let's move on to the second expression ¬p ∨ ¬q

Therefore we can say the two expressions are equivalent.

Attached  the truth table to verify the first De Morgan's law ¬(p ∧ q) ≡ ¬p ∨ ¬q:

As you can see from the attached truth table, the truth values for ¬(p ∧ q) and ¬p ∨ ¬q are the same for all combinations of p and q, confirming the validity of the first De Morgan's law.

De Morgan's law is a fundamental principle in propositional logic.

It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the individual propositions.

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Answer:

Can you make friend with me ?

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.