Solution :
Given :
radius of perigee, [tex]$r_p$[/tex] = 10,000 km
radius of apogee, [tex]$r_a$[/tex] = 100,000 km
a). Eccentricity of the orbit
[tex]$e=\frac{|r_p-r_a|}{r_p+r_a}$[/tex]
[tex]$e=\frac{|10,000-100,000|}{10,000+100,000}$[/tex]
[tex]$e=\frac{9}{11}$[/tex]
or e = 0.818
b). Semi major axis of the orbit
[tex]$a=\frac{r_p+r_a}{2}$[/tex]
[tex]$a=\frac{10,000+100,000}{2}$[/tex]
= 55,000 km
c). period of orbit
[tex]$T=\frac{2\pi}{\sqrt{\mu}}\times a^{3/2}$[/tex]
Replacing μ with [tex]$398600 \ km^3/s^2$[/tex]
[tex]$T=\frac{2\pi}{\sqrt{398600}}\times (55,000)^{3/2}$[/tex]
[tex]$T=128304.04 \ s \left(\frac{1 \ hr}{3600 \ s}\right)$[/tex]
T = 35.64 hr
d). Specific energy of the orbit
[tex]$\varepsilon = -\frac{\mu}{2a}$[/tex]
[tex]$\varepsilon = -\frac{398600}{2 \times 55000}$[/tex]
[tex]$\varepsilon = -3.62 \ km^2/s^2$[/tex]
e). the equation of the distance to the focus
[tex]$\theta = \cos^{-1}\left(\frac{a(1-e^2)}{r}-\frac{1}{e}\right)$[/tex]
[tex]$\theta = \cos^{-1}\left(\frac{55000(1-(0.818)^2)}{(1000+6378)}-\frac{11}{9}\right)$[/tex]
[tex]$\theta = \cos^{-1}\left(\frac{55000(0.33)}{(7378)}-\frac{11}{9}\right)$[/tex]
[tex]$\theta = \cos^{-1}\left(2.4-1.2\right)$[/tex]
[tex]$\theta = \cos^{-1}\left(1.2\right)$[/tex]
θ = 1.002°
f).Calculating the angular momentum
[tex]$r_p=\frac{h^2}{\mu(1+e)}$[/tex]
or [tex]$h=\sqrt{r_p \mu(1+e)}$[/tex]
Now calculate the radial velocity
[tex]$v_r=\frac{\mu}{h} e \sin \theta$[/tex]
Substituting for h,
[tex]$v_r=\frac{\mu}{h}e \sin \theta$[/tex]
[tex]$v_r=\frac{e\mu \sin \theta}{\sqrt{r_p \mu(1+e)}}$[/tex]
[tex]$v_r=\frac{\frac{9}{11}\sqrt{398600} \sin 20}{\sqrt{10,000 (1+0.818)}}$[/tex]
[tex]$v_r= 1.30 \ km/s$[/tex]
Now calculating the azimuthal velocity
[tex]$v_{\perp}=\frac{\mu}{h}(1+e \cos \theta)$[/tex]
[tex]$v_{\perp}=\frac{\mu (1+e \cos \theta)}{\sqrt{r_p \mu(1+e)}}$[/tex]
[tex]$v_{\perp}=\frac{\sqrt{398600} (1+0.818 \cos 20)}{\sqrt{10000(1+0.818)}}$[/tex]
[tex]$v_{\perp}=7.58 \ km/s$[/tex]
g). Velocity at perigee
[tex]$v_p=\frac{h}{r_p}$[/tex]
[tex]$v_p=\frac{\sqrt{r_p \mu (1+e)}}{r_p}$[/tex]
[tex]$v_p=\frac{\sqrt{10000 (398600) (1+0.818)}}{10000}$[/tex]
[tex]$v_p=8.52 \ km/s$[/tex]
Now calculate the velocity of the apogee
[tex]$v_a=\frac{h}{r_a}$[/tex]
[tex]$v_a=\frac{\sqrt{r_p \mu (1+e)}}{r_a}$[/tex]
[tex]$v_p=\frac{\sqrt{10000 (398600) (1+0.818)}}{100000}$[/tex]
[tex]$v_a= 0.85 \ km/s$[/tex]
At 800K, a plot of ln[cyclobutane] vs t gives a straight line with a slope of -1.6 s-1. Calculate the time needed for the concentration of cyclobutane to fall to 1/16 of its initial value.
Answer:
hmmm.........
Explanation:
Anyone help me please ?
Answer:
I can help but I need to know what it looking for
In which phase and for what purpose does a construction manager work with various consultants? In the [blank] phase, a construction manager works closely with architects, civil engineers, electrical engineers, and other consultants to prepare a [blank].
Choices for first [blank]
A. design and planning
B. Construction
C. Post construction
Choices for second [blank]
A. draft
B. Sketch
C. Blueprint
Answer:
a and c
Explanation:
4. The instant the ignition switch is turned to the start position,
A. The starter motor starts to rotate before energizing the starter p
O B. Only the pull-in winding is energized.
C. Only the hold-in winding is energized.
D. Both pull-in and hold-in windings are energized.
Answer:
D. Both pull-in and hold-in windings are energized.
Explanation:
The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.
The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.
Explain the difference between the connection of a cumulative compound and a differential compound motor
Answer:
Explanation:
A motor is a device that directs current in electrical energy form to mechanical energy, which is known as direct current (DC) motors.
DC motors are of three types: (a) The series motor, (b) The shunt motor, and (c) the compound motor. Our main focus here is the Compound motor, which is further sub-divided into:
i) The cumulative compound motors
ii) The differential compound motors
The difference between these two are:
Cumulative compound motors Differential compound motors
In cumulative compound motors, In differential compound motors,
both the series and shunt windings both series and shunt are
are connected in a way that, connected in a way that the
production of fluxes through them production of fluxes via them
assist each other i.e. they aid each always opposes each other i.e.
other in the production of magnetism they oppose each other in the
production of magnetism.
For proper function hydraulics systems need a reservoir of which of the following?
A.) Compressible fluid
B.) Non-compressible fluid C.) Non-compressible air
An oxygen–nitrogen mixture consists of 35 kg of oxygen and 40 kg of nitrogen. This mixture is cooled to 84 K at 0.1 MPa pressure. Determine the mass of the oxygen in the liquid and gaseous phase.
Answer:
The mass of oxygen in liquid phase = 14.703 kg
The mass of oxygen in the vapor phase = 20.302 kg
Explanation:
Given that:
The mass of the oxygen [tex]m_{O_2}[/tex] = 35 kg
The mass of the nitrogen [tex]m_{N_2}[/tex] = 40 kg
The cooling temperature of the mixture T = 84 K
The cooling pressure of the mixture P = 0.1 MPa
From the equilibrium diagram for te-phase mixture od oxygen-nitrogen as 0.1 MPa graph. The properties of liquid and vapor percentages are obtained.
i.e.
Liquid percentage of [tex]O_2[/tex] = 70% = 0.70
Vapor percentage of [tex]O_2[/tex] = 34% = 0.34
The molar mass (mm) of oxygen and nitrogen are 32 g/mol and 28 g/mol respectively
Thus, the number of moles of each component is:
number of moles of oxygen = 35/32
number of moles of oxygen = 1.0938 kmol
number of moles of nitrogen = 40/28
number of moles of nitrogen = 1.4286 kmol
Hence, the total no. of moles in the mixture is:
[tex]N_{total} = 1.0938+1.4286[/tex]
[tex]N_{total} = 2.5224 \ kmol[/tex]
So, the total no of moles in the whole system is:
[tex]N_f + N_g = 2.5224 --- (1)[/tex]
The total number of moles for oxygen in the system is
[tex]0.7 \ N_f + 0.34 \ N_g = 1.0938 --- (2)[/tex]
From equation (1), let N_f = 2.5224 - N_g, then replace the value of N_f into equation (2)
∴
0.7(2.5224 - N_g) + 0.34 N_g = 1.0938
1.76568 - 0.7 N_g + 0.34 N_g = 1.0938
1.76568 - 0.36 N_g = 1.0938
1.76568 - 1.0938 = 0.36 N_g
0.67188 = 0.36 N_g
N_g = 0.67188/0.36
N_g = 1.866
From equation (1)
[tex]N_f + N_g = 2.5224[/tex]
N_f + 1.866 = 2.5224
N_f = 2.5224 - 1.866
N_f = 0.6564
Thus, the mass of oxygen in the liquid and vapor phases is:
[tex]m_{fO_2} = 0.7 \times 0.6564 \times 32[/tex]
[tex]m_{fO_2} = 14.703 \ kg[/tex]
The mass of oxygen in liquid phase = 14.703 kg
[tex]m_{g_O_2} = 0.34 \times 1.866 \times 32[/tex]
[tex]m_{g_O_2} = 20.302 \ kg[/tex]
The mass of oxygen in the vapor phase = 20.302 kg
The pressure at the bottom of an 18 ft deep storage tank for gasoline is how much greater than at the top? Express your answer in the units of psi.
The pressure at the bottom of an 18 ft deep storage tank for gasoline is 5.625 psi.
What is pressure?Pressure is defined as the force exerted on a surface's unit area. The mass of the air molecules above is what exerts pressure on the atmosphere. The enormous quantities of air molecules that make up the layers of our atmosphere collectively have a tremendous amount of weight, which bears down on whatever is below.
The storage tank's top pressure is 14.7 psi, or p atm, or 2116.8 lb/ft3.
Gasoline has a density of 45 lb/ft3.
Tank depth, h = 18 foot
The storage tank's bottom pressure is 2926.8 pounds per square foot (Pb = 2116.8 + 45 x 18).
As a result, the pressure difference between the tank's bottom and top is as follows: P = Pb - P atm = 20.325 - 14.7 = 5.625 psi
Thus, the pressure at the bottom of an 18 ft deep storage tank for gasoline is 5.625 psi
To learn more about pressure, refer to the link below:
https://brainly.com/question/12971272
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A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as shown. Determine the velocity v of the motorcycle when s=200 m. At this point, also determine the value of the derivative dv/ds.
Answer:
Follows are the solution to this question:
Explanation:
Calculating the area under the curve:
A = as
[tex]=\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}[/tex]
Calculating the kinematics equation:
[tex]\to v^2 = v^2_{o} + 2as\\\\[/tex]
[tex]=0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}[/tex]
Calculating the value of acceleration:
[tex]\to a= \frac{dv}{dt}[/tex]
[tex]=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}[/tex]
[tex]\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\[/tex]
[tex]=\frac{0.092}{s}[/tex]
If the specific surface energy for soda–lime glass is 0.30 J/m2 and its modulus of elasticity is (69 GPa), compute the critical stress required for the propagation of an internal crack of length 0.8 mm.
Answer: the critical stress for the sodalime glass = 5.7MPa
Explanation:
Ist Step
we Calculate half length of internal crack as
2a =0.8 mm
a = 0.8/2 = 0.4 mm
Changing to meters becomes = 0.4 / 1000 =0.0004m
2nd Step
Now the critical stress required for the propagation of the internal crack can be calculated using the formulae
Critical Stress (σc) = (2 E γs/ πa) 1/2
where E= modulus of elasticity
γs= specific surface energy for soda–lime glass
a= Length
= (2 x 69 x 10 ^9 x 0.30/ π x 0.0004)1/2
=[tex]\sqrt{ 32,940,802,036,919}[/tex]
= 5,739,407.8= 5.7 x 10^6 N/m^2
= 5.7MPa
If ice homogeneously nucleates at -44.6°C, calculate the critical radius given values of -3.1 × 10^8 J/m3 and 25 × 10^-3 J/m^2, respectively, for the latent heat of fusion and the surface free energy.
Answer:the critical radius for the homogeneous nucleating ice is 0.986 nm
Explanation:
Using the formulae below to calculate the critical radius for homogeneous nucleation,
We Have that
Critical radius ( r *) = [2γ Tm/ΔHf]{ 1/ Tm- T]
where γ = surface free energy =25 × 10^-3 J/m^2
Tm= solidification temperature at equilibrium =273K
Hf= latent heat of fusion = -3.1 × 10^8 J/m3
Temperature , T = -44.6°C
Critical radius ( r *) = (2 X 25 × 10^-3 J/m^2 x 273)/ (-3.1 × 10^8 J/m3 ) X ( 1/ 273 - ( -44.6 +273)
4.40x 10^-8X ( 1/273-228.4)
4.40x 10^-8 X 1/44.6
4.40x 10^-8 x 0.0224
9.86x 10 ^-10m =0.986 x 10 ^-9m = 0.986nm
Consider a circuit element, with terminals a and b, that has vab= -12V and iab= 3A. Over a period of 2 seconds, how much charge moves through the element? If electrons carry the charge, which terminal do they enter? How much energy is transferred? Is it delivered to the element or taken from it?
Answer:
a) 6 coulombs
b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current
c) = -72 joules
Energy is taken from element
Explanation:
Given data:
V ab = -12 v
I ab = 3A
period ( t ) = 2 seconds
a) determine how much charge moves through the element
q = I * t
= 3 * 2 = 6 coulombs
b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current
c) determine how much energy is transferred
= Vab * Iab * t
= -12 * 3 * 2
= -72 joules
Energy is taken from element
An electric circuit is made up of a 100 m long manganin wire with a section of I mm^2; this wire constitutes 4/5 of the total resistance of the circuit itself and the intensity of the current circulating there is 2.5 A. Calculate the voltage applied to the terminals of the manganin wire, the energy dissipated on this wire in 30 minutes and the voltage applied by the generator across the circuit.
Answer:
a.dont know e
Explanation:
because d q tlga ammu
A 8-core machine has 4 times the performance of a single-core machine of the same frequency. Performance is proportional to frequency. Voltage decreases proportionally to frequency. To achieve the same performance, how much (in percentage) dynamic power would the 8-core system save?
Answer: The 8-core machine saves 87.5% of the dynamic power.
Explanation:
Let Fold = f , Vold = V , Cold = Capacitance
so
Old Dynamic power = Cold × (Vold × Vold) × f
therefore for the 8-core machine
Fnew / Fold = 1/4
Fnew = Fold/4
we were told that Voltage decreases proportional to frequency,
so
Vnew / Vold = 1/4
Vnew = V / 4
So New Capacitance will be;
Cnew = Cold
Thus, New Dynamic power = 8 × Cnew × ( Vnew × Vnew ) × Fnew
= 8 × Cold × (Vold × Vold/16) × ( f/4 )
= 8 × ( Cold ) × ( Vold × Vold ) × ( f ) / 64
= (Old Dynamic Power) / 8
therefore
Old Dynamic Power / New Dynamic Power = 8
Thus, Percentage of power saved will be;
Percentage power saved = 100 × ( Old Dynamic Power - New Dynamic Power ) / Old Dynamic Power
= 100 × (8-1) / 8
= 87.5 %
Therefore The 8-core machine saves 87.5% of the dynamic power.
A MBR treatment plant has a flow of 0.3 mgd with 220 mg/l BOD (the soluble portion is 120 mg/l) and 230 mg/l suspended solids (184 mg/l is volatile). Effluent soluble BOD is 2 mg/l. The design calls for 5000 mg/l MLSS and sludge age of 20 days with a Y = 0.8, kd = 0.4. Calculate the aeration basin volume, detention time, BOD loading, ratio, and waste solids, both VSS and TSS.
Answer:
attached below is the detailed solution
A) 8288.77 cu.ft
B) 4.96 hours
C) Vss = 131.21 IbVss/day
Tss = 164 IbTss/day
D) attached below
E ) 0.2
F) 287.23 Ib/day
Explanation:
A) Determine the aeration basin volume
Given
∅c = 20 days
Y = 0.8Ib VSS/Ib BOD
Q = 0.3 mgd
So = 120 mg/l
Se = 2 mg/l
X = 5000 mg/l
Kd = 0.04 per day
attached below is the detailed solution
B) Determine the detention time using this relation
t = ( V / Q )* 24
= ( 0.062 / 0.3 ) * 24 = 4.96 hours
C ) Determine Vss and Tss
we first calculate the excess biomass Px then assuming Vss ratio to be 80% for sludge age of 20 days
Vss = 131.21 IbVss/day
Tss = 164 IbTss/day
D ) determine BOD loading
Q = 0.3 mgd , BOD = 220mg/l , V = 8288.77 cu.ft
solution attached below
e) food to microorganism ratio
F/M = 0.2
solution attached below
f) determine the waste solid
waste solid = Q * SS * % removal of suspended solids
where : Q = 0.3 , SS = 220mgl , % = 50 %
waste solids = 0.3 * 230 * 0.5 * 8.34 = 287.23 Ib/day
Which of these words was first used during the 1970s economic crisis?
influx
stagflation
deficit
programs
Answer:
stagflation
Explanation:
it was used in the article
A word which was first used during the 1970s economic crisis is stagflation.
The economic crisis of the 1970s.In the 1970s, an energy crisis took place in the United States of America due to the oil embargo that was imposed on it by OPEC. This oil embargo was imposed on the United States of America by the Organization of Petroleum Exporting Countries (OPEC) in 1973 because of its role in the Arab-Israeli War.
Consequently, the economy of the United States of America experienced stagflation in the following ways:
Slow economic growth.Relatively high unemployment.Read more on stagflation here: https://brainly.com/question/25505087
the reaction of 4A+3B→2C+D is studied. Unknown masses of the reactants were mixed . After a reaction time of 1 hour the analysis of the mixture showed 2 kmol, 1 kmol of B and 4 kmol of C. product D was present in the mixture but could not be analysed. what is the mole fraction of D in the mixture?
Answer: the mole fraction of D in the mixture is 0.2222
Explanation:
Given that;
mixture analysis shows 2 kmol A, 1 kmol B, 4 kmol C and some unknown kmol of D was present.
4A+3B→2C+D
As from reaction stoichiometry, for every 2 kmol of C produced, kmol of D produced = 1 kmol
so, for 4 kmol C, kmol of D produced = 4/2 × 1 kmol = 2 kmol
Now our mixture has 2 kmol A, 1 kmol B, 4 kmol C and also 2 kmol of D
so, total moles in mixture, we have (2 + 1 + 4 + 2) kmol = 9 kmol
mole fraction of D in mixture will be;
( Kmol of D) / (total moles in mixture) = 2 / 9 = 0.2222
Therefore the mole fraction of D in the mixture is 0.2222
Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can be tripped to a turbulent state by adding roughness to the leading edge of the plate. For a particular situation, experimental results show that the local heat transfer coefficients for laminar and turbulent conditions are
h_lam(x)= 1.74 W/m^1.5. Kx^-0.5
h_turb(x)= 3.98 W/m^1.8 Kx^-0.2
Calculate the average heat transfer coefficients for laminar and turbulent conditions for plates of length L = 0.1 m and 1 m.
Answer:
At L = 0.1 m
h⁻_lam = 11.004K W/m^1.5
h⁻_turb = 7.8848K W/m^1.8
At L = 1 m
h⁻_lam = 3.48K W/m^1.5
h⁻_turb = 4.975K W/m^1.8
Explanation:
Given that;
h_lam(x)= 1.74 W/m^1.5. Kx^-0.5
h_turb(x)= 3.98 W/m^1.8 Kx^-0.2
conditions for plates of length L = 0.1 m and 1 m
Now
Average heat transfer coefficient is expressed as;
h⁻ = 1/L ₀∫^L hxdx
so for Laminar flow
h_lam(x)= 1.74 . Kx^-0.5 W/m^1.5
from the expression
h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5 dx
= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L
= 1.74k/L = [ (x^0.5)/0.5)]⁰^L
= 1.74K × L^0.5 / L × 0.5
h⁻_lam= 3.48KL^-0.5
For turbulent flow
h_turb(x)= 3.98. Kx^-0.2 W/m^1.8
form the expression
1/L ₀∫^L 3.98 . Kx^-0.2 dx
= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L
= (3.98K/L) × (L^0.8 / 0.8)
h⁻_turb = 4.975KL^-0.2
Now at L = 0.1 m
h⁻_lam = 3.48KL^-0.5 = 3.48K(0.1)^-0.5 W/m^1.5
h⁻_lam = 11.004K W/m^1.5
h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2
h⁻_turb = 7.8848K W/m^1.8
At L = 1 m
h⁻_lam = 3.48KL^-0.5 = 3.48K(1)^-0.5 W/m^1.5
h⁻_lam = 3.48K W/m^1.5
h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2
h⁻_turb = 4.975K W/m^1.8
Therefore
At L = 0.1 m
h⁻_lam = 11.004K W/m^1.5
h⁻_turb = 7.8848K W/m^1.8
At L = 1 m
h⁻_lam = 3.48K W/m^1.5
h⁻_turb = 4.975K W/m^1.8
Indicate similarities between a nucleus and a liquid droplet; why small droplets are stable and very big droplets are not?
Answer:
There are several similarities between the nucleus and a liquid droplet.
Explanation:
A droplet of liquid simply is is very small or tiny drop of liquid. It is also considered as a tiny column of liquid that is surrounded by surfaces that have zero shear stress.
A nucleus on the other hand is an assembly between protons and neutrons. The latter is electrically charged whilst the former is positively charged. The number of protons present in an element is very crucial to the qualities of an element.
The main similarities between a nucleus and a liquid droplet are:
1. a nucleus consists of a large amount of neutrons and protons in the same volume as would a liquid which contains large numbers of molecules in the same volume;
2. both the nucleus and the droplet are similar for their homogeneity in electric charge and density;
3. the molecules exert the same amount for forces towards one another as would the nuclear forces in the nucleons.
4. both of them cannot be compressed
5. both molecules and nucleus are can be subject to nuclear fission which simply mean the breaking apart into smaller units (in the case of the nucleus) or the breaking apart into smaller droplets in the case of the liquid molecule.
6. There are two types of phenomena which occurs in both the liquid droplet and the nucleus which are similar to one another. They are:
Evaporation (in the case of the liquid molecule) and reaction emission (in the case of the nucleus). In evaporation, particles are lost, in Atomic transmutation, particles are lost as well.
B) the forces which determine the stability of droplets are surface tension and gravitation. The smaller the area, the stronger the surface tension available to keep the drops from going out of shape.
Cheers
Calculate the LER for the rectangular wing from the previous question if the weight of the glider is 0.0500 Newton’s.
Answer:
0.2
Explanation:
Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.
Let the span of the rectangular wing be 0.225 m
Let the chord of the rectangular wing be 0.045 m.
Then, the area of any rectangular chord is
A = chord * span
A = 0.045 * 0.225
A = 0.010 m²
And using the weight of the glider given to us from the question, we can find the LER for the wing.
LER = Area / weight.
LER = 0.010 / 0.05
LER = 0.2.
Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2
Please mark brainliest...
Answer: 0.2025
Explanation: I got it correct
(3 points) One end of a 48 cm long copper rod with a diameter of 2.0 cm is kept at 360 ° C, and the other is immersed in water at 32° C. Calculate the heat conduction rate along the rod. The thermal conductivity for copper is 386 MK
Answer:
hi tommoro i have phisics exam i needd help only for 20 min how wants to help messege me instgram meeraalk99
Explanation:
please
You are the project manager assigned to construct a new 10-story office building. You are trying to estimate the costs for this project. You start by assigning the costs associated with each of the project activity. Then you sum up all the individual costs into a final cost estimate. Which type of cost estimation technique did you use?
Answer:
Bottom-up Estimation
Explanation:
Bottom-up estimation is a type of project cost estimation that considers the cost of individual project activities and finally sums them up or finds the aggregates. The summation gives an idea of what the entire project will cost.
This is an effective way of estimating the cost of a project as it evaluates the costs on a wholistic basis. It also considers the tiniest details during the estimation process. The process moves from the simpler details to the more complicated details.
It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less common contexts are listed toward the bottom. According to O*NET, what are common work contexts for Construction Carpenters? Check all that apply. Spend time sitting wear common protective or safety equipment face-to-face discussions exposed to hazardous equipment spend time using hands to handle, control, or feel objects, tools, or controls public speaking
Answer:
bcde!!
Explanation:
They also tend to be traditional, which means that they enjoy working in structured environments and are typically organized and detail-oriented. Thus option B,C,D, E is correct.
What are the characteristics of Construction Carpenters?Carpenters continue to have a bright future in their profession. According to Job Outlook, the carpentry industry is expanding quickly. The advantages of carpentry lead to employment security and a long-term career with this kind of industrial growth.
Carpentry is a physically demanding line of work that calls for endurance. You frequently spend the most of your shift standing, moving slowly, and crouching.
Therefore, As well as using hand tools to shape and cut wood, lifting big objects, moving heavy beams, furniture, or equipment are all possible.
Learn more about Construction Carpenters here:
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Rear defrosters generally have a relay with a timer. This allows ___.
draw afd,sfd and bmd of frame
Answer:
uh, i dont understand?
Explanation:
Steam enters an adiabatic nozzle at 1 MPa, 250°C, and 30 m/s. At one point in the nozzle the enthalpy dropped 40 kJ/kg from its inlet value. Determine velocity at that point. (A) 31 m/s (B) 110 m/s (C) 250 m/s (D) 280 m/s
Answer:
284.4 m/s
Explanation:
At the inlet of the nozzle P =1 atm.
Temperature T = 250° C
Velocity of the steam at the inlet V_1 = 30 m/s
Change in enthalpy Δh = 40 KJ/kg
let V_2 be the final velocity
then
[tex]V_2 =\sqrt{2\Delta h+V_1^2} \\=\sqrt{2\times40+30^2}\\= 284.4 m/s[/tex]
For a 3-Phase, Wye connected system the Line to Line Voltage measures 12,470 Volts, the Phase current measures 120 Amps.
Required:
a. What will the Line-to-Neutral/Phase voltage be?
b. What will the Line current be?
Answer:
A. 7199.55 volts
B. 120A
Explanation:
In this question we have the
line voltage = VLL = 12470volts
Phase current = Iph = 120 amps
A.)
We are to calculate the line-to-neutral/phase voltage here
VLL = √3VL-N
VL-N = VLL/√3
VL-N = 12470/√3
This gives a line to neutral phase/voltage of 7199.55 volts.
B.
We are to calculate the line current here:
In this connection, the line current and the phase current are equal
ILL = Iph = 120A
A laboratory furnace wall is constructed of 0.2 m thick fireclay brick having a thermal conductivity of 1.82 W/m-K. The wall is covered on the outer surface with insulation of thermal conductivity of 0.095 W/m-K. The furnace inner brick surface is at 950 K and the outer surface of the insulation material is at 300 K. The maximum allowable heat transfer rate through the wall of the furnace is 830 W/m^2. Determine how thick in cm the insulation material must be.
Answer:
The appropriate solution will be "6.4 cm".
Explanation:
The given values are:
Length,
l = 0.2 m
Thermal conductivity,
K₁ = 1.82 W/m-K
K₂ = 0.095 W/m-K
Temperature,
T = 950 K
T = 300 K
Heat transfer rate,
Q = 830 W/m²
Now,
⇒ [tex]Q = \frac{\Delta T}{\frac{L_1}{K_1 A} +\frac{L_2}{K_2 A} }=\frac{A \Delta T}{\frac{L_1}{K_1 } +\frac{L_2}{K_2 } }[/tex]
⇒ [tex]\frac{Q}{A} =\frac{\Delta T}{\frac{L_1}{K_1} +\frac{L_2}{K_2} }[/tex]
On substituting the above given values in the equation, we get
⇒ [tex]830=\frac{(980-300)}{\frac{0.2}{1.82} +\frac{x}{0.095} }[/tex]
On applying cross-multiplication, we get
⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{950-300}{830}[/tex]
⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{650}{830}[/tex]
⇒ [tex]x =0.639 \ m[/tex]
⇒ [tex]x=6.345 \ i.e., 6.4 \ m[/tex]
In a CS amplifier, the resistance of the signal source Rsig = 100 kQ, amplifier input resistance (which is due to the biasing network) Rin = 100kQ, Cgs = 1 pF, Cgd = 0.2 pF, gm = 5 mA/V, ro = 25 kΩ, and RL = 20 kΩ. Determine the expected 3-dB cutoff frequency.
Answer:
406.140 KHz
Explanation:
Given data:
Rsig = 100 kΩ
Rin = 100kΩ
Cgs = 1 pF,
Cgd = 0.2 pF, and etc.
Determine the expected 3-dB cutoff frequency
first find the CM miller capacitance
CM = ( 1 + gm*ro || RL )( Cgd )
= ( 1 + 5*10^-3 * 25 || 20 ) ( 0.2 )
= ( 11.311 ) pF
now we apply open time constant method to determine the cutoff frequency
Th = 1 / Fh
hence : Fh = 1 / Th = [tex]\frac{1}{(Rsig +Rin) (Cm + Cgs )}[/tex]
= [tex]\frac{1}{( 200*10^3 ) ( 12.311 * 10^{-12} )}[/tex] = 406.140 KHz
Calculate the Lee for the same wing if we increase the span to 0.245 m. By increasing the span we also increase the glider weight to 0.0523
Answer:
0.21
Explanation:
This would have been a fairly easy one, except for that the first part of the question is missing, and as such, I'd assume a value.
We need to use chord, so, I'm assuming the length of the chord to be 0.045 m
The Area is given by the formula
Area = span * chord
Area = 0.245 * 0.045
Area = 0.011 m²
This area gotten, is what we then divide the glider weight by to get our answer.
Lee = area / weight
Lee = 0.011 / 0.0523
Lee = 0.21
Therefore, using the values of the chord I'd assumed, the Lee of the same wing is 0.21
Answer: 0.2108
Explanation:got it correct