Answer:
The answer for the question is true
Explanation:
If you get a virus or get hacked you will still have it saved
People tend to self-disclose to others that are in age, social status, religion, and personality.
Answer:people tend to do this when they are in a different environment they lose something or just have something going on in their life
Explanation:
A fluid has a mass of 5 kg and occupies a volume of 1 m3 at a pressure of 150 kPa. If the internal energy is 25000 kJ/kg, what is the total enthalpy?
Answer:
155 KJ
Explanation:
The total enthalpy is given by
ΔH=ΔU + PV
Where;
ΔH = enthalpy
ΔU = internal energy = 25000 kJ/kg/ 5 kg = 5000 KJ
P = 150 kPa = 150,000 Pa
V = 1 m3
ΔH= 5000 + (150,000 * 1)
ΔH= 155 KJ
Based on the pattern, what are the next two terms of the sequence? 9,94,916,964,9256,... A. 91024,94096 B. 9260,91028 C. 9260,9264 D. 91024,91028
Answer:
The answer is "Option A".
Explanation:
Series:
[tex]9, 94, 916, 964, 9256, ........[/tex]
Solving the above series:
[tex]\to 9\\ \to 9(4) =94\\\to 9 (4^2) = 9(16) =916\\\to 9 (4^3) = 9(64) =964\\\to 9 (4^4) = 9(256) =9256\\\to 9 (4^5) = 9(1024) =91024\\\to 9 (4^6) = 9(4096) =94096\\[/tex]
So, the series is: [tex]9, 94, 916, 964, 9256, 91024, 94096, .................[/tex]
Calculate the number of vacancies per cubic meter for some metal, M, at 783°C. The energy for vacancy formation is 0.95 eV/atom, while the density and atomic weight for this metal are 6.10 g/cm^3 (at 783°C) and 43.41 g/mol, respectively.
Answer:
Following are the solution to this question:
Explanation:
The number of vacancies by the cubic meter is determined.
[tex]N_V =N exp(\frac{Q_v}{kT})[/tex]
[tex]= \frac{N_A \rho}{A} exp (\frac{Q_v}{kT})[/tex]
[tex]= \frac{6.022 \times 10^{23} \times 6.10}{43.41} \exp(\frac{-0.95}{8.62\times 10^{-5} \times (783+273)})\\\\= \frac{36.7342 \times 10^{23}}{43.41} \exp(\frac{-0.95}{0.0313626})\\\\= 0.846215158 \times 10^{23} \exp(-30.290856)\\\\[/tex]
[tex]=1.57 \times 10^{25} \ cm^{-3}[/tex]
How would you achieve the linear convolution of a 100 sample time series and a 20 tap filter in the frequency domain?
Answer:
divide then add XD my guy this is easy
Explanation:
The linear convolution is a mathematical operation that finds the output of the linear time-variant system that is given its impulse and linear time-invariant responses.
The convolution for a 100 sample of time series and a 20 tap filter in the freq domain can be represented as y(n) = x(n) . h(n).Learn more about the achieve the linear convolution of a 1.
brainly.com/question/24452045.
How many snaps points does an object have?
Answer:
what do you mean by that ? snap points ?
An insulated 40 ft^3 rigid tank contains air at 50 psia and 580°R. A valve connected to the tank is now opened, and air is allowed to escape until the pressure inside drops to 25 psia. The air temperature during this process is kept constant by an electric resistance heater placed in the tank. Determine the electrical work done during this process in Btu assuming variable specific heats.
Answer:
The answer is "[tex]\bold{W_{in} = 645.434573 \ Btu}[/tex]".
Explanation:
Its air enthalpy is obtained only at a given temperature by A-17E.
The solution from the carbon cycle is acquired:
[tex]\to \Delta U= W_{in}-m_{out}h_{out}=0\\\\\to W_{in} = (m_1 - m_2)h[/tex]
[tex]=\frac{Vh}{RT}(P_1- P_2)\\\\= \frac{40 \times 138.66}{0.3704\times 580}(50-25) Btu\\\\= \frac{5,546.4}{214.832}(25) Btu\\\\= 25.8173829(25) Btu\\\\ =645.434573 Btu[/tex]
[tex]W_{in} = 645.434573 \ Btu[/tex]
The 240-ft structure is used to provide various support services to launch vehicles prior to liftoff. In a test, a 12-ton weight is suspended from joints F and G, with its weight equally divided between the two joints. Determine the forces in members GJ and GI. What would be your path of joint analysis for members in the vertical tower, such as AB or KL?
Answer:
hello your question lacks the required question attached below is the missing diagram
Forces in GJ = -4.4444 i.e. 4.4444 tons
Forces in IG = 15.382 tons ( T )
Explanation:
Forces in GJ = -4.4444 i.e. 4.4444 tons
Forces in IG = 15.382 tons ( T )
attached below is the detailed solution
Air ows steadily in a thermally insulated pipe with a constant diameter of 6.35 cm, and an average friction factor of 0.005. At the pipe entrance, the air has a Mach number of 0.12, stagnation pressure 280 kPa, and stagnation temperature 825 K. Determine:
a. The length of pipe required to reach the sonic state
b. The static pressure and temperature at the exit if the pipe is 25 m long
Solution:
Given :
D = 6.35 cm
[tex]$\bar f = 0.005$[/tex]
[tex]$P_s = 280 \ kPa$[/tex]
[tex]$T_s= 825 K[/tex]
a). From fanno flow table (γ = 1.4)
At [tex]$M_1 = 0.12$[/tex] , [tex]$\left(\frac{4 \bar f L_{max}}{D}\right)_1 = 45.408$[/tex]
At [tex]$M_2 = 1$[/tex] , [tex]$\left(\frac{4 \bar f L_{max}}{D}\right)_2 = 0$[/tex]
∴ [tex]$\left(\frac{4 \bar f L_{max}}{D}\right)_1 - \left(\frac{4 \bar f L_{max}}{D}\right)_2 = \frac{4 \bar f L}{D}$[/tex]
[tex]$45.408 - 0 = \frac{4 \times 0.005 \times L}{0.0635}$[/tex]
[tex]$45.408 = \frac{4 \times 0.005 \times L}{0.0635}$[/tex]
L = 144.17 m
b). If L = 25 m
[tex]$\frac{4 \bar f L}{D}=\frac{4 \times 0.005 \times 25}{0.0635} = 7.874$[/tex]
From fanno flow , (γ = 1.4)
[tex]$At, M_2 = 0.26 , \frac{4 \bar f L}{D} = 7.874$[/tex]
[tex]$\frac{P_s}{P_1}=\frac{T_s}{T_1}^{\frac{\gamma}{\gamma - 1}} = (1+\frac{\gamma-1}{2}M_1^2)^{\frac{\gamma}{\gamma - 1}}$[/tex]
[tex]$\frac{280}{P_1}=\frac{825}{T_1}^{\frac{1.4}{1.4 - 1}} = (1+\frac{1.4-1}{2}(0.12)^2)^{\frac{1.4}{1.4 - 1}}$[/tex]
[tex]$\frac{280}{P_1}=\left(\frac{825}{T_1}\right)^{3.5} =1.0101$[/tex]
[tex]$P_1 = 277.2 \ kPa$[/tex]
[tex]$T_1=822.63 \ K$[/tex]
[tex]$\frac{T_2}{T_1}=\frac{1+\frac{\gamma -1}{2}M_1^2}{1+\frac{\gamma -1}{2}M_2^2}$[/tex]
[tex]$\frac{T_2}{822.63}=\frac{1+\frac{1.4 -1}{2}(0.12)^2}{1+\frac{1.4 -1}{2}(0.26)^2}$[/tex]
[tex]$\frac{T_2}{822.63}=\frac{1.00288}{1.01352}$[/tex]
[tex]$T_2=814 \ K$[/tex]
[tex]$\frac{P_2}{P_1}=\frac{M_1}{M_2}\left(\frac{1+\frac{\gamma -1}{2}M_1^2}{1+\frac{\gamma -1}{2}M_2^2}\right)^{1/2}$[/tex]
[tex]$\frac{P_2}{P_1}=\frac{0.12}{0.26}\left(\frac{1+\frac{1.4 -1}{2}(0.12)^2}{1+\frac{1.4 -1}{2}(0.26)^2}\right)^{1/2}$[/tex]
[tex]$\frac{P_2}{277.2}=0.459$[/tex]
[tex]$P_2=127.27 \ kPa$[/tex]
please help me make a lesson plan. the topic is Zigzag line. and heres the format.
A. Objective
B. Subject matter
C. Learning activities.
D. Assessment.
E. Reinforcement
Explanation:
D. B. C. A. E. Is this a good idea
Which of the following is an example of someone who claims that the media has a shooting blanks effect?
A. "Along with parents, peers, and teachers, the media socializes children about how boys and girls are supposed to behave."
B. "My kid saw a cigarette ad in a magazine and now he's smoking. It's the magazine's fault!"
C. "The media doesn't affect me at all because I'm smart enough to know the difference between right and wrong."
D. "There is no definitive evidence that the media affects our behavior"
Answer:
the answer would be d its d
Answer:
Pretty sure the answer is "C"
Explanation:
"The media doesn't affect me at all because I'm smart enough to know the difference between right and wrong."
A rectangular channel 3-m-wide carries 12 m^3/s at a depth of 90cm. Is the flow subcritical or supercritical? For the same flowrate, what depth will five critical flow?
Answer:
Super critical
1.2 m
Explanation:
Q = Flow rate = [tex]12\ \text{m}^3/\text{s}[/tex]
w = Width = 3 m
d = Depth = 90 cm = 0.9 m
A = Area = wd
v = Velocity
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
[tex]Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}[/tex]
Froude number is given by
[tex]Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5[/tex]
Since [tex]F_r>1[/tex] the flow is super critical.
Flow is critical when [tex]Fr=1[/tex]
Depth is given by
[tex]d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}[/tex]
The depth of the channel will be 1.2 m for critical flow.
How much energy in joule is added to a 12 g of sample of aluminum (c=0.897 J/g ◦C) to raise the temperature from 20 ◦C to 45 ◦C?
Answer:
269.1J
Explanation:
m = 12g
c = 0.897J/g°C
∆T = 45 - 20 = 25°C
H = mc∆T = 12 × 0.897 × 27 = 269.1J
A single phase inductive load draws 10 MW at 0.6 power factor lagging. Draw the power triangle and determine the reactive power of a capacitor to be connected in parallel with the load to raise the power factor to 0.85.
Answer: attached below is the power triangles
7.13589 MVAR
Explanation:
Power ( P1 ) = 10 MW
power factor ( cos ∅ ) = 0.6 lagging
New power factor = 0.85
Calculate the reactive power of a capacitor to be connected in parallel
Cos ∅ = 0.6
therefore ∅ = 53.13°
S = P1 / cos ∅ = 16.67 MVA
Q1 = S ( sin ∅ ) = 13.33 MVAR ( reactive power before capacitor was connected in parallel )
note : the connection of a capacitor in parallel will cause a change in power factor and reactive power while the active power will be unchanged i.e. p1 = p2
cos ∅2 = 0.85 ( new power factor )
hence ∅2 = 31.78°
Qsh ( reactive power when power factor is raised to 0.85 )
= P1 ( tan∅1 - tan∅2 )
= 10 ( 1.333 - 0.6197 )
= 7.13589 MVAR
Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclined plane (b) principal and maximum shear stresses and identify all the terms in the expression including units.
Answer:
a) Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
b) principal stress :
б1 = ( бx + бy ) / 2 - [tex]\sqrt{}[/tex]( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
Explanation:
Combined normal stress and shear stress sketches attached below
The terms in the sketch are :
бx = tensile stress in x direction
бy = tensile stress in y direction
Txy = y component of shear stress acting on the perpendicular plane to x axis
бn = Normal stress acting on the inclined plane EF
Tn = shear stress acting on the inclined plane EF
A) Normal and shear stresses on inclined plane
Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
B) principal and maximum shear stresses
principal stress :
б1 = ( бx + бy ) / 2 - [tex]\sqrt{}[/tex]( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
A production line manufactures 10-liter gasoline cans with a volume tolerance of up to 5%. The probability that any one is out of tolerance is 0.03. If five cans are selected at random. a) What is the probability that they are all out of tolerance? b) What is the probability that exactly two are out of tolerance?
Answer:
In the case of the production Line, we know that,
No of gasoline cans = 5
probability that 1st can is out of tolerance = 0.03
probability that 2nd can is out of tolerance = 0.03
.
.
probability that the 5th can is out of tolerance = 0.03
Therefore,
probability of 1st can out of tolerance + probability of 1st can not out of tolerance = 1
Probability of 1st can not out of tolerance = 1 -- 0.03 = 0.97
probability of 2nd can not out of tolerance = 0.97
.
.
probability of 5th can not out of tolerance = 0.97
Question A:
Probability that they are all out of tolerance
= P(1st can out of tolerance) * P(2nd can out of tolerance) * P(3rd can out of tolerance) * P(4th can out of tolerance) * P(5th can out of tolerance)
= (0.03 ) * (0.03) * (0.03) * (0.03) * (0.03) = 2.43 E⁻⁸ (2.43 ˣ 10⁻⁸)
Question B:
Probability that exactly two are out of tolerance
= P(1st can is out of tolerance) * P(2nd can is out of tolerance) * P(3rd can is not out of tolerance) * P(4th can is not out of tolerance) * P(5th can is not out of tolerance)
= (0.03) * (0.03) * (0.97) * (0.97) * (0.97) = 0.0008214057
Explanation:
A three-phase motor rated 25 hp, 480 V, operates with a power factor of 0.74 lagging and supplies the rated load. The motor efficiency is 96%. Calculate the motor input power, reactive power and current.
Answer:
the motor input power is 19.42 KW
the Reactive power is 17.65 KVAR
Current is 31.56 A
Explanation:
Given that;
V = 480V
h.p = 25 hp
p.f = 0.74 lagging
n_motor = 96%
so output = 25hp
and we know that;
1hp = 746 watt
watt = hp × 1hp
so output in watt = 25 × 746 = 18650 Watt = 18.65 KW
n_motor = (output / input) × 100
96 = 1865 / Input
96Input = 1865
Input = 1865 / 96
Input = 19.42 KW
Therefore the motor input power is 19.42 KW
P = √( 3 × V × I × cos∅)
19.42 = √( 3 ×480 × I × 0.74)
I = 31.56 A
Therefore Current is 31.56 A
Q = √( 3 × V × I × sin∅)
we know that
cos∅ = 0.74
so ∅ = cos⁻¹(0.74) = 42.26
so we substitute
Q = √( 3 × 480 × 31.56 × sin(42.26))
= 17.65 KVAR
Therefore the Reactive power is 17.65 KVAR
A battery with a nominal voltage of 200-V with a resistance of 10 milliohms to be charged at a constant current of 20 amps from a 3-phase semi-converter with a 220-V (line-to-line) Y-connected 60 Hs supply. Determine:
a. The firing angle of the thyristors for the charging process.
b. The displacement power factor and the supply power factor.
Answer:
a) ( ∝ ) = 69.6548
b) supply power factor = 0.6709
displacement power factor = 0.8208
Explanation:
Given data:
Nominal voltage ( E ) = 200-V
resistance (r) = 10 milliohms
constant current ( I ) = 20 amps
Phase ; 3-phase
semi-converter with 220-v ( line-to-line ) , 220√2 ( phase voltage )
frequency ; 60 Hz
a) determine the firing angle of thyristors
Vo = E + I*r
= 200 + 20*10*10^-3
= 200.2 v
attached below is the remaining part of the solution
firing angle of thyristors for charging process ( ∝ ) = 69.6548
b) determine displacement power factor and supply power factor
attached below is the detailed solution
Displacement power factor ( Dpf ) = cos ( ∝ /2 ) = 0.8208
displacement power factor = g * Dpf
= 0.81747 * 0.8208 = 0.6709
Products exit a combustor at a rate of 100 kg/sec, and the air-fuel ratio is 9. Determine the air flow rate. a. 9 kg/sec b. 90 kg/sec c. 100 kg/sec d. 10 kg/sec
Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer
Explanation:
Given that;
product of combustor flow rate m = 100 kg/s
air-fuel = 9
Airflow rate = ?
⇒We know that in the combustor, air fuel are mixed and then ignited,
⇒air fuel products are exited at the combustor
let air and fuel be a and b respectively
⇒ a + b = 100 kg/sec ----- let this be equation 1
now
⇒ air / fuel = 9
a / b = 9
a = 9b -----------let this be equation 2
now input a = 9b in equation 1
9b + b = 100 kg/sec
10b = 100 kg/sec
b = 10 kg/sec
we know that
a = 9b
so a = 9 × 10 = 90 kg/sec
Therefore the air flow rate a is 90 kg/sec
A semicircular loop of radius a in free space carries a current I. Determine the magnetic flux density at the center of the loop.
Answer: the magnetic flux density at the center of the loop is μ₀I / 4πα
Explanation:
Taking a look at the diagram;
we draw an imaginary curve to complete it as a circle.
we can now apply amperes law to write;
flux density B at centre;
∫B.dl = μ₀I
now since;
∫dl = 2πα
and field direction at centre is perpendicular to the screen, so it add up , and hence constant in magnitude.
so we can be taken out of the integral ,
B( 2πα ) = μ₀I
hence ;
B_circle = B = μ₀I / 2πα
so if we remove the half part of this;
we get a semicircle, which is what we are looking for;
Now
B_semi = 1/2.B = 1/2 × μ₀I / 2πα
B_semi = μ₀I / 4πα
Therefore the magnetic flux density at the center of the loop is μ₀I / 4πα
Contrast the electron and hole drift velocities through a 10 um (micro meter) layer of intrinsic silicon across which a voltage of 5V is imposed. Let up = 480cm2/Vs and un=1350cm2/Vs.
Answer:
Explanation:
Since we are considering electron and hole drift velocities, then electric field E will have to be taken into consideration as well.
Where E = V/d...... 1
Drift velocity (u) = -μE. For electron.... 2
Drift velocity (v) = μE. For hole...... 3
Given that : V = 5V and d = 10 um (micro meter)
From equation 1
E = V/d
E = 5V/10×10^-4cm
E = 5V ÷1/1000
E = 5×1000
E = 5000v/cm
From equation 2
Un = -μE.
Un = - 1350cm^2/vs × 5000
= -6750000cm/s
From equation 3
Vp = μE
= 480cm^2/vs × 5000
= 2400000cm/s
Since it was stated in the question that we should contrast between hole drift and electron drift.
6750000/2400000
= 2.8125
Hence the electron drift velocity is 2.8 times that of hole drift velocity indicating that the speed of the electron through the silicon was faster.
An unmanned satellite orbits the earth with a perigee radius of 10,000 km and an apogee radius of 100,000 km. Calculate:
a. the eccentricity of the orbit
b. the semimajor axis of the orbit(km)
c. the period of the orbit(hours)
d. the specific energy of the orbit(km^2/s^2)
e. the true anomaly at which the altitude is 1000km (degrees)
f. Vr and V(perpendicular) at the points found in part (e) (km/s)
g. the speed at the perigee and apogee (km/s)
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]
Which of following are coding languages used in controlling a robot? *
A. Scratch
B. B/B--
C. C/C++
D. Robot Z
Answer:
C/C++
Explanation:
C/C++
A cylindrical tank, 0.42 m in diameter, drains through a hole in its bottom. At the instant when the water depth is 0.73 m, the flow rate from the tank is observed to be 5 kg/s. Determine the rate of change of water level at this instant.
Answer:
The rate of change of water level at this instant is approximately -0.036 meters per second.
Explanation:
From this problem we know that outflow mass rate of water, current depth of water within the tank and the diameter of the cylindrical tank. From Physics we get the equation for the current mass of water inside the tank:
[tex]m = \rho \cdot \left(\frac{\pi}{4}\cdot D^{2}\cdot h \right)[/tex] (1)
Where:
[tex]m[/tex] - Current mass of water inside the tank, measured in kilograms.
[tex]D[/tex] - Diameter of the cylindrical tank, measured in meters.
[tex]h[/tex] - Depth of the water in the tank, measured in meters.
By differentiation in time, we derive the expression for the mass outflow of water ([tex]\frac{dm}{dt}[/tex]), measured in kilograms per second, by considering geometric characteristics of the cylinder ([tex]\frac{dD}{dt} = 0[/tex]) and the fact that water is an incompressible fluid ([tex]\frac{d\rho}{dt} = 0[/tex]):
[tex]\frac{dm}{dt} = \rho \cdot \frac{\pi}{4} \cdot D^{2}\cdot \frac{dh}{dt}[/tex] (2)
Where [tex]\frac{dh}{dt}[/tex] is the rate of change of water level, measured in meters per second.
If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]D = 0.42\,m[/tex] and [tex]\frac{dm}{dt} = - 5\,\frac{kg}{s}[/tex], then the rate of change of water level is:
[tex]\frac{dh}{dt} = \frac{4}{\pi \cdot \rho\cdot D^{2}} \cdot \frac{dm}{dt}[/tex]
[tex]\frac{dh}{dt} = \left[\frac{4}{\pi\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot (0.42\,m)^{2}} \right]\cdot (-5\,\frac{kg}{s} )[/tex]
[tex]\frac{dh}{dt} \approx -0.036\,\frac{m}{s}[/tex]
The rate of change of water level at this instant is approximately -0.036 meters per second.
Copy bits 3..0 in $s1 to 6..3 in $s2. Bits 6..3 in $s2 are already set to 0. Registers$s0 0..01111$s1 0..0101$s3 0
Answer:
Following are the solution to this question:
Explanation:
To copy 3.0 bits in 50 dollars or run at 50 dollars, it takes just 3.0 bits as well as other bits but masks, and 50 dollars.
Instead of shifting the $ 50 by 3 bits to 6...3 bits of [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex] This procedure instead took place at $53 and $50
AND [tex]\$ \ 50,\$ \ 50,0*0000 000f[/tex], take 3..0 bits
SLL [tex]\$ \ 50, \$ \ 50,0*0000 0003,[/tex]Shifts the bits to 6..3
O R [tex]\$ \ 53,\$ \ 53,\$ \ 50 ,[/tex] coping to [tex]\$ \ 53[/tex]
What is the basic molecular mechanism for polymer plasticity, and how does it differ from that of ductile crystalline metals?
Answer:
In Crystalline metals or materials, plasticity is examined from the perspective of the motion of linear defects or dislocations within the polymer chains.
Explanation:
When a temperature range below and near the glass transition temperature is reached, there is warping or contortion of structureless or malformed polymers. This warping happens as the polymer chains move over one another.
Unlike elasticity when requires or enables an object to resume its original dimensions, ductility is the quality of an element or material to change form albeit permanently.
Cheers
A roadway with a rough-asphalt pavement has a cross slope of 2%, a longitudinal slope of 2.5%, a curb height of 8 cm, and a 90-cm-wide concrete gutter. If the flow rate in the gutter is 0.07 m/s, determine the size (W XL, in mm) and interception capacity (m/s) of a reticuline grate that should be used to intercept as much of the flow as possible.
a. Reticuline grate size?
b. Interception capacity?
Answer:
b
Explanation:
Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.
Answer:
β = [tex]\frac{c}{\sqrt{km} }[/tex] = 0.7071 ≈ 1 ( damping condition )
closed-form expression for the response is attached below
Explanation:
Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s
computing a solution :
M = 1,
c = 2,
k = 2,
Wn = [tex]\sqrt{\frac{k}{m} }[/tex] = [tex]\sqrt{2}[/tex]
next we determine the damping condition using the damping formula
β = [tex]\frac{c}{\sqrt{km} }[/tex] = 0.7071 ≈ 1
from the condition above it can be said that the damping condition indicates underdamping
attached below is the closed form expression for the response
A 40Ω resistor, a 5 mH inductor, and a 1.25μF capacitor are connected in series. The series-connected elements are energized by a sinusoidal voltage source whose voltage is 600 cos(8000t + 20°) V.
Required:
a. Draw the frequency-domain equivalent circuit.
b. Reference the current in the direction of the voltage rise across the source, and find the phasor current.
c. Find the steady-state expression for i(t).
Solution :
Given
From the voltage expression, w = 8000 rad/s
Inductive reactance, XL = jwL = j 40 Ω
Capacitive reactance, XC = -j/wC
= - j 100 Ω
Now, [tex]$I=\frac{Vs}{Z}$[/tex]
Z = 40 + j40 - j100
= 46 - j 60 Ω
= [tex]$72.1 \angle -56.3^\circ$[/tex]
So, [tex]$I=\frac{Vs}{Z}$[/tex]
[tex]$=\frac{600 \angle 20^\circ}{72.1 \angle -56.3^\circ}$[/tex]
[tex]$=8.32\ \angle 76.3^\circ$[/tex]
Therefore,
[tex]$i(t) = 8.32 \ \cos(8000t+76.3^\circ)$[/tex]
Calculate the molar volume of ammonia at 92oC and 310 bar. What phase is the ammonia in?
Answer:
The ammonia is still in the gas phase
Explanation:
Given that 1 bar is approximately = 1 atm
From;
PV=nRT
P= 310 atm
V= the unknown
n= 1
R = 0.082atm LK-1mol-1
T = 92oC + 273 = 365 K
V= nRT/P
V= 1 * 0.082 * 365/310
V = 0.0965 L = 96.5 mL
molar mass of NH3 = 17 g/mol
Molar density of NH3 = 17g/ 96.5mL = 0.176g/mL
The ammonia is still in the gas phase