Answer:
the inertia of the crate is (49.67 kg)r²
Explanation:
Given the data in the question;
First; we will use the law of conservation of momentum to determine the mass of the crate;
m₁v₁ - m₂v₂ = 0
given that; m₁ = 0.60 kg and v₂ = 0.058 m/s
we substitute
0.60 × v₁ = m₂ × 0.058 = 0
m₂ = 0.60v₁ / 0.058 ----------- EQU 1
Next, we use the energy conservation relation to find the velocity
According to conservation of energy;
1/2m₁v₁² + 1/2m₂v₂² = 7 J
we substitute
1/2×0.60×v₁² + 1/2×m₂×(0.058)² = 7 J
0.3v₁² + 0.001682m₂ = 7 J ----- EQU 2
substitute value of m₂ form equ 1 into equ 2
0.3v₁² + 0.001682(0.60v₁ / 0.058) = 7 J
0.3v₁² + 0.0174v₁ = 7 J
0.3v₁² + 0.0174v₁ - 7 J = 0
we solve the quadratic equation;
{ x = [-b±√( b² - 4ac)] / 2a }
v₁ = [-0.0174 ±√( 0.0174² - 4×0.3×-7)] / 2×0.3
= [-0.0174 ±√(8.4003)] / 0.6
= [-0.0174 ± 2.8983 ] / 0.6
= -4.8595 or 4.8015 but{ v₁ ≠ - }
so v₁ = 4.8015 m/s ≈ 4.802 m/s
next we input value of v₁ into equation 1
m₂ = (0.60×4.8015) / 0.058
m₂ = 2.8809 / 0.058
m₂ = 49.67 kg
So, the moment of inertia of the crate will be;
I₂ = m₂r²
we substitute value of m₂
I₂ = (49.67 kg)r²
Therefore, the inertia of the crate is (49.67 kg)r²
PLZ FAST!!
Compare and contrast microscopic and macroscopic energy transfer. Give at least three comparisons for each. THX
Answer:
Macroscopic energy is energy at a level of system while microscopic energy is energy at the level of atoms and molecules
Explanation:
1. Macroscopic energy is possessed by a system as whole while microscopic energy is possessed by its constituents’ atoms or molecules.
2. The common form of macroscopic energy is Kinetic and potential energy while the microscopic form of energy are atomic forces due its random, disordered motion and due to intermolecular forces
3. At microscopic level we consider behaviour of every molecule and in macroscopic approach we consider gross or average effects of various molecular infractions
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a) Determine the forces and and the couple (b) Determine the sum of the moments about the right end of the beam. (c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the
This question is incomplete, the complete question is;
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.
(a) Determine the forces and and the couple
(b) Determine the sum of the moments about the right end of the beam.
(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?
Answer:
a)
the x-component of the force at A is [tex]A_{x}[/tex] = 0
the y-component of the force at A is [tex]A_{y}[/tex] = 400 N
the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m
b)
the sum of the momentum about the right end of the beam is; ∑[tex]M_{R}[/tex] = 0
c)
the equivalent force acting at the left end is; F = -400J ( N)
the couple acting at the left end is; M = - 146 N-m
Explanation:
Given that;
The sum of the forces acting on the beam is zero ∑f = 0
Sum of the moments about the left end of the beam is also zero ∑[tex]M_{L}[/tex] = 0
Vector force acting at A, [tex]F_{A}[/tex] = [tex]A_{x}i[/tex] + [tex]A_{y}j[/tex]
Now, From the image, we have;
a)
∑f = 0
[tex]F_{A}[/tex] - 600j + 200j = 0i + 0j
[tex]A_{x}i[/tex] + [tex]A_{y}j[/tex] - 600j + 200j = 0i + 0j
[tex]A_{x}i[/tex] + ([tex]A_{y}[/tex] - 400)j = 0i + 0j
now by equating i- coefficients'
[tex]A_{x}[/tex] = 0
so, the x-component of the force at A is [tex]A_{x}[/tex] = 0
also by equating j-coefficient
[tex]A_{y}[/tex] - 400 = 0
[tex]A_{y}[/tex] = 400 N
hence, the y-component of the force at A is [tex]A_{y}[/tex] = 400 N
we also have;
∑[tex]M_{L}[/tex] = 0
[tex]M_{A}[/tex] - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0
[tex]M_{A}[/tex] - 30 N-m - 228 N-m + 112 Nm = 0
[tex]M_{A}[/tex] - 146 N-m = 0
[tex]M_{A}[/tex] = 146 N-m
Therefore, the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m
b)
The sum of the moments about right end of the beam is;
∑[tex]M_{R}[/tex] = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)([tex]A_{y}[/tex] ) + [tex]M_{A}[/tex]
∑[tex]M_{R}[/tex] = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m
∑[tex]M_{R}[/tex] = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m
∑[tex]M_{R}[/tex] = 0
Therefore, the sum of the momentum about the right end of the beam is; ∑[tex]M_{R}[/tex] = 0
c)
The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.
The equivalent force at the left end will be;
F = -600j + 200j (N)
F = -400J ( N)
Therefore, the equivalent force acting at the left end is; F = -400J ( N)
Also couple acting at the left end
M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)
M = -(30 N-m) + (112 N-m) - ( 228 N-m))
M = 112 N-m - 258 N-m
M = - 146 N-m
Therefore, the couple acting at the left end is; M = - 146 N-m
A compact car has a mass of 1310 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs.
Required:
a. What is the spring constant of each spring if the empty car bounces up and down 2.0 times each second?
b. What will be the car’s oscillation frequency while carrying four 70 kg passengers?
Answer:
a) k= 3232.30 N / m, b) f = 4,410 Hz
Explanation:
In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.
The expression for the angular velocity is
w = √k/m
the angular velocity is related to the period
w = 2π / T
we substitute
T = 2[tex]\pi[/tex] √m/ k
a) empty car
k = 4π² m / T²
k = 4 π² 1310/2 2
k = 12929.18 N / m
This is the equivalent constant of the short springs
F1 + F2 + F3 + F4 = k_eq x
k x + kx + kx + kx = k_eq x
k_eq = 4 k
k = k_eq / 4
k = 12 929.18 / 4
k= 3232.30 N / m
b) the frequency of oscillation when carrying four passengers.
In this case the plus is the mass of the vehicle plus the masses of the passengers
m_total = 1360 + 4 70
m_total = 1640 kg
angular velocity and frequency are related
w = 2pi f
we substitute
2 pi f = Ra K / m
in this case the spring constant changes us
k_eq = 12929.18 N / m
f = 1 / 2π √ 12929.18 / 1640
f = π / 2 2.80778
f = 4,410 Hz
Bill is not only good at riding a bike with no hands, but he can also ride the bike with no hands and facing backwards while the bike goes forward. Bill is riding his bike in such a manner while playing catch with Betty who is stationary on the ground and facing Bill. Bill is on the bike (facing backwards toward Betty) and traveling away from Betty at a speed of 12.0 m/s when he throws a ball to Betty at a speed of 17.8 m/s relative to the bike. What is the velocity of the ball as measured by Betty
Answer:
5.8 m/s
Explanation:
Let v = velocity of bike relative to Betty = -12.0 m/s (since the bike is moving away from betty).
u = velocity of ball relative to bike = + 17.8 m/s
and V = velocity of ball relative to Betty.
So, by Galilean relativity,
V = v + u
V = -12.0 m/s + 17.8 m/s
V = 5.8 m/s
So, the velocity of the ball as measured by Betty is 5.8 m/s
A cheerleader of mass 55 kg stand on the shoulders of a football player of mass 86 kg. The football player is standing in a soft, thin layer of mud that does not permit air under his shoes. If each of his shoes has an area of 264 cm2, calculate the absolute pressure exerted on the surface underneath one of the shoes. Answer in Pascal, assuming g = 9.80 m/s2 and atmospheric pressure is 101,000 Pa.
While investigating Kirchhoff's Laws, you begin observing a blackbody, such as a star, from Earth using advanced technology that can analyze spectra. While pointing it at the star with nothing between you and the star, you observe a full spectrum. You come back and repeat this same experiment a year later using the same star, except this time you observe an absorption spectrum. What is the most likely explanation for this
Answer:
the second time there is a gas between you and the star,
Explanation:
When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.
When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,
The information obtained from the two spectra is the same, the type of atoms that make up the star
state four law of photoelectric effect
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.
---------------------------------------------
LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.
-----------------------------------------------
LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.
-----------------------------------------------
LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.
Explanation:
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.
A long copper bar of rectangular cross-section, whose width w is much greater than its thickness L, is maintained in contact with a heat sink at its lower surface, and the temperature throughout the bar is approximately equal to that of the sink, To. Suddenly, an electric current is passed through the bar and an airstream of temperature T is passed over the top surface, while the bottom surface continues to be maintained at To. Obtain the differential equation and the top surface boundary condition that could be solved to determine the temperature as a function of position and time in the bar.
How is energy transferred when the
torch is switched on?
Which of the following is a vector quantity?
speed
distance
acceleration
When monochromatic light passes through the interface between two unknown materials at an angle θ where 0∘<θ<90∘, no changes in the direction of propagation of light are observed. What can be said about the two materials? Check all that apply. View Available Hint(s) Check all that apply. The two materials have matching indexes of refraction. The second material through which light propagates has a lower index of refraction. The second material through which light propagates has a higher index of refraction. The two materials are identical.
Answer:
the correct one is the first, the refractive index of the two materials must be the same
Explanation:
When a beam of light passes through two materials, it must comply with the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of each medium.
In this case, it indicates that the light does not change direction, so the input and output angle of the interface must be the same,
θ₁ = θ₂ = θ
substituting
n₁ = n₂
therefore the refractive index of the two materials must be the same
When reviewing the answers, the correct one is the first
Compare and Contrast the Following:
1. Pitch and Loudness
2. Infrasonic and Ultrasonic
3. Luminous and Nonluminous
It's okay if you answer only one of these.
Thanks in Advance
Answer:
pitch and loudnesss
thanks for your question
Welding requires extensive training.
True
False
Answer:
True.
Explanation:
Welding requires extensive training because welding involves fire and we need to use fire safety measurements. A normal man can't just simply go and weld so a person must require extensive training for welding.
during a typical afternoon thunderstorm in the summer, an area of 66.0 km2 receives 9.57 108 gal of rain in 18 min. how many inches of rain fell during this 18 min period
Answer:
2.16 inch
Explanation:
area under water = 66 km²
= 66 x ( 3280.84 x 12 )² inch²
= 1.023 x 10¹¹ sq inch
volume of rain = 9.57 x 10⁸ gallon = 9.57 x 10⁸ x 231 inch³
= 2.21 x 10¹¹ inch³
If depth of rainfall be t
volume of rain = surface area x depth
= 1.023 x 10¹¹ x t
So ,
1.023 x 10¹¹ x t = 2.21 x 10¹¹
t = 2.16 inch
All charged objects exert a force that can cause other charges to move. What is the force that
charged objects give off called? What else can it do?
Answer:
exerts force
Explanation:
The accumulation of excess electric charge on an object is called static electricity. ... An electric field surrounds every electric charge and exerts the force that causes other electric charges to attract or repel. Electric fields are represented by arrows showing the electric field would make a positive charge move.
Passage of an electric current through a long conducting rod of radiusriand thermalconductivitykrresults in uniform volumetric heating at a rate ofq. The conducting rodis wrapped in an electrically nonconducting cladding material of outer radiusroandthermal conductivitykc, and convection cooling is provided by an adjoining fluid. Forsteady-state conditions, write appropriate forms of the heat equations for the rod andcladding. Express appropriate boundary conditions for the solution of these equations.
Answer:
a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface; [tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0
b) For constant temperature; [tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])
c) The heat transfer in the conducting rod and the cladding material is the same, i.e; [tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] = [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]
d) The convection surface conduction by cooling fluid will be;
[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )
Explanation:
Given the data in question;
we write the general form of the heat conduction equation equation in cylindrical coordinates with internal heat generation.
1/r[tex]\frac{d}{dr}[/tex]( kr[tex]\frac{dT}{dr}[/tex] ) + 1/r² [tex]\frac{d}{d\beta }[/tex]( ( k[tex]\frac{dT}{dr}[/tex] ) + [tex]\frac{d}{dz}[/tex]( k[tex]\frac{dT}{dr}[/tex]) + q = 0
where radius of cylinder is r, thermal conductivity of the cylinder is k, and q is heat generated in cylinder.
Now, Assume one dimensional heat conduction
lets substitute the condition for conducting rod with steady state condition.
[tex]k_{y}[/tex]/r [tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{y} }{dr}[/tex] ) + q = 0
Apply the conditions for cladding by substituting 0 for q
[tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{r} }{dr}[/tex] ) = 0
Apply the following boundary conditions;
a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface;
[tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0
b) For constant temperature
[tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])
c) The heat transfer in the conducting rod and the cladding material is the same, i.e
[tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] = [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]
d) The convection surface conduction by cooling fluid will be;
[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )
What is Ex(P), the value of the x-component of the electric field produced by by the line of charge at point P which is located at (x,y) = (a,0), where a = 8.7 cm?
Answer:
The answer is below
Explanation:
We are going to use Gauss’ law to find the electric field equation. Since electric field is coming from an infinite line of charge, hence it is going out in a radial direction.
Therefore we use the area of the electric field which passes through, forming a Gaussian cylinder. We neglect the ends of the area.
Hence:
[tex]\int\limits {E} \, dA=\frac{Q_{enc}}{\epsilon_o}\\\\E(2\pi rL)= \frac{\lambda L}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r\epsilon_o} \\\\Given \ that:\\\\r=a=8.7\ cm=0.087\ m, \lambda=-2.3 \mu C/cm=-2.3*10^{-4}\ C/m,\epsilon_o=8.85*10^{-12}F/m.\\\\Hence:\\\\E=\frac{-2.3*10^{-4}}{2\pi *0.087*8.85*10^{-12}}=-4.75*10^7\ N/C[/tex]
The value of the x-component of the electric field is -475213.968 newtons per coulomb.
Procedure - Determination of the magnitude of an electric field at a given pointIn this question we shall apply Gauss' Law to determine the magnitude of the electric field ([tex]E_{x}[/tex]), in newtons per coulomb, rapidly and based on the assumptions of uniform charge distribution and cylindrical symmetry.
[tex]\frac{Q_{enc}}{\epsilon_{o}} = \oint\,\vec E\,\bullet d\vec A[/tex] (1)
Where:
[tex]Q_{enc}[/tex] - Enclosed charge, in coulombs.[tex]\epsilon_{o}[/tex] - Vacuum permitivity, in quartic second-square amperes per kilogram-cubic meter.[tex]\vec E[/tex] - Electric field vector, in newtons per coulomb.[tex]\vec A[/tex] - Area vector, in square meters.Based on all assumptions, we simplify (1) as follows:
[tex]\frac{\lambda\cdot l}{\epsilon_{o}} = E \cdot (2\pi\cdot r\cdot l)[/tex]
And the equation of the x-component of the electric field is:
[tex]E = \frac{\lambda}{2\pi\cdot \epsilon_{o}\cdot r}[/tex] (2)
Where [tex]\lambda[/tex] is the linear charge density, in coulomb per meter.
If we know that [tex]\lambda = -2.3\times 10^{-6}\,\frac{C}{m}[/tex] and [tex]a = 0.087\,m[/tex], then the electric field produced by the line of charge at point P is:
[tex]E = \frac{\left(-2.3\times 10^{-6}\,\frac{C}{m} \right)}{2\pi\cdot \left(8.854\times 10^{-12}\,\frac{s^{4}\cdot A^{2}}{kg\cdot m^{3}} \right)\cdot (0.087\,m)}[/tex]
[tex]E_{x} = -475213.968 \,\frac{N}{C}[/tex]
The value of the x-component of the electric field is -475213.968 newtons per coulomb. [tex]\blacksquare[/tex]
RemarkThe figure is missing, we present the corresponding image in the file attached below.
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the luminous flux of a torch of intensity 50 cd is?
Answer:
i dont know i am right but here Luminous intensity is defined as dI=dΨλ / dΩ, where dΨλ is the luminous flux (light energy flux in watts per m2) emitted within a solid angle dΩ. The light energy flux may be expressed in terms of the incident x-ray energy flux and the x-ray absorption and conversion properties of the scintillator(7,8,9).
Explanation:
What energy store is in the torch
BEFORE it gets switched on?
Answer:
Chemical energy
Explanation:
The energy in the torch is stored as chemical energy before the torch gets switch on.
The chemical energy energy in the battery of cell will power the cell and allows it to produce light.
Chemical energy is a form of potential energy. The electrolytes within the battery are capable of producing electric current. So the chemical energy is transformed into electrical energy which is used to produce the light of the torch.2. Mrs. Stern is standing still on rollerblades on a frictionless floor in the middle of the A-gym while
carrying heavy textbooks. How can she use the textbooks to get herself moving?
Answer:
If she bends forward
Explanation:
because the equilibrium of gravity will not stay the same causing her to move forward
A three-phase line, which has an impedance of (2 + j4) ohm per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30 + j40) ohm per phase, and the other is connected with an impedance of (60 - j45) ohm per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 120 √3V (rms, line-to-line).
Determine:
a. the current, real power and reactive power delivered by the sending-end source
b. the line-to-line voltage at the load
c. the current per phase in each load
d. the total three-phase real and reactive powers absorbed by each load and by the
Answer:
hello your question has a missing information
The other is Δ-connected with an impedance of (60 - j45) ohm per phase.
answer : A) 5A ∠0° ,
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) 193.64 v
C) current at load 1 = 2.236 A , current at load 2 = 4.472 A
D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )
Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )
Explanation:
First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution
A) determine the current, real power and reactive power delivered by the sending-end source
current power delivered (Is) = 5A ∠0°
complex power delivered ( s ) = 3vs Is
= 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )
also s = p + jQ ------ ( 2 )
comparing equation 1 and 2
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) determine Line-to-line voltage at the load
Vload = √3 * 111.8
= 193.64 v
c) Determine current per phase in each load
[tex]I_{l1} = Vl1 / Zl1[/tex]
= [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A hence current at load 1 = 2.236 A
[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]
= [tex]\frac{111.8<-10.3}{25<-36.87}[/tex] = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A
D) Determine the Total three-phase real and reactive powers absorbed by each load
For load 1
3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR
for load 2
3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex] = 1200 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR
The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.
(a) The impedance per phase of the equivalent Y,
[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]
The phase voltage,
[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]
Total impedance from the input terminals,
[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]
The three-phase complex power supplied [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]
P =1800 W and Q = 0 VAR delivered by the sending-end source.
(b) Phase voltage at load terminals will be,
[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]
The line voltage magnitude at the load terminal,
[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]
(c) The current per phase in the Y-connected load,
[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} }[/tex]
The phase current magnitude,
[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]
(d) The three-phase complex power absorbed by each load,
[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]
The three-phase complex power absorbed by the line is
[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]
Since, the sum of load powers and line losses,
[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]
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A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 77.0-kg water-skier has an initial speed of 6.3 m/s. Later, the speed increases to 10.9 m/s. Determine the work done by the net external force acting on the skier.
Answer:
the work done by the net external force acting on the skier is 3046.12 J.
Explanation:
Given;
initial speed of the water skier, u = 6.3 m/s
final speed of the water skier, v = 10.9 m/s
mass of the water skier, m = 77 kg
The work done by the net external force is calculated as;
W = ΔK.E
[tex]W = \frac{1}{2} m(v^2 - u^2)\\\\W = \frac{1}{2} \times \ 77.0(10.9^2 - 6.3^2)\\\\ W= 3046.12 \ J[/tex]
Therefore, the work done by the net external force acting on the skier is 3046.12 J.
Select the Moon and use the Info view to determine which of the following statements is correct. The Last Quarter Moon.... rises near noon and sets near midnight. rises at about 6am and sets at about 6pm. rises near midnight and sets near midday. rises and sets at the same time as the Sun. Submit Your Answer
Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).
Explanation:
Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.
There are 8 main types of the moon phases these includes:
--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.
--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.
--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.
--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.
--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.
--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker
--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.
--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.
The diagram shows two balls before they collide.
2 balls with grey arrows pointing to them from the outside. The left ball has below it m subscript 1 = 0.6 kilograms v subscript 1 = 0.5 meters per second. The right ball has below it m subscript 2 = 0.5 kilograms v subscript 2 = negative 0.2 meters per second.
What is the momentum of the system after the collision?
1. 0.0 kg • m/s
2. 0.2 kg • m/s
3. 0.3 kg • m/s
4. 0.4 kg • m/s
Answer:
The Answer is B)0.2 kg • m/s
Explanation:
I made a 100 on my test. Sorry if I'm late but hope I helped.
Answer:
B. 0.2 kg x m/s
Explanation:
At an airport, two business partners both walk at 1.5 m/sm/s from the gate to the main terminal, one on a moving sidewalk and the other on the floor next to it. The partner on the moving sidewalk gets to the end in 60 ss, and the partner on the floor reaches the end of the sidewalk in 90s.
Required:
What is the speed of the sidewalk in the Earth reference frame?
Answer:
[tex]v=0.8m/s[/tex]
Explanation:
From the question we are told that
Distance [tex]d=1.5m/sm/s[/tex]
Time [tex]t_1=60s[/tex]
Time [tex]t_2=90s[/tex]
Generally the the equation for the distance traveled is mathematically given as
[tex]d=vt[/tex]
[tex]d=1.5*90[/tex]
[tex]d=138m[/tex]
Generally equation for speed of side walk is mathematically given as
[tex]d=(v+u)t[/tex]
[tex]v=\frac{d}{t}-u[/tex]
[tex]v=\frac{138}{60}-1.5[/tex]
[tex]v=0.8m/s[/tex]
A flat circular mirror of radius 0.100 m is lying on the floor. Centered directly above the mirror, at a height of 0.920 m, is a small light source. Calculate the diameter of the bright circular spot formed on the 2.70 m high ceiling by the light reflected from the mirror.
Answer:
the diameter of the bright circular spot formed is 0.787 m
Explanation:
Given that;
Radius of the flat circular mirror = 0.100 m
height of small ight source = 0.920 m
high ceiling = 2.70 m
now;
Diameter(mirror) = 2×r = 2 × 0.100 = 0.2 m
D(spot) = [Diameter(mirror) × ( 2.70m + 0.920 m)] / 0.920 m
so
D(spot) = 0.2m × 3.62m / 0.920 m
D(spot) = 0.724 m / 0.920 m
D(spot) = 0.787 m
Therefore, the diameter of the bright circular spot formed is 0.787 m
what measurement do geologists use to find absolute age
Answer:
see below :)
Explanation:
Radiometric dating.
a wooden block is cut into two pieces, one with three times the mass of the other. a depression is made in both faces of the cut so that a fire cracker can be placed in it and the block is reassembled. the reassembled block is set on rough surface and the fuse is lit. when the fire cracker explodes, the two blocks separate. what is the ratio of distances traveled by blocks?
Answer:
1/9
Explanation:
Let A denote the bigger piece and let B denote the smaller piece.
We are told that one with three times the mass of the other.
Therefore, we have;
M_a = 3M_b
Firecracker is placed in the block and it explodes and thus, momentum is conserved.
Thus;
V_ai = V_bi = 0
Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.
Since initial momentum equals final momentum, we have;
P_i = P_f
Thus;
0 = (M_a × V_af) + (M_b × V_bf)
Since M_a = 3M_b, we have;
(3M_b × V_af) + (M_b × Vbf) = 0
Making V_af the subject, we have;
V_af = -⅓V_bf
The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;
W_f = -½M_b•(v_bf)²
Now, let's express the work is in terms of the force and the distance.
Thus;
W_f = F_f × Δx × cos 180°
Frictional force is also expressed as μmg
Thus;
W_f = -μM_b × g × Δx
Earlier, we saw that;
W_f = -½M_b•(v_bf)²
Thus;
-½M_b•(v_bf)²= -μM_b × g × Δx
Δx = (v_bf)²/2μg
Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b
Thus;
Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)
Δx_a/Δx_b = ((v_af)²/((v_bf)²)
Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²
Δx_a/Δx_b = 1/9
An automobile follows a circular road whose radius is 50 m. Let x and y respectively denote the eastern and northern directions, with origin at the center of the circle. Suppose the vehicle starts from rest at x = 50 m heading north, and its speed depends on the distance s it travels according to v = 0.5s − 0.0025s 2 , where s is measured in meters and v is in meters per second. It is known that the tires will begin to skid when the total acceleration of the vehicle is 0.6g. Where will the automobile be and how fast will it be going when it begins to skid? Describe the position in terms of the angle of the radial line relative to the x axis.
Answer:
The automobile is running at speed of 23.806 meters per second.
Explanation:
From Kinematic we remember that acceleration ([tex]a[/tex]) can be defined by this ordinary differential equation in terms of distance:
[tex]a = v\cdot \frac{dv}{ds}[/tex] (1)
Where:
[tex]v[/tex] - Speed of the automobile, measured in meters per second.
[tex]s[/tex] - Distance travelled by the automobile, measured in meters.
If we know that [tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex], then the equation of acceleration is:
[tex]a = (0.5\cdot s - 0.0025\cdot s^{2})\cdot \left(0.5-0.0050\cdot s\right)[/tex]
[tex]a = s\cdot (0.5-0.0025\cdot s)\cdot (0.5-0.0050\cdot s)[/tex]
[tex]a = s\cdot (0.0025\cdot s - 0.5)\cdot (0.0050\cdot s-0.5)[/tex]
But distance covered by the vehicle is defined by the following formula:
[tex]s = \theta \cdot r[/tex] (2)
Where:
[tex]\theta[/tex] - Arc angle, measured in radians.
[tex]r[/tex] - Radius, measured in radians.
Then, we expand (1) by means of this result:
[tex]a = \theta\cdot r \cdot (0.0025\cdot \theta\cdot r -0.5)\cdot (0.0050\cdot \theta \cdot r-0.5)[/tex]
[tex]a = \theta\cdot r \cdot (1.25\times 10^{-5}\cdot \theta^{2}\cdot r^{2}-3.75\times 10^{-3}\cdot \theta\cdot r +0.25)[/tex]
[tex]a = 1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r[/tex]
And finally we get the following third order polynomial:
[tex]1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r - a = 0[/tex] (3)
If we know that [tex]r = 50\,m[/tex], [tex]a = 0.6\cdot g[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the polynomial becomes into this:
[tex]1.5625\cdot \theta^{3}-9.375\cdot \theta^{2} +12.5\cdot \theta - 5.886 = 0[/tex] (3b)
This polynomial can be solved analytically by Cardano's Method or by numerical methods. The roots of the polynomial are, respectivelly:
[tex]\theta_{1} \approx 4.365\,rad[/tex], [tex]\theta_{2} \approx 0.818+i\,0.441\,rad[/tex], [tex]\theta_{3}\approx 0.818 -i\,0.441\,rad[/tex], [tex]\theta_{4} \approx 1.563\,rad[/tex]
Both first and fourth roots are physically reasonable solution, but the latter represents the angle where automobile begins to skid first. Then, the automobile begins to skid at an angle of 1.563 radians relative to x axis.
The distance travelled by the automobile is: ([tex]r = 50\,m[/tex], [tex]\theta \approx 1.563\,rad[/tex])
[tex]s = (1.563\,rad)\cdot (50\,m)[/tex]
[tex]s = 78.15\,m[/tex]
Lastly, the speed of the automobile at this location is: ([tex]s = 78.15\,m[/tex])
[tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex] (4)
[tex]v = 0.5\cdot (78.15)-0.0025\cdot (78.15)^{2}[/tex]
[tex]v = 23.806\,\frac{m}{s}[/tex]
The automobile is running at speed of 23.806 meters per second.
A parallel-plate capacitor with plate area 2.6 cm^2 and air-gap separation 0.25 mm is connected to a 30 V battery, and fully charged. The battery is then disconnected.
I Solved most of the parts of this q's but not this one:
The plates are now pulled to a separation of 0.55 mm. What is the charge on the capacitor now?
note:
I found 125.5pC and the answer unit is pC
Answer:
276.12 pC
Explanation:
We are given that
Area,A=[tex]2.6 cm^2=2.6\times 10^{-4}m^2[/tex]
Where [tex]1 cm^2=10^{-4} m^2[/tex]
[tex]d=0.25 mm=0.25\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
Potential difference, V=30 V
We have to find the charge on the capacitor when the plates are pulled to a separation of 0.55 mm.
We know that
Charge ,[tex]Q=\frac{\epsilon_0 A V}{d}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
Using the formula
[tex]Q=\frac{8.85\times 10^{-12}\times 2.6\times 10^{-4}\times 30}{0.25\times 10^{-3}}[/tex]
[tex]Q=2.7612\times 10^{-10} C[/tex]
[tex]Q=276.12p C[/tex]
[tex]1 pC=10^{-12} C[/tex]
When the plates are now pulled to a separation of 0.55 mm.Then, the charge on the plates remain same because the battery has been disconnected.
Therefore, charge on the capacitor=276.12 pC