Answer:
yes you can find a motor in a circuit
Which reverses the flow of current through
an electric motor?
Answer:
a commutator
Explanation:
Assignment: Ultrasound and Infrasound Research Exploration Ultrasound and infrasound are categories of sound with different frequencies. These categories of sound can be used for a variety of different applications. In this research assignment, you will take a look at the industrial applications of ultrasound and infrasound. Your essay should do the following things: • Define and describe in detail (and in your own words) ultrasound and infrasound • Describe how ultrasound and infrasound are used in specific industrial applications and provide detailed examples. When conducting your research, remember to gather information from multiple sources. Start your research by defining ultrasound and infrasound. Then find ideas about industrial applications. The essay should be about 350-450 words, which is 1-1.5 typed pages, double-spaced, using 12-pt. font. Check the rubric to review how you will be graded for this assignment. plz no funny answers
Answer:
Infrasound vs. Ultrasound: Infrasound is sound that is below the lower limit of human hearing, below 20 Hz, and ultrasound is above the upper limit of human hearing, above 20,000 Hz. Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes, graphing rock and oil developments underneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.
For instance, a few creatures, for example, whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, water falls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.
Explanation:
idk how many words this is but its a start for u to add on to and i hope this helps and its in my own words - pls mark me brainiest
Answer:
Ultrasound vs. Infrasound Research Exploration
Beyond the limit of human hearing, ultrasound is above 20,000 Hz. Under the limit of human hearing, infrasound is below 20 Hz. Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes as well as graphing rock and oil developments beneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.
For instance, creatures like whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, waterfalls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.
(Not turned in yet, but this is what i have so far. Good luck 8th graders <33)
-Sav xx
There is a Full Moon on September 14th. On which date will the New Moon occur? A.September 21st B.September 28th C.13th D.October 31st
Answer:
the answer B
Explanation:
(c) An object of mass 100 kg is accelerated uniformly from a velocity of 15m/s to 20 m/s in 10 s.
Calculate the initial and final momentum of the object.
Also, find the magnitude of the force exerted on the object.
Explanation:
momentum = mass x velocity
initial momentum = 100 x 15 = 1500kgm/s
after momentum = 100 x 20 = 2000kgm/s
a =(v-u)/t
a = (20-15)/10
a = 5/10
a = 0.5m/s²
f = ma
f = 100 x 0.5
f = 50N
Question 21 of 25
Barry slides across an icy pond. The coefficient of kinetic friction between his
shoes and the ice is 0.15. If his mass is 83 kg, what is the force of friction
acting on him
O A 352 N
O B. 122 N
C. 1150 N
O D. 813N
Answer:
B. 122N
Explanation:
f = 0.15 x 83 x 9.8 = 122.01
f = 122N
1. Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse.
2. Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude.
Answer:
1) a α, m I, W=F.d W =τ . θ,
2) a = v²/r
Explanation:
1) The amounts of rotational and translational motion are related
acceleration is
a = d²x / dt²
linear displacement is equivalent to angular rotation, therefore angular acceleration is
α = d²θ / dt²
force in linear motion is equivalent to moment in endowment motion
F = m a
τ = I α
the mass is the inertia of the translation, in rotational motion the moment of inertia is the rotational inertia
I = m r²
Work is defined by W = F. d
in rotation it is defined by W = τ . θ
The linear momentum is p = mv
the angular momentum L = I w
momentum the linear motion is I = F dt
in the rotation it is I = τ dt
2) The velocity is a vector therefore it has modulus and direction, linear acceleration changes the modulus of velocity, whereas circular motion changes the direction (the other element of the vector).
[tex]a_{c}[/tex]Ac = v²/r
Did to gravity, the moon has a much smaller acceleration than earth. How do you think that would affect the period of pendulum?
I think any pendulum would swing slower on the moon than it would on Earth.
The time it takes a pendulum to go through a complete back and forth swing is:
Time period = 2 π √(length/gravity)
You can see that 'gravity' is in the denominator of the fraction, so the smaller gravity gets, the longer the period gets.
To be a little bit more technical, the period is inversely proportional to the square root of gravity.
So the period for a complete swing on the moon would be √(9.8/1.6) times as long as the complete swing of the same pendulum on Earth.
That number is roughly 2.47 .
So, for every 1 second that a pendulum takes to swing back and forth once on Earth, the same pendulum would take 2.47 seconds to do it on the moon.
Answer:
based on my opinion....
as we know that gravity in moon are less than gravity in earth.. since the force of gravity is less on the moon, the pendulum would swing slower at the same length, angle
and the frequency would be less.
I hope this helps
As light from a star spreads out and weakens, do gaps form between the photons?
Answer:
There are no gaps in space between the photons as they travel. If you look at light as a wave, then there no gaps unless specifically placed there on purpose. ... The light from a distance star indeed spreads out and weakens as it travels, but this just reduces the wave strength and does not introduce gaps.
Which statement describes how work and power are similar?
A: You must know time and energy to calculate both.
B: You must know velocity and time to calculate both.
C: You must know force and distance to calculate both.
D: You must know energy and force to calculate both.
Answer:
C: You must know force and distance to calculate both.
Explanation:
A force is said to have done work, when it has succeeded in causing displacement in an object from its starting point.
Power can be defined as the rate of doing work.
The work can be calculated using the following formula:
Work done = Force × Displacement
The power can be calculated using the following formula:
Power = Work \ Time
In order to calculate the work and power, the force and distance should be known. The force and distance will help in the calculating the work, which will in turn help in the calculation of the power of the object.
Hence, the answer is you must know force and distance to calculate both.
formula of minimmum pressure
Answer:
pressure=force/area
The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is the kinetic energy of the protons when they are ejected from the cyclotron
Answer:
1.92 x 10⁻¹²J
Explanation:
The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e
F = ma
Where;
a = [tex]\frac{v^2}{r}[/tex] [v = linear velocity, r = radius of circular path]
=> F = m[tex]\frac{v^2}{r}[/tex] ------------(i)
We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;
F = q v B sin θ ----------(ii)
Where;
q = charge of the particle
v = velocity of the particle
B = magnetic field
θ = the angle between the velocity and the magnetic field.
Combining equations (i) and (ii) gives
m[tex]\frac{v^2}{r}[/tex] = q v B sin θ [θ = 90° since the proton is orbiting at the maximum orbital radius]
=> m[tex]\frac{v^2}{r}[/tex] = q v B sin 90°
=> m[tex]\frac{v^2}{r}[/tex] = q v B
Divide both side by v;
=> m[tex]\frac{v}{r}[/tex] = qB
Make v subject of the formula
v = [tex]\frac{qBr}{m}[/tex]
From the question;
B = 1.25T
m = mass of proton = 1.67 x 10⁻²⁷kg
r = 0.40m
q = charge of a proton = 1.6 x 10⁻¹⁹C
Substitute these values into equation(iii) as follows;
v = [tex]\frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}[/tex]
v = 4.79 x 10⁷m/s
Now, the kinetic energy, K, is given by;
K = [tex]\frac{1}{2}[/tex]mv²
m = mass of proton
v = velocity of the proton as calculated above
K = [tex]\frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )[/tex]
K = 1.92 x 10⁻¹²J
The kinetic energy is 1.92 x 10⁻¹²J
Part E Now convert the time from seconds to years. This value is the approximate age of the universe. Write the age in scientific notation. Use the conversion, 1 year = 3.154 × 107 seconds.
Explanation:
The age of the universe is 13.8 billion years.
We know that,
1 billion years = 10⁹ years
So,
[tex]13.8\ \text{billion years}=13.8\times 10^9\ \text{years}[/tex]
We need to convert the age of the universe to the scientific notation.
Since, [tex]1\ \text{year}=3.154\times 10^7\ s[/tex]
So,
[tex]13.8\times 10^9\ \text{years}=13.8\times 10^9\times 3.154 \times 10^7\\\\=4.35\times 10^{17}\ s[/tex]
So, the age of the universe is [tex]4.35\times 10^{17}\ s[/tex].
Answer:
So the universe is approximately 1.34 x 10^10 years old
Explanation:
Edmentum
What does Electromagnetic induction mean?
Transformers are of two types: Step up and Step down.
What is step up transformer?
What is step down transformer?
What is the difference between them?
Please I really need help.
Don't answer the question for points if you don't know what it means!
Answer:
Electromagnetic introduction is the production of an electromotive force (voltage) across an electrical conductor in a changing magnetic field.
Step up transformers is a transformer in which the output (secondary) voltage is greater than its input (primary) voltage is called a step-up transformer. The step-up transformer decreases the output current for keeping the input and output power of the system equal.
Step down transformer is a transformer in which the output (secondary) voltage is less than its input (primary) voltage is called a step-down transformer. The number of turns on the primary of the transformer is greater than the turn on the secondary of the transformer.
The difference between them:
A transformer is a static device which transfers a.c electrical power from one circuit to the other at the same frequency, but the voltage level is usually changed. For economical reasons, electric power is required to be transmitted at high voltage whereas it has to be utilized at low voltage from a safety point of view. This increase in voltage for transmission and decrease in voltage for utilization can only be achieved by using a step-up and step-down transformer.
Hopefully this helped.
On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the same material. The mass is displaced by a small displacement Δy such that equal tensions T exist in the two strings, as shown in the figure. The mass is released and begins to oscillate back and forth. Assume that the displacement is small enough so that the tensions do not change appreciably. (a) Show that the restoring force on the mass can be given by F = -(2T∆y)/l (for small angles) (b) Derive an expression for the frequency of oscillation.
Answer:
(a) By small angle approximation, we have;
F = -2×T×Δy/l
(b) [tex]The \ frequency \ of \ oscillation, \ f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]
Explanation:
(a) The diagram shows the mass, M, being restored by two equal tension, T acting on the elastic strings l, such the restoring force, F acts along the path of motion of the mass, with distance Δy
Therefore, the component of the tension T that form part of the restoring force is given as follows;
Let the angle between the line representing the extension of the elastic strings T and the initial position of the string = ∅
Then we have;
String force, [tex]F_{string}[/tex] = T×sin∅ + T×cos∅ + T×sin∅ - T×cos∅ = 2×T×sin∅
Whereby the angle is small, we have;
sin∅ ≈ tan∅ = Δy/l
Which gives;
[tex]F_{string}[/tex] = 2×T×sin∅ = 2×T×Δy/l (for small angles)
Restoring force F = [tex]-F_{string}[/tex] = -2×T×Δy/l
F = -2×T×Δy/l
(b) Given that the the tensions do not change appreciably as the mass, M, oscillates from Δy we have;
By Hooke's law, F = -k×x
Whereby Δy corresponds to the maximum displacement of the mass, M from the rest position, which gives;
Which gives;
F = M×a = -k×Δy
a = -k×Δy/M
d²(Δy)/dt² = -k×Δy/M
When we put angular frequency as follows;
ω² = k/M
We get;
d²(Δy)/dt² = -ω²×Δy
Which gives;
Δy(t) = A×cos(ωt + Ф)
The angular frequency is thus, ω = √(k/M)
Period of oscillation = 2·π/ω = 2·π/√(k/M)
The frequency of oscillation, f = 1/T = √(k/M)/(2·π)
Where:
k = 2·T/l, we have;
f = √(k/M)/(2·π) = √(2·T/l)/m)/(2·π)
The frequency of oscillation is given as follows;
[tex]f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]
What is at the very center of our galaxy? This question is difficult to answer because the center is completely obscured to our eyes behind a large amount of dust. On the basis of what you know about dust from previous chapters, what can be done to improve our view of the galaxy’s center?
Answer:
The answer is "Observe it in the infrared"
Explanation:
The infrared observatory is a device, which helps to track but also fix thermal light from sources out and the front of Earth's atmosphere like nebulae, young stars, and dust and gas in several other galaxies.
The very center of our galaxy is like a star cluster is situated in the central of our universe across the Milky Way supermassive, and the central black hole, around 4 million times our sun's mass.
The main super black hole named the Sagittarius A-star has been located in the nucleus of the galaxy, that's why in this question "Observe it in the infrared" is the correct answer.
The total mass of eight identical
building blocks is 31.52 kg. Find the
mass of 1 block.
Answer:
3.94
Explanation:
divide total mass by the number of blocks since they are identical
Answer:
3.94
Explanation:
You want to find the mass of one block. Since we know there is 8 blocks with the same mass, you can divide the total mass by 8 since the mass is equally distributed within the 8 blocks
Ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. How many ozone molecules are present in 2.00 L of air under the stratospheric ozone conditions of 275 K temperature and 1.89 × 10−3 atm pressure?
Answer:
1.01×10^20 molecules of ozone.
Explanation:
Data obtained from the question include:
Volume (V) = 2 L
Temperature (T) = 275 K
Pressure (P) = 1.89×10¯³ atm
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) of ozone =.?
Using the ideal gas equation, we can obtain the number of mole of ozone as follow:
PV = nRT
1.89×10¯³ x 2 = n x 0.0821 x 275
Divide both side by 0.0821 x 275
n = (1.89×10¯³ x 2) /(0.0821 x 275)
n = 1.67×10¯⁴ mole.
Therefore the number of mole of ozone in 2 L of air is 1.67×10¯⁴ mole.
Finally, we shall determine the number of molecules present in 1.67×10¯⁴ mole of ozone.
This can be obtained as follow:
From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules. This implies that 1 mole of ozone contains 6.02×10²³ molecules.
If 1 mole of ozone contains 6.02×10²³ molecules,
therefore, 1.67×10¯⁴ mole of ozone will contain = 1.67×10¯⁴ x 6.02×10²³ = 1.01×10^20 molecules.
Therefore, 1.01×10^20 molecules of ozone are present in 2 L of air.
A massless, rigid board is placed across two bathroom scales that are separated by a distance of 1.89 m. A person lies on the board. The scale under his head reads 470 N, and the scale under his feet reads 334 N. (a) Find the weight of the person. (b) Locate the center of gravity of the person relative to the scale beneath his head.
Answer:
A. 119kg
B.0.53m from head
Explanation:
A. Weight = 730+340.
=1170N
F= Wg
W = 1170/9.8
= 119kg
If x is distance from head to CG then 1.92–x is the other distance.
Moments must equal
470x = 330(1.89–x)
470x = 623.7 – 330x
1170x = 623.7
x = 0.53m from head
The large-scale distribution of galaxies in the universe reveals Group of answer choices a smooth, continuous, and homogenous arrangement of clusters large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of universe a central void with walls of galaxies at the edge of the universe
Question
The large-scale distribution of galaxies in the universe reveals
A) a smooth, continuous, and homogenous arrangement of clusters
B) large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of the universe
C) a central void with walls of galaxies at the edge of the universe
Group of answer choices
Answer:
The correct answer is B)
Explanation:
The universe is arranged in a filamentary structure. Filamentary structures are very large. They are the largest kind of structures in the universe and comprise mostly of galaxies that are held together by gravity.
The structures found within Galaxy filaments have thread-like qualities spanning 52 to 78.7 megaparsecs h⁻¹ in lenght.
Other phenomena associated with the nature fo the universe is the existence of void spaces.
Cheers!
Find the mass. 10 points. Will give brainliest.
Answer:
3.94 kgExplanation:
Given,
Force ( f ) = 30 N
Acceleration(a) = 7.6 m/s
Now, Let's find the mass of the ball
Using the Newton's second law of motion:
We get:
[tex]force \: = mass \: \times acceleration[/tex]
plug the value
[tex]30 \: = m \: \times 7.6[/tex]
Use the commutative property to reorder the terms
[tex] 30 = 7.6 \: m[/tex]
Swap the sides of the equation
[tex]7.6m = 30[/tex]
Divide both sides of the equation by 7.6
[tex] \frac{7.6 \: m}{7.6} = \frac{30}{7.6} [/tex]
Calculate
[tex]m = 3.94 \: kg[/tex]
Hope this helps..
Best regards!!
Answer:
[tex]\displaystyle \boxed{\mathrm{3.95 \: kg }}[/tex]
Explanation:
[tex]\mathrm{force \: (N) = mass \: (kg) \times acceleration \: (m/s^2)}[/tex]
[tex]\mathrm{force = 30N}[/tex]
[tex]\mathrm{acceleration = 7.6 \: m/s^2 }[/tex]
[tex]\mathrm{Find \: the \: mass.}[/tex]
[tex]\mathrm{30 = m \times 7.6}[/tex]
[tex]\displaystyle \mathrm{m =\frac{30}{7.6} }[/tex]
[tex]\displaystyle \mathrm{m = 3.947... }[/tex]
You stand at the top of a tall building with a stopwatch. You drop a rock off the side of the building, and it takes the rock 3.2 seconds to hit the ground. (Assume no air resistance.) solved for: [please help check!] the building is 50.176 m tall the impact speed is 31.36 m/s If you were standing at the bottom of this building, at what speed would you have to throw the rock for it to reach the top of the building?
Answer:
50.176m ; 31.36m/s ; 31.36m/s
Explanation:
Given the following :
Time (t) = 3.2s
Height(s) of the building
The initial velocity (u) will be 0
Using the equation:
S = ut + 0.5at^2
Acceleration due to gravity (a) = g = +9.8m/s^2 (downward)
S = 0*t + 0.5(9.8)(3.2^2)
S = 0 + 50.176
S = 50.176m
B.) speed impact when it touches the ground:
v^2 = u^2 + 2aS
where v is the final Velocity
v^2 = 0^2 + 2(9.8)(50.176)
v^2 = 0 + 983.4496
v = √983.4496
v = 31.36m/s
C) Speed of throw from the ground to reach the top of the building
Here we need the initial velocity
Height (s) = 50.176m
Acceleration due to gravity g = a = - 9.8m/s ( upward)
Using the third equation of motion:
S = ut + 0.5at^2
50.176 = u*3.2 + 0.5(-9.8)(3.2^2)
50.176 = 3.2u - 50.176
50.176 + 50.176 = 3.2u
100.352 = 3.2u
u = 100.352 / 3.2
u = 31.36m/s
(a) In electron-volts, how much work does an ideal battery with a 17.0 V emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If 3.88 × 1018 electrons pass through each second, what is the power of the battery?
Answer:
(a) 17.0eV
(b) 10.55W
Explanation:
(a) The amount of work done (W) on an electron by an ideal battery of emf value of V as it moves from the positive to the negative terminal is given by;
W = q x V --------(i)
Where;
q = charge on the electron = 1e
From the question;
V = 17.0 V
Substitute the values of q and V into equation (i) as follows;
W = 1e x 17.0
W = 17.0eV
Therefore, the work done in electron volts is 17.0
(b) The power (P) of the battery as some electrons (n) pass through it at time t, is given as;
P = (n q V) / t --------------(ii)
Where;
n = number of electrons = 3.88 x 10¹⁸
t = 1s
q = 1.6 x 10⁻¹⁹C
V = 17.0V
Substitute these values into equation (ii) as follows;
P = (3.88 x 10¹⁸ x 1.6 x 10⁻¹⁹ x 17.0) / 1
P = 10.55W
Therefore the power of the battery is 10.55W
Which of the following object is in dynamic equilibrium?
Answer:
A car driving in a straight line 20 m/s
Explanation:
ayepecks silly
HELP me pleaseeee somebody
an object is placed 30cm from a mirror of focal length 15 cm the object is 7.5cm tall. where is the image located? how tall is the image??
Explanation:
It is given that,
Object distance from the mirror, u = -30 cm
Focal length of the mirror, f = +15 cm
Size of the object, h = 7.5 cm
We need to find the image distance and the size of the image.
Mirror's formula, [tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
v is image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(15)}-\dfrac{1}{(-30)}\\\\v=10\ cm[/tex]
Let h' is the size of the image. So,
[tex]\dfrac{h'}{h}=\dfrac{-v}{u}\\\\h'=\dfrac{-vh}{u}\\\\h'=\dfrac{-10\times 7.5}{-30}\\\\h'=2.5\ cm[/tex]
So, the image is located at a distance of 10 cm and the size of the image is 2.5 cm.
explain why energy is important to us?
Answer:
we need it to work and without it we dont have strength to do anything
Answer:
energy is important to all living organisms. energy for producers comes from the sun, and energy for consumers comes from other living organisms. the abundance of energy available for organisms impacts the population.
A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. The lower end of the string is suddenly displaced horizontally. How long does it take the resulting wave pulse to travel to the upper end
Answer: 0.0180701 s
Explanation:
Given the following :
Length of string (L) = 10 m
Weight of string (W) = 0.32 N
Weight attached to lower end = 1kN = 1×10^3
Using the relation:
Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity
g = acceleration due to gravity = 9.8m/s^2
Weight of string = 0.32N
Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]
Time = √3.2 / 9800
= √0.0003265
= 0.0180701s
Show that energy dissipated due to motion of a conductor in the magnetic field is due to mechanical energy.
Explanation:
let us use the explanation below to get the intuition so desired;
According to Faraday's law of electro magnetic induction, when ever a coil/conductor is made to rotate in a magnetic field, voltage or emf is created and current is produced, in the long run energy has be produced or converted.
The conversion of this energy is made possible by the motion of the coil/conductor is the magnetic field, just by the motion of the conductor cutting through the magnetic field, thus creating electro motive force(E.M.F) hence producing current, and ultimately energy is created
A 140-Hz sound travels through pure carbon dioxide. The wavelength of the sound is measured to be 1.92 m. What is the speed of sound in carbon dioxide?
Answer:
V = 268.8 m/s
Explanation:
The speed of a wave in general is given by the following formula:
V = fλ
where,
V = Speed of that wave
f = Frequency of the wave
λ = wavelength of the wave
In this case we have a sound wave, travelling across carbon dioxide. The properties of sound wave are as follows:
V = Speed of Sound in Carbon dioxide = ?
f = frequency of sound wave = 140 Hz
λ = wavelength of sound wave = 1.92 m
Therefore,
V = (140 Hz)(1.92 m)
V = 268.8 m/s
A carpenter measured the lengeth of a small piece of timber as 24.6cm .Calculate the relative error in the measurement if the true length is 24.5cm
ANSWER:
0.4081%
Explanation:
Difference=24.6-24.5=0.1
Relative error = 0.1/24.5*100=0.4081%
Relative error is equal to the = difference between both the values/The true value *100
The fan on a personal computer draws 0.3 ft3/s ofair at 14.7 psia and 708F through the box containing the CPU and other components. Air leaves at 14.7 psia and 838F.Calculate the electrical power, in kW, dissipatedby the PCcomponents
Answer:
0.12 kW
Explanation:
Given that
The flow rate of air (V)=0.3 ft³/s
V=0.008 m³/s
Pressure, P=14.7 psia
P=1.013529 atm=101.325 kPa
Inlet temperature = 70° F=294.261 K
Exit temperature = 83° F=301.483 K
We know that , specific heat capacity of the air
Cp=1.005 kJ/kg.K
The mass flow rate of air is given as
[tex]\dot{m}=\dfrac{P\times V}{R\times T}\\\dot{m}=\dfrac{101.325\times 0.008}{0.287\times 294.261}\\\dot{m}= 0.0095\ kg/s[/tex]
By using energy conservation
[tex]Electric\ power =\dot{m}\times C_p\times (T_2-T_1)\\Electric\ power =0.0095\times 1.005\times (83-70)=0.12\ kW[/tex]
Therefore electric power dissipate by components will be 0.12 kW.