This sis a brief guide to the topic of energy balance. Specifically to energy without reaction balance. There are some solved examples. But all of these are in spanish.
Ultrasound and Ozone processing techniques in cassava starch and Sago industr...Krishnakumar T
A higher percentage recovery of starch from tropical tuber crops could be a better way to get higher economic yield for developing countries like India. Ultrasound-Assisted Extraction (UAE) is considered as an emerging green technique and found suitable alternative to conventional techniques, gaining notable attention in recent years because of reduction in solvent usage, low extraction time, increase in extraction yield and improve quality of extracts.
This sis a brief guide to the topic of energy balance. Specifically to energy without reaction balance. There are some solved examples. But all of these are in spanish.
Ultrasound and Ozone processing techniques in cassava starch and Sago industr...Krishnakumar T
A higher percentage recovery of starch from tropical tuber crops could be a better way to get higher economic yield for developing countries like India. Ultrasound-Assisted Extraction (UAE) is considered as an emerging green technique and found suitable alternative to conventional techniques, gaining notable attention in recent years because of reduction in solvent usage, low extraction time, increase in extraction yield and improve quality of extracts.
With this mantra success is sure to come your way. At APEX INSTITUTE we strive our best to realize the Alchemist's dream of turning 'base metal' into 'gold'.
CHAPTER 18
PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY (Sections 18.1-18.4)
Objectives:
a) Define the various ways a force and couple do work.
b) Apply the principle of work and energy to a rigid body.
APPLICATIONS
The work of the torque developed by the driving gears on the two motors on the mixer is transformed into the rotational kinetic energy of the mixing drum.
The work done by the compactor's engine is transformed into the translational kinetic energy of the frame and the translational and rotational kinetic energy of
its roller and wheels
2. Rotation: When a rigid body is rotating about a fixed axis passing through point O, the body has both translational and rotational kinetic energy:
T = 0.5m(vG)2 + 0.5IGw2
Since
vG = rGw, T = 0.5(IG + m(rG)2)w2 = 0.5IOw2
With this mantra success is sure to come your way. At APEX INSTITUTE we strive our best to realize the Alchemist's dream of turning 'base metal' into 'gold'.
CHAPTER 18
PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY (Sections 18.1-18.4)
Objectives:
a) Define the various ways a force and couple do work.
b) Apply the principle of work and energy to a rigid body.
APPLICATIONS
The work of the torque developed by the driving gears on the two motors on the mixer is transformed into the rotational kinetic energy of the mixing drum.
The work done by the compactor's engine is transformed into the translational kinetic energy of the frame and the translational and rotational kinetic energy of
its roller and wheels
2. Rotation: When a rigid body is rotating about a fixed axis passing through point O, the body has both translational and rotational kinetic energy:
T = 0.5m(vG)2 + 0.5IGw2
Since
vG = rGw, T = 0.5(IG + m(rG)2)w2 = 0.5IOw2
Younes Sina ,Student Poster Competition ,The Oak Ridge Chapter of ASM, the Ma...Younes Sina
Poster for ASM Oak Ridge Chapter competition Younes Sina, Dr. Carl.J. McHargue
Student Poster Competition (The Oak Ridge Chapter of ASM, the Materials Information Society and the Smoky Mountain Chapter of the Society of Plastic Engineers)
(May 29th, 2024) Advancements in Intravital Microscopy- Insights for Preclini...Scintica Instrumentation
Intravital microscopy (IVM) is a powerful tool utilized to study cellular behavior over time and space in vivo. Much of our understanding of cell biology has been accomplished using various in vitro and ex vivo methods; however, these studies do not necessarily reflect the natural dynamics of biological processes. Unlike traditional cell culture or fixed tissue imaging, IVM allows for the ultra-fast high-resolution imaging of cellular processes over time and space and were studied in its natural environment. Real-time visualization of biological processes in the context of an intact organism helps maintain physiological relevance and provide insights into the progression of disease, response to treatments or developmental processes.
In this webinar we give an overview of advanced applications of the IVM system in preclinical research. IVIM technology is a provider of all-in-one intravital microscopy systems and solutions optimized for in vivo imaging of live animal models at sub-micron resolution. The system’s unique features and user-friendly software enables researchers to probe fast dynamic biological processes such as immune cell tracking, cell-cell interaction as well as vascularization and tumor metastasis with exceptional detail. This webinar will also give an overview of IVM being utilized in drug development, offering a view into the intricate interaction between drugs/nanoparticles and tissues in vivo and allows for the evaluation of therapeutic intervention in a variety of tissues and organs. This interdisciplinary collaboration continues to drive the advancements of novel therapeutic strategies.
Richard's aventures in two entangled wonderlandsRichard Gill
Since the loophole-free Bell experiments of 2020 and the Nobel prizes in physics of 2022, critics of Bell's work have retreated to the fortress of super-determinism. Now, super-determinism is a derogatory word - it just means "determinism". Palmer, Hance and Hossenfelder argue that quantum mechanics and determinism are not incompatible, using a sophisticated mathematical construction based on a subtle thinning of allowed states and measurements in quantum mechanics, such that what is left appears to make Bell's argument fail, without altering the empirical predictions of quantum mechanics. I think however that it is a smoke screen, and the slogan "lost in math" comes to my mind. I will discuss some other recent disproofs of Bell's theorem using the language of causality based on causal graphs. Causal thinking is also central to law and justice. I will mention surprising connections to my work on serial killer nurse cases, in particular the Dutch case of Lucia de Berk and the current UK case of Lucy Letby.
Cancer cell metabolism: special Reference to Lactate PathwayAADYARAJPANDEY1
Normal Cell Metabolism:
Cellular respiration describes the series of steps that cells use to break down sugar and other chemicals to get the energy we need to function.
Energy is stored in the bonds of glucose and when glucose is broken down, much of that energy is released.
Cell utilize energy in the form of ATP.
The first step of respiration is called glycolysis. In a series of steps, glycolysis breaks glucose into two smaller molecules - a chemical called pyruvate. A small amount of ATP is formed during this process.
Most healthy cells continue the breakdown in a second process, called the Kreb's cycle. The Kreb's cycle allows cells to “burn” the pyruvates made in glycolysis to get more ATP.
The last step in the breakdown of glucose is called oxidative phosphorylation (Ox-Phos).
It takes place in specialized cell structures called mitochondria. This process produces a large amount of ATP. Importantly, cells need oxygen to complete oxidative phosphorylation.
If a cell completes only glycolysis, only 2 molecules of ATP are made per glucose. However, if the cell completes the entire respiration process (glycolysis - Kreb's - oxidative phosphorylation), about 36 molecules of ATP are created, giving it much more energy to use.
IN CANCER CELL:
Unlike healthy cells that "burn" the entire molecule of sugar to capture a large amount of energy as ATP, cancer cells are wasteful.
Cancer cells only partially break down sugar molecules. They overuse the first step of respiration, glycolysis. They frequently do not complete the second step, oxidative phosphorylation.
This results in only 2 molecules of ATP per each glucose molecule instead of the 36 or so ATPs healthy cells gain. As a result, cancer cells need to use a lot more sugar molecules to get enough energy to survive.
Unlike healthy cells that "burn" the entire molecule of sugar to capture a large amount of energy as ATP, cancer cells are wasteful.
Cancer cells only partially break down sugar molecules. They overuse the first step of respiration, glycolysis. They frequently do not complete the second step, oxidative phosphorylation.
This results in only 2 molecules of ATP per each glucose molecule instead of the 36 or so ATPs healthy cells gain. As a result, cancer cells need to use a lot more sugar molecules to get enough energy to survive.
introduction to WARBERG PHENOMENA:
WARBURG EFFECT Usually, cancer cells are highly glycolytic (glucose addiction) and take up more glucose than do normal cells from outside.
Otto Heinrich Warburg (; 8 October 1883 – 1 August 1970) In 1931 was awarded the Nobel Prize in Physiology for his "discovery of the nature and mode of action of the respiratory enzyme.
WARNBURG EFFECT : cancer cells under aerobic (well-oxygenated) conditions to metabolize glucose to lactate (aerobic glycolysis) is known as the Warburg effect. Warburg made the observation that tumor slices consume glucose and secrete lactate at a higher rate than normal tissues.
This pdf is about the Schizophrenia.
For more details visit on YouTube; @SELF-EXPLANATORY;
https://www.youtube.com/channel/UCAiarMZDNhe1A3Rnpr_WkzA/videos
Thanks...!
What is greenhouse gasses and how many gasses are there to affect the Earth.moosaasad1975
What are greenhouse gasses how they affect the earth and its environment what is the future of the environment and earth how the weather and the climate effects.
Deep Behavioral Phenotyping in Systems Neuroscience for Functional Atlasing a...Ana Luísa Pinho
Functional Magnetic Resonance Imaging (fMRI) provides means to characterize brain activations in response to behavior. However, cognitive neuroscience has been limited to group-level effects referring to the performance of specific tasks. To obtain the functional profile of elementary cognitive mechanisms, the combination of brain responses to many tasks is required. Yet, to date, both structural atlases and parcellation-based activations do not fully account for cognitive function and still present several limitations. Further, they do not adapt overall to individual characteristics. In this talk, I will give an account of deep-behavioral phenotyping strategies, namely data-driven methods in large task-fMRI datasets, to optimize functional brain-data collection and improve inference of effects-of-interest related to mental processes. Key to this approach is the employment of fast multi-functional paradigms rich on features that can be well parametrized and, consequently, facilitate the creation of psycho-physiological constructs to be modelled with imaging data. Particular emphasis will be given to music stimuli when studying high-order cognitive mechanisms, due to their ecological nature and quality to enable complex behavior compounded by discrete entities. I will also discuss how deep-behavioral phenotyping and individualized models applied to neuroimaging data can better account for the subject-specific organization of domain-general cognitive systems in the human brain. Finally, the accumulation of functional brain signatures brings the possibility to clarify relationships among tasks and create a univocal link between brain systems and mental functions through: (1) the development of ontologies proposing an organization of cognitive processes; and (2) brain-network taxonomies describing functional specialization. To this end, tools to improve commensurability in cognitive science are necessary, such as public repositories, ontology-based platforms and automated meta-analysis tools. I will thus discuss some brain-atlasing resources currently under development, and their applicability in cognitive as well as clinical neuroscience.
2. Work (W): Work is defined as the product of parallel force and
distance. W = F ll x .
F ll denotes the component of force F that is parallel to
displacement , x .
The SI unit for work is N.m called " Joule", and the American unit is
" lb-ft "
If the direction of F is parallel to that of x, the work done is simply
W = F x.
3. Example :
Find the work done by force F = 25N in
pushing the block a distance of 14m .
25 N force applied
Block moved by 14 m
4. 25 N
14 m
W=F . x
W=(25N)(14 m)
W=350 Joules
Solution
5. If an object placed on a horizontal surface is pushed or
pulled by force F as shown below, the component of F that
is parallel to x does useful work .
The component perpendicular to x does no work.
6.
7. Example : A block is pulled a distance of x = 24m
from A to B as shown via force F = 45N that makes
a 30 .0 o angle with the horizontal surface . Calculate
the work done by F.
x = 24m
30 .0 o
8. Solution: For θ = 30.0o, and F = 45N
F ll = 45N cos (30 .0 ) = 39N
W = F ll x
W = (39N)(24m) = 940 Nm
W = 940 J
9. Example : In the following figure, find
(a) the magnitude of F such that the block slides at constant
speed to the right .
(b) Find the work done by this force if the displacement is 38 m .
10.
11. Example: In previous example, find the work done by
the frictional force, F k , within the same distance.
12. Solution: F k acts opposite to x; therefore, it does negative
work.
W friction = F k x
W = ( -45N )( 38m ) = -1700 J
13. Energy: Energy is defined as the ability to do work. Energy and
work are expressed in same units. Typical units are: J, cal, kcal, and eV.
Types of Energy:
Energy exists in different forms such as mechanical, electric, nuclear,
light, chemical, etc... .
In this chapter, mechanical energy will be discussed only.
Heat energy is also a will be discussed in Chapter 12 .
Mechanical Energy:
Kinetic energy, Gravitational potential energy, and elastic (spring)
potential energy are forms of mechanical energy and will be
discussed under this topic.
14. I ) Kinetic Energy (K.E.):
Kinetic energy is the energy that a mass (an object) has because of
its motion. As long as an object moves or has some speed, it has
kinetic energy.
K.E. is proportional to mass ( M) and proportional to the square of
velocity ( v 2 ) .
K.E. = (1/2) Mv 2
II ) Gravitational Potential Energy (P.E.):
This energy is the type that an object has due to its elevation with
respect to a reference level. This energy is significant when objects
are in the vicinity of planets, stars and other heavenly masses .
Gravitational P.E. is proportional to the acceleration of gravity ( g) of
the planet or star, the mass of the object, M, and its elevation ( h)
from a reference level.
P.E. = Mgh
15. Example :
Calculate the K .E . of a 2000-kg car that is moving
(a) at 10.0m/s
(b) 20.0 m/s
(c) 30.0 m/s
19. Example :
In the following figure, a 95-kg crate is pushed up on an incline that
is practically frictionless. The incline is 8 .0m long and makes a 31 o
angle with the horizontal floor. Calculate the P .E . Of the crate with
respect to the floor when it reaches the top of the incline.
20.
21. Example : A swimming pool is on the top of a hill at an average elevation of 150m
from a certain ground level . The swimming pool has dimensions:
15 m X 25 m X 2 .4 m and is full. The mass density of water is 1.000 ton/m 3.
Determine
(a) the P .E . Of the water in the pool with respect to that level .
If the pool is allowed to empty to the ground level,
(b) how much energy is at most available for use?
(c) how much electric energy will become available? If a generator uses this energy
for electricity production, and the overall efficiency is 62%
22. Solution:
(a) The pool's volume is V = w l h
V = 900m 3
r = M / V
M = r V
M = (1000 kg/m 3)(900 m3) = 900,000 kg
P.E. = Mgh
P.E. = ( 900,000 kg )( 9 .8 m/s 2 )( 150m ) = 1.3 × 10 9 J
(b) 1.3 × 10 9 J
(c) 0.62 (1.3 × 10 9 J) = 8.2 × 10 8 J
100 W
2278 hr
23. Example : A 750-kg car is traveling at a velocity of 72 km/h eastward
and on a level road . Determine
(a) its initial kinetic energy (K .E .) i
If this car is slowed down to a speed of 36 km/h, calculate
(b) its final (K.E.) f .
(c) How much is the change in its kinetic energy and where does this
energy go ?
Vi=72 km/h Vf=36 km/h
24. Solution:
(a) (K.E.) i = (1/2)Mv i
2
(K.E.) i = (1/2)( 750 kg)( 20 . m/s)2 = 150,000 J
(b) (K.E.) f = (1/2)Mv f
2
(K.E.) f = (1/2)( 750 kg)( 10 . m/s)2 = 38,000 J
(c) Δ(K.E.) = (K.E.) f - (K.E.) i
Δ(K.E.) = 38,000 J - 150,000 J = - 112,000 J
This energy is consumed by force of friction (brakes force, for
example) that acts opposite to the direction of motion . In fact, it is
the work done by force of friction. Most of this energy converts to
heat and warms up the brakes. Friction does negative work .
25. Example: If the work done by force of friction is -112,000J .
Use the work formula to calculate the force applied by friction if brakes were used
within a distance of 56m .
Wf=-112,000J
56m
27. Work-Kinetic Energy Theorem:
The work done by the net force acting on a mass is equal to the
change in the kinetic energy of that mass.
(ΣF ) x = Δ(K.E.)
(ΣF ) x = (K.E.) f - (K.E.) i
28. Example : A 900-kg car traveling at 15m/s changes its speed to
25m/s in a distance of 50 m due to a net force . Calculate
(a) the net force
(b) the engine force if the frictional forces add up to 1400N .
V=15m/s V=25m/s
X=50 m
M= 900 kg
29. Solution:
(a) Work-K.E. theorem :
(Σ F) ∙ x = (K.E.) f - (K.E.) i
Σ F (50 m) = (1/2)(900)(25) 2 - (1/2)(900)(15) 2
Σ F = 3600 N
(b) Σ F = F e - Ff
3600N = Fe - 1400N
F e = 5000 N
30. Example : A boy pulls a 45.0-kg sled (including his friend in it)
from rest for a distance of 12.0 m with a horizontal and constant
force of 165.0 N. The frictional force between the sled and snow
is 55.0 N. Calculate
(a) the work done by the boy on the sled
(b) the work done by the frictional force on the sled
(c) the work done by the net force on the sled
(d) the speed of the sled at the end of the 12.0 m distance .
12.0m
45.0 kg
165.0 N
Wf=55.0 N
v=0 m/s v=? m/s
31. Solution:
(a) Wboy = F ll x
W = (165 N)(12 .0 m) = 1980 J
(b) Wfriction = F k x
W = (55 .0 N)(12.0 m) = 660.0 J
(Work of friction is negative)
(c) Wnet = Fnet x
(Σ F) x = (165N - 55.0 N)(12.0 m) = 1320 J
(d) (Σ F) x = (K.E.) f - (K.E.) i
1320 J = (1/2)(45 .0kg) Vf
2 - (1/2)(45 .0kg) (0)2
Vf = + 7.66 m/s
32. The Law of Conservation of Energy
This law states that:
"Energy is conserved.
It is neither created nor destroyed.
It converts from one form to another".
33. Conservation of Mechanical Energy
Example : In the following figure, neglecting friction, find the speed
Of the 750-kg car at the bottom of the hill . Suppose the car is put in
neutral and starts from rest from the top of the hill .
V=0, M=750 kg
45 m
34. Conservation of mechanical energy:
Total Energy at A = Total Energy at B
(K .E .)A + (P .E .)A = (K .E .)B + (P .E .)B
(1/2)MVA
2 + M g hA = ( 1/2)MVB
2 + M g hB
VA = 0 and hB = 0
M g hA = ( 1/2)MVB
2
g hA = (1/2)VB
2
2 g hA = VB
2
VB = 30 m/s
A
BhA
hB = 0
VB
35. Example :
In the following figure, if 120,000 J of energy is consumed by
frictional forces, find the speed of the 750-kg car at the bottom
of the hill . Suppose the car is put in neutral and starts from rest
from the top of the hill .
frictional Energy=120,000 J
In this case, part of the available energy will be wasted by friction
36. Conservation of mechanical energy:
Total Energy at A - W friction= Total Energy at B
(K .E .)A + (P .E .)A - Wfriction = (K .E .)B + (P .E .)B
(1/2)M A
2 + M g hA -120,000J = ( 1/2)MV B
2 + M g hB
V A = 0 and h B = 0
M ghA -120,000J = ( 1/2)MV B
2
(750)(9 .8)(45) J - 120,000 J = 0 .5(750 kg)VB
2
562(m/s) 2 = VB
2
VB = 24 m/s
V=0, M=750 kg
37. Example: In the figure shown, find
(a) the work done by the frictional force on the car as it coasts
down the hill in neutral
(b) the energy loss due to friction
(c) the speed of the 750-kg car (VB) at the bottom of the hill .
400 m
38. Solution:
(a) The work done by friction is
W friction = F k x
W friction = ( -300N)(400m) = -120,000 J
(b) -120,000 J
(c) Conservation of mechanical energy:
Total Energy at A - Wfriction = Total Energy at B
(K .E .)A + (P .E .)A - Wfriction= (K .E .)B + (P .E .)B
(1/2)MVA
2 + M g h A - 120,000 J = (1/2)MVB
2 + M g hB
VA = 0 and hB = 0
M ghA -120,000 J = (1/2)MVB
2
(750)(9.8)(45) J - 120,000 J = 0.5(750)VB
2
[562(m/s)]2 = VB
2
VB = 24 m/s
39. Power is defined as the work done per units of time.
Mathematically:
The (SI) unit for power is :
The industrial unit: hp
Power
t
W
P
=
Wattcalled
s
J
s
ftlb
hphP
== 5501andwatts7461
40. Example : An electric motor is capable of delivering 7 .2 Million
Joules of work in one hour and 400.0 seconds. Find the power
of the motor in watts, kilowatts, and hp .
Solution:
P = W/t
P = (7 .2 x10 6 J )/(4000 s) = 1800 J/s = 1800 watts
P = 1.8 kilowatts (kw)
P = (1800/746 ) hp
P = 2.4 hp
41. Example:
Calculate the amount of work or energy that a 4.50 hp
electric motor can deliver in 10.0 hours.
Solution:
P = 4 .5hp = (4 .50)(746 watts) = 3360 watts or ( J /s )
P = W / t
W = Pt
W = (3360 J/s)(36,000s) = 1.21 x 10 8 J
42. Efficiency: When a device receives power from a source, it does not deliver all of
it in the intended form and converts a portion of it to other undesired forms. This
makes a device to be less than 100 percent efficient.
Most power plants have efficiencies of under 50 percent. A good nuclear power
plant is about 45% efficient . Early coal plants were hardly 20% efficient .
Eff. = Pout /Pin
43. Example : A 1.25 hp electric pump with an efficiency of 92.5% is used to perform
1 .40 Mega Joules of work.
(a) How long does it take for the electric pump to do the job ?
(b) If it was 100% efficient , how long would it take?
Solution:
A 1.25 hp electric pump means it draws 1 .25 hp of electric power from the
source.
It generally cannot give back all of it .
Make sure that you write the formulas with horizontal fraction bars .
Eff. = P out / P in
0 .925 = P out / (1 .25 hp)
P out = 0 .925(1 .25hp) = 1.16 hp
(a) P = W / t
t = W / P
t = (1 .40 x 10 6 J) / ( 1 .16 x 746 watts ) = 1620 sec
(b) P = W/t
t = W/P
t =(1 .40 x 10 6 J)/( 1 .25 x 746 watts ) = 1500 sec
44. Example: Calculate
(a) the necessary power to empty a swimming pool with dimensions
(12m)(25m)(2 .4m) by pumping its water to an average height of 3.5m in 2.5hrs .
The mass density of water is 1000 kg/m 3.
(Hint: The power you calculate is the power that the electric pump has to deliver
to water, Pout) .
If the efficiency of the electric pump used is 88% , calculate
(b) the electric power that the electric motor pulls from the power source, Pin .
45. Solution:
V = w. l. h
V = 720m3
r = M / V
M = r V
M = (1000 kg/m 3)(720 m3) = 720,000 kg
(a) P = W / t
W = F x
F = the weight of water= Mg
x = height
P = (F x) / t
P = ( Mg x) / t
P = (720,000kg )( 9 .8 m/s 2 )( 3 .5m ) / (2 .5 X 3600s ) = 2700 watts
P = 2.7kw
P = (2700 / 746) hp = 3.6 hp
(b) Eff. = P out / P in
0 .88 = 2700watts / P in
P in = 2700watts / 0 .88 = 3100 watts
46. Another Version of Power Formula:
P = W / t
W = F x
P = (F x) / t
(x / t )= v
P = F v
This means that power is equal to force times velocity (if both of
force and velocity are in the same direction).
As you see the product of velocity and force is a constant if a
constant power is available . When P is constant, the product F ∙v
is constant . When a greater F is needed, a smaller v is attainable
and vice versa .
47. Example :
A car is using 56 hp of its total power. When it is going up a steep hill, a force of
4200 N is needed for constant velocity motion of it up that hill. When it is
traveling along a level and horizontal road, a force of 1200 N is needed for its
constant velocity motion. Find the velocity of the car in each case .
P = F v
56 hp
56 hp
F= 4200 N
F=1200 N
48. Solution:
P = (56)(746 watts) = 42000 watts
(a) P = F v
v = P / F
v = 42000 watts / 4200N = 10 m/s
(b) P = F v
v = P / F
v = 42000watts / 1200N = 35 m/s