The document provides information about uniformly accelerated motion along a straight line. It defines key terms like velocity, acceleration, displacement and equations of motion. Several examples are presented to demonstrate the use of equations to solve problems involving uniformly accelerated motion. Examples include calculating acceleration, distance traveled, time taken and velocities given information about an object's motion under constant acceleration along a straight path.
Unit 6, Lesson 5 - Newton's Laws of Motionjudan1970
Unit 6, Lesson 5 - Newton's Laws of Motion
Lesson Outline:
1. Law of Inertia
2. Law of Acceleration
3. Law of Interaction
4. Momentum and Impulse: An Overview
Unit 6, Lesson 5 - Newton's Laws of Motionjudan1970
Unit 6, Lesson 5 - Newton's Laws of Motion
Lesson Outline:
1. Law of Inertia
2. Law of Acceleration
3. Law of Interaction
4. Momentum and Impulse: An Overview
Unit 2- mechanisms, Kinematics of machines of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Unit-3 - Velocity and acceleration of mechanisms, Kinematics of machines of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Unit 1-introduction to Mechanisms, Kinematics of machines of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Richard's entangled aventures in wonderlandRichard 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.
Professional air quality monitoring systems provide immediate, on-site data for analysis, compliance, and decision-making.
Monitor common gases, weather parameters, particulates.
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.
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.
Slide 1: Title Slide
Extrachromosomal Inheritance
Slide 2: Introduction to Extrachromosomal Inheritance
Definition: Extrachromosomal inheritance refers to the transmission of genetic material that is not found within the nucleus.
Key Components: Involves genes located in mitochondria, chloroplasts, and plasmids.
Slide 3: Mitochondrial Inheritance
Mitochondria: Organelles responsible for energy production.
Mitochondrial DNA (mtDNA): Circular DNA molecule found in mitochondria.
Inheritance Pattern: Maternally inherited, meaning it is passed from mothers to all their offspring.
Diseases: Examples include Leber’s hereditary optic neuropathy (LHON) and mitochondrial myopathy.
Slide 4: Chloroplast Inheritance
Chloroplasts: Organelles responsible for photosynthesis in plants.
Chloroplast DNA (cpDNA): Circular DNA molecule found in chloroplasts.
Inheritance Pattern: Often maternally inherited in most plants, but can vary in some species.
Examples: Variegation in plants, where leaf color patterns are determined by chloroplast DNA.
Slide 5: Plasmid Inheritance
Plasmids: Small, circular DNA molecules found in bacteria and some eukaryotes.
Features: Can carry antibiotic resistance genes and can be transferred between cells through processes like conjugation.
Significance: Important in biotechnology for gene cloning and genetic engineering.
Slide 6: Mechanisms of Extrachromosomal Inheritance
Non-Mendelian Patterns: Do not follow Mendel’s laws of inheritance.
Cytoplasmic Segregation: During cell division, organelles like mitochondria and chloroplasts are randomly distributed to daughter cells.
Heteroplasmy: Presence of more than one type of organellar genome within a cell, leading to variation in expression.
Slide 7: Examples of Extrachromosomal Inheritance
Four O’clock Plant (Mirabilis jalapa): Shows variegated leaves due to different cpDNA in leaf cells.
Petite Mutants in Yeast: Result from mutations in mitochondrial DNA affecting respiration.
Slide 8: Importance of Extrachromosomal Inheritance
Evolution: Provides insight into the evolution of eukaryotic cells.
Medicine: Understanding mitochondrial inheritance helps in diagnosing and treating mitochondrial diseases.
Agriculture: Chloroplast inheritance can be used in plant breeding and genetic modification.
Slide 9: Recent Research and Advances
Gene Editing: Techniques like CRISPR-Cas9 are being used to edit mitochondrial and chloroplast DNA.
Therapies: Development of mitochondrial replacement therapy (MRT) for preventing mitochondrial diseases.
Slide 10: Conclusion
Summary: Extrachromosomal inheritance involves the transmission of genetic material outside the nucleus and plays a crucial role in genetics, medicine, and biotechnology.
Future Directions: Continued research and technological advancements hold promise for new treatments and applications.
Slide 11: Questions and Discussion
Invite Audience: Open the floor for any questions or further discussion on the topic.
Introduction:
RNA interference (RNAi) or Post-Transcriptional Gene Silencing (PTGS) is an important biological process for modulating eukaryotic gene expression.
It is highly conserved process of posttranscriptional gene silencing by which double stranded RNA (dsRNA) causes sequence-specific degradation of mRNA sequences.
dsRNA-induced gene silencing (RNAi) is reported in a wide range of eukaryotes ranging from worms, insects, mammals and plants.
This process mediates resistance to both endogenous parasitic and exogenous pathogenic nucleic acids, and regulates the expression of protein-coding genes.
What are small ncRNAs?
micro RNA (miRNA)
short interfering RNA (siRNA)
Properties of small non-coding RNA:
Involved in silencing mRNA transcripts.
Called “small” because they are usually only about 21-24 nucleotides long.
Synthesized by first cutting up longer precursor sequences (like the 61nt one that Lee discovered).
Silence an mRNA by base pairing with some sequence on the mRNA.
Discovery of siRNA?
The first small RNA:
In 1993 Rosalind Lee (Victor Ambros lab) was studying a non- coding gene in C. elegans, lin-4, that was involved in silencing of another gene, lin-14, at the appropriate time in the
development of the worm C. elegans.
Two small transcripts of lin-4 (22nt and 61nt) were found to be complementary to a sequence in the 3' UTR of lin-14.
Because lin-4 encoded no protein, she deduced that it must be these transcripts that are causing the silencing by RNA-RNA interactions.
Types of RNAi ( non coding RNA)
MiRNA
Length (23-25 nt)
Trans acting
Binds with target MRNA in mismatch
Translation inhibition
Si RNA
Length 21 nt.
Cis acting
Bind with target Mrna in perfect complementary sequence
Piwi-RNA
Length ; 25 to 36 nt.
Expressed in Germ Cells
Regulates trnasposomes activity
MECHANISM OF RNAI:
First the double-stranded RNA teams up with a protein complex named Dicer, which cuts the long RNA into short pieces.
Then another protein complex called RISC (RNA-induced silencing complex) discards one of the two RNA strands.
The RISC-docked, single-stranded RNA then pairs with the homologous mRNA and destroys it.
THE RISC COMPLEX:
RISC is large(>500kD) RNA multi- protein Binding complex which triggers MRNA degradation in response to MRNA
Unwinding of double stranded Si RNA by ATP independent Helicase
Active component of RISC is Ago proteins( ENDONUCLEASE) which cleave target MRNA.
DICER: endonuclease (RNase Family III)
Argonaute: Central Component of the RNA-Induced Silencing Complex (RISC)
One strand of the dsRNA produced by Dicer is retained in the RISC complex in association with Argonaute
ARGONAUTE PROTEIN :
1.PAZ(PIWI/Argonaute/ Zwille)- Recognition of target MRNA
2.PIWI (p-element induced wimpy Testis)- breaks Phosphodiester bond of mRNA.)RNAse H activity.
MiRNA:
The Double-stranded RNAs are naturally produced in eukaryotic cells during development, and they have a key role in regulating gene expression .
This presentation explores a brief idea about the structural and functional attributes of nucleotides, the structure and function of genetic materials along with the impact of UV rays and pH upon them.
2. The motion of a particle along a straight
line at a constant acceleration is called
the "Uniformly Accelerated Motion."
a= constant Uniformly Accelerated Motion
3. Velocity is defined as the change in displacement per unit of time.
t=t0
t=t
Δx
4.
5. V=0 km/h
V=260 km/h
A value of 9 (km/h)/s means the velocity changes by 9.0 km/h
during each second of the motion
Example
/
t= 29 s
6.
7. V=0 km/s V= 260 km/s
V=0 m/s V= 72 m/s
An acceleration of 2.5 m/s2 is read as “2.5 meters per second per second” (or “2.5
meters per second squared”) and means that the velocity changes by 2.5 m/s
during each second of the motion.
8. Negative Acceleration and Decreasing Velocity
A drag racer crosses the finish line, and the driver deploys a
parachute and applies the brakes to slow down (above Figure).
The driver begins slowing down when t0 =9.0 s and the car’s
velocity is v0= 28 m/s. When t= 12.0 s, the velocity has been
reduced to v= 13 m/s. What is the average acceleration of the
dragster?
v
11. V=Δx/Δt=350 mi/5.0hr=70.0 mi/hr
Example : A car's speed changes from 15m/s to 25m/s in
along a straight road while moving in one way.
Find the magnitude of the acceleration of the car
during 5 seconds.
13. The Equation of Motion of Uniformly Accelerated Motion:
x=(1/2) at2+vit
where x and t are variables, and a and vi, the constant
14. Example : A car traveling along a straight road at 8.0m/s
accelerates to a speed of 15.0m/s in 5.0s. Determine
(a) its acceleration,
(b) the distance it travels during this period,
(c) its equation of motion,
(d) the interval of validity of this equation, and
(e) the distance already traveled at t = 2.0 s.
v= 8.0 m/s v= 15.0 m/s
x
15. x = (1/2) a t2 + vi t
Calculation of acceleration using final and initial
velocities and time:
a = (vf -vi ) /t
Calculation of traveled distance using
acceleration, initial velocity and time:
Any object that moves along a straight line and at constant
acceleration, has an equation of motion in the form:
16. Part (e) x(2.0) = 0.7 (2.0)2 + 8.0 (2.0) = 18.8m ≈ 19m.
Part (a):
a = (15.0 m/s - 8.0 m/s) / 5.0s = 1.4m/s2.
Part (b):
x = (1/2) (1.4m/s2) (5.0s)2 + (8.0m/s)(5.0s) = 58m (rounded to 2 sig. figs.)
The equation of motion of this car can be found by substituting the constants in the equation.
The constants are: vi = 8.0m/s and a = 1.4m/s2.
The equation of motion becomes:
Part (c) x(t) = (1/2)(1.4)t2 + 8.0t or x(t) = 0.7t2 + 8.0t
Part (d) This equation is valid for 0 ≤ t ≤ 5.0s only.
(The statement of the problem gives us information for a 5.0s period only).
17. Example: A cyclist traveling at 30.0m/s along a straight road
comes uniformly to stop in 5.00s. Determine the stopping
acceleration, the stopping distance, and the equation of
motion.
v= 30.0 m/s v= 0 m/s
t=5.00 s
18. Solution:
a = (vf - vi ) / t
a= ( 0 - 30.0m/s) / 5.00s = -6.00 m/s2.
x = (1/2) a t2 + vi t
x = (1/2)(-6.00m/s2)(5.00s)2 + (30.0m/s)(5.00s)
x= +75m.
x = (1/2) (-6.00m/s2)t2 + (30.0m/s)t
x(t) = -3.00t2 + 30.0t.
(a relation between x and t ).
19. Example : The equation of motion of a lady skiing along a steep
and straight ramp is x = 3.40t2 + 2.10t where x is in meters
and t in seconds. Determine
(a) her distance from the starting point at the end of
1.0s, 2.0s, 3.0s, 4.0s, and 5.0s. periods.
(b) Determine the distances she travels during the 1st, 2nd, 3rd,
4th, and 5th seconds.
(c) Are the answers in Part (b) equal? If not why?
What should be the pace of motion for such distances to be equal?
22. (c) As can be seen from Part (b), the distances traveled in
subsequent seconds become greater and greater.
What is this telling us?
It is because the motion is an accelerated one.
In accelerated motion, unequal distances are traveled in
equal time intervals.
24. An Equation With No Apparent Time Element
a = (vf -vi) / t → t=(vf -vi) /a
x = (1/2)at2 + vit
a
vv
v
a
vv
ax
if
i
if
.
)(
.)2/1( 2
2
a
vvvivv
x
ifif
2
)(2)( 2
2ax= vf
2-2vfvi+vi
2+2vivf-2vi
2
vf
2 - vi
2 = 2ax ( explicitly independent of time ).
25. Example : During take off, an airplane travels 960m along a straight runway to
reach a speed of 65m/s before its tires leave the ground. If the motion is
uniformly accelerated, determine
(a) its acceleration,
(b) the elapsed time
(c) its equation of motion
(d) its midway speed.
x=960m
V=0 m/s V= 65 m/s
a= constant midway
26. Solution:
(a) vf
2 - vi
2 = 2ax ;
(65m/s)2 - (0m/s)2 = 2a(960m)
4225 = 1920a
a = 2.2 m/s2
(b) a = (vf -vi) / t
solving for ( t ): t = (vf -vi) / a
t = (65m/s - 0m/s) / ( 2.2m/s2) = 30. sec
(c) x(t) = (1/2)( 2.2m/s2) t2 + (0) t
x(t) = 1.1 t2
(d) vf
2 - vi
2 = 2ax
vf
2 - (0m/s)2 = 2( 2.2 m/s2)[(960/2)m]
vf
2 = 2113
vf = 46 m/s.
27. Freely Falling of Objects:
(1) g = (vf -vi ) / t
(2) y = (1/2) g t2 + vi t
(3) vf
2 - vi
2 = 2 g y
28. Example: A rock is released from a height of 19.6m. Determine
(a) the time it spends in air
(b) its speed just before striking the ground.
19.6m
t=?
vf=?
vi= 0 m/s
(1) g = (vf -vi ) / t
(2) y = (1/2) g t2 + vi t
(3) vf
2 - vi
2 = 2 g y
(+ y)
29. Solution:
Since the rock is not thrown and is only released from rest,
the initial speed, vi = 0.
If the (+ y) axis is taken to be downward, then
g = + 9.8m/s2, and the equation of motion,
y = (1/2) g t2 + vi t
becomes: y = 4.9t2 .
Substituting the given 19.6m for y, we get:
19.6 = 4.9t2
4 = t2 ; t = 2.0s. (The positive root is acceptable).
To find the velocity at which it hits the ground, we may use
vf
2 - vi
2 = 2 g y
vf
2 - 02 = 2 (9.8)(19.6)
vf = 19.6 m/s. (Again the positive answer is acceptable).
30. Example: A rifle fired straight upward at
the ground level and the bullet leaves
the barrel at an initial speed of 315 m/s.
Determine
(a) the highest it reaches
(b) the time to reach the highest point
(c) the return time to the ground, and
its speed just before striking the
ground. Neglect air resistance.
vi= 315 m/s
vf= 0 m/s
h
32. Example: A motorcycle ride consists of two segments, as shown in the following
Figure. During segment 1, the motorcycle starts from rest, has an acceleration of
2.6 m/s2, and has a displacement of 120 m. Immediately after segment 1, the
motorcycle enters segment 2 and begins slowing down with an acceleration of -
1.5 m/s2 until its velocity is 12 m/s. What is the displacement of the motorcycle
during segment 2?
a=2.6 m/s2
vi=0 m/s
x=120 m
vf
2 - vi
2 = 2ax
a=-1.5 m/s2
v= 12 m/s
vf1= ? m/s
vf
2 - vi
2 = 2ax
Answer: x2=160 m
33. Example: Using the time and position intervals indicated in the
drawing, obtain the velocities for each segment of the trip.