The fraction of the kinetic energy of the ball lost during the collision is [tex]\frac{1}{3}[/tex].
The given parameters;
initial speed of the ball, = vifinal momentum of the ball, Pf = ¹/₃PiThe initial and final momentum of the ball is calculated as;
[tex]P_i = m_ivi[/tex]
[tex]P_f = m_fv_f = \frac{1}{3} m_iv_i[/tex]
The initial and final kinetic energy of the ball is calculated as;
[tex]K.E_i = \frac{1}{2} m_iv_i^2 = \frac{1}{2} P_iv_i\\\\K.E_f = \frac{1}{2} m_fv_f^2= \frac{1}{2} (\frac{1}{3} P_iv_i)= \frac{1}{6} P_iv_i[/tex]
The change in the kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f - K.E_i \\\\\Delta K.E = \frac{1}{6} P_iv_i - \frac{1}{2} P_iv_i = \frac{1}{3} P_iv_i[/tex]
Thus, the fraction of the kinetic energy of the ball lost during the collision is [tex]\frac{1}{3}[/tex].
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is putting water in a water bottle increasing its mass?
Answer:
No
Explanation:
Answer:
yes
Explanation:
A balance and a graduated cylinder are used to determine the density of a mineral sample. The sample has a mass of 14. 7 g and a volume of 2. 2 cm3. What is the density of the mineral sample?.
sports photographers often use large aperture, long focal length lenses. what limitations do these lenses impose on the photographs?
Answer:
The depth of focus achievable with those lenses is small.
Explanation:
A larger aperture makes it much harder to focus on more than one object. When using a telephoto lens (the ones the question is referring to), the depth of focus is very small. For example, using a telephoto lens to take a photo of a runner might get the runner in focus, but certainly not the track, or the audience behind them. If you look at photos, especially older photos, of Olympians in almost any sport you can see this.
Hope this helps!
how does amplitude determine loudness?
Answer:
Explanation:
Amplitude is a measure of the size of sound waves. It depends on the amount of energy that started the waves. Greater amplitude waves have more energy and greater intensity, so they sound louder.
Answer:
Amplitude determines the loudness of a wave. Greater the amplitude, greater is the loudness.
Explanation:
Amplitude is a measure of the size of sound waves. It depends on the amount of energy that started the waves. Greater amplitude waves have more energy and greater intensity, so they sound louder. ... The same amount of energy is spread over a greater area, so the intensity and loudness of the sound is less.
I hope this helps have a great day :)
If you double the pressure of a constant amount of gas at a constant temperature, what happens to the volume
Answer: if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times.
Explanation:
What feature of a planet's orbit does Kelper's first law of motion describe?
A. the time required for each orbit
B. the shape
C. the distance from the aphelion to the perihelion
D. the distance to the sun
please help I'm stuck on this question
Answer:
increase
decrease
Explanation:
using formula
Vt=mg/6πηr
so if m increases V increases
r is the denominator so if r increases V decreases
a person weighing 500 N climbs 3 m how much power. is needed to make the climb in 5 s?
Answer:
300watts
Explanation:
Answer:
300 watts
lol hoped this helped
:)
1. A 12 kilogram block is sitting on a platform 24 m high. How much Potential energy does
it contain?
The amount of potential energy the block contains is 2,822.4 Joules
Given the following data:
Mass of block = 12 kgHeight of platform = 24 meters.We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex].
To determine the amount of potential energy the block contains:
Mathematically, potential energy (P.E) is given by the formula;
[tex]P.E = mgh[/tex]
Where:
m is the mass of object.g is the acceleration due to gravity.h is the height of an object.Substituting the parameters into the formula, we have;
[tex]P.E = 12 \times 9.8 \times 24[/tex]
Potential energy (P.E) = 2,822.4 Joules
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Составьте правильное соответствие:
1. А. Лавуазье
А. Он говорил: «Знание – сила»
2. Д. Дальтон
Б. Изучал теорию газов.
3. Д.И. Менделеев
В. Опубликовал труд «Методы наименования химических элементов»
4. Р. Броун
Г. Работы посвящены морфологии и систематике растений.
5. М. Смолуховский
Д. Работы посвящены броуновскому движению.
6. Р. Бойль
Е. Ввел понятие «атомный вес»
7. Ф. Бэкон
Ж. Открыл периодический закон химических элементов.
1. good morning mam going to be a long day for me to be a long day for me to be a long day for me to be a long day for me to be a long day for me to be a long day for
An example of a conversation killer is?
making eye contact.
encouraging another person
talking about others
sharing your ideas and feelings
Answer: talking about others
How does the radius of a string affect centripetal force.
Answer:
because a raduis is half of 25% of a cicrle.
Explanation:
Calculate the acceleration of a bottle rocket that is traveling at a speed of 74 m/s, and then descends at a speed of 89 m/s. It takes the rocket 5 seconds to change its speed.
a) 1 m/s/s
b) 2 m/s/s
c) 3 m/s/s
d) 4 m/s/s
Answer:Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.
Explanation:
Suppose cars start at a car race. In how many ways can the top 3 cars finish the race?.
Answer:
there is six ways the top three cars can finish the race
To solve a problem using the equation for Kepler’s third law, Enrico must convert the average distance of Mars from the Sun from meters into astronomical units. How should he make the conversion? multiply the average distance by 1. 5 × 1011 multiply 1. 5 × 1011 by the average distance divide the average distance by 1. 5 × 1011 divide 1. 5 × 1011 by the average distance.
The conversion from meters to astronomical units is done by dividing the average distance by 1. 5 × 10^11.
Just as different areas of human endeavor has different units, the unit of length in astronomy is astronomical units. The astronomical unit is defined as the distance between the center of the earth and the sun.
To solve a problem using the equation for Kepler's third law, Enrico must convert the average distance of Mars from the Sun from meters into astronomical units. This is done by dividing the average distance by 1. 5 × 10^11.
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Is mixing mixing sugar and cinnamon a chemical or physical change?
Answer:
It is a physical
Explanation:
What happens when you want to move the boat forward? You pull the oars toward yourself.Explain why you do this.
Answer:
You pull on the oars. By the third law, the oars push back on your hands, but that’s irrelevant to the motion of the boat. The other end of each oar (the blade) pushes against the water. By the third law, the water pushes back on the oars, pushing the boat forward.
What is the net force on a 4,000-kg car
that doubles its velocity from 15 m/s west
to 30 m/s west over 10 seconds?
Answer:
6000 N toward west
Explanation:
F = ma
a = 30-15/10 = 1.5 m/s^2 towards west
m = 4000 kg
F = 4000 x 1.5 = 6000 N towards west
given a force of 88N and an acceleration of 4 m/s 2, what is the mass?
Answer:22kg
Explanation:
from F=ma, 88=4m, m=22kg
A marble is launched at by dropping it through a marble launcher. The marble took 0.22s to land 0.31m from
where it was launched. Find the horizontal velocity of the marble as it rolls off the table.
A. 1.4m/s
B. 0.53m/s
C. 0.71 m/s
D. 0.07m/s
Hi there!
For projectile motion, the horizontal and vertical components are SEPARATE.
We can use the kinematic equation to solve:
dₓ = vₓt
We can rearrange to solve for vₓ:
dₓ/t = vₓ
0.31/0.22 = vₓ
vₓ = 1.4 m/s ⇒ A
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Why there has to be a driving force to keep the car moving at a constant speed?
Explanation:
For example, when a car travels at a constant speed, the driving force from the engine is balanced by resistive forces such as air resistance and friction in the car's moving parts. ... an object falling at terminal velocity experiences the same air resistance as its weight.
A 2 kg object has a specific heat capacity of 1,700 J/(kg \cdot⋅oC)
To raise this object's temperature from 15 Celsius to 25 Celsius, the object must absorb _______ Joules of heat.
The amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.
HOW TO CALCULATE SPECIFIC HEAT CAPACITY:
The amount of heat absorbed by an object can be calculated by using the following expression:Q = m.c.∆TWhere;
Q = amount of heat absorbed or released (J)m = mass of objectc = specific heat capacity (J/g°C)∆T = change in temperature (°C)According to this question, 2 kg object has a specific heat capacity of 1,700J/kg°C and was raised from a temperature of 15 Celsius to 25 Celsius. The heat absorbed is calculated as follows:Q = 2 × 1700 × {25 - 15}Q = 3400 × 10Q = 34000JTherefore, the amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.Learn more about how to calculate heat absorbed at: https://brainly.com/question/11194034?referrer=searchResults
Using your outline and the materials you’ve gathered, write a 500- to 750-word paper using word processing software. Be sure to proofread and revise your writing to catch any errors in grammar, spelling, logic, or organization. Add a works cited page at the end to give credit to your sources. Submit your completed paper along with this activity to your teacher for evaluation.
please do it based off of Werner Heisenberg the scientist.
Answer:
im half new how does this work
Explanation:
Answer:Werner Heisenberg was born on 5th December, 1901, at Wurzburg. He was the son of Dr. August Heisenberg and his wife Annie Wecklein. His father later became Professor of the Middle and Modern Greek languages in the University of Munich. It was probably due to his influence that Heisenberg remarked, when the Japanese physicist Yukawa discovered the particle now known as the meson and the term “mesotron” was proposed for it, that the Greek word “mesos” has no “tr” in it, with the result that the name “mesotron” was changed to “meson”.Heisenberg went to the Maximilian school at Munich until 1920, when he went to the University of Munich to study physics under Sommerfeld, Wien, Pringsheim, and Rosenthal. During the winter of 1922-1923 he went to Göttingen to study physics under Max Born, Franck, and Hilbert. In 1923 he took his Ph.D. at the University of Munich and then became Assistant to Max Born at the University of Göttingen, and in 1924 he gained the venia legendi at that University.From 1924 until 1925 he worked, with a Rockefeller Grant, with Nlels Bohr, at the University of Copenhagen, returning for the summer of 1925 to Göttingen.In 1926 he was appointed Lecturer in Theoretical Physics at the University of Copenhagen under Nlels Bohr and in 1927, when he was only 26, he was appointed Professor of Theoretical Physics at the University of Leipzig.In 1929 he went on a lecture tour to the United States, Japan, and India.In 1941 he was appointed Professor of Physics at the University of Berlin and Director of the Kaiser Wilhelm Institute for Physics there.At the end of the Second World War he, and other German physicists, were taken prisoner by American troops and sent to England, but in 1946 he returned to Germany and reorganized, with his colleagues, the Institute for Physics at Göttingen. This Institute was, in 1948, renamed the Max Planck Institute for Physics.In 1948 Heisenberg stayed for some months in Cambridge, England, to give lectures, and in 1950 and 1954 he was invited to lecture in the United States. In the winter of 1955-1956 he gave the Gifford Lectures at the University of St. Andrews, Scotland, these lectures being subsequently published as a book.During 1955 Heisenberg was occupied with preparations for the removal of the Max Planck Institute for Physics to Munich. Still Director of this Institute, he went with it to Munich and in 1958 he was appointed Professor of Physics in the University of Munich. His Institute was then renamed the Max Planck Institute for Physics and Astrophysics.
An object of mass 10 kg is moving along positive x- axis with velocity 5 m/s. At some point it breaks into two parts of 6 kg and 4 kg. 6 kg moves making 600 with x –axis and 4 kg moves making -300 with axis. What is velocity of their centre of mass after breaking?
Explanation:
Correct option is
B
−3m/s
2
Given,
m=10kg
u=10m/s
v=−2m/s
t=4sec
The acceleration is rate of change of velocity,
a=
t
v−u
a=
4
−2−10
a=
4
−12
=−3m/s
2
The correct option is B.
A student tested a variety of interacting forces to determine how they would result in motion of an object. If the student used the forces as shown in the lettered choices above, predict which results would occur. Select ALL that apply.A) The variable shown by letter B would result in a movement of the object to the left. B) The variable shown by letter B would result in a movement of the object to the right. C) The variable shown by letter A would result in a movement of the object to the right. D) The variable shown by letter C would result in a movement of the object to the right. E) The variable shown by letter D would result in a movement of the object
Answer:
A, D, C
Explanation:
If you drag a 50kg block across the floor which has a coefficient of friction of .30, what is
the force needed to accelerated it at 2.0m/s^2 ?
Answer:
750 Newton
Explanation:
force=mass(acceleration)
When is there no atmospheric pressure pjysics.
Which is larger, the sun's pull on earth or earth's pull on the sun?.
Answer:
the suns pull
Explanation:
The Sun's pull on Earth is twice as large as Earth's pull on the much larger Sun.
What is the mass (in kg) of an object which accelerates at 3 m/s^2 when acted upon by a 15 N force?
[tex]\purple{\longrightarrow \sf Force = Mass \times Acceleration}[/tex]
[tex] \longrightarrow \sf Mass = \dfrac{Force}{Acceleration }[/tex]
[tex]\longrightarrow \sf Mass = \dfrac{15}{3}[/tex]
[tex]\longrightarrow \sf Mass = \cancel{\dfrac{15}{3}} [/tex]
[tex]\longrightarrow {\pink{\underline{\underline{\sf{ Mass = 5 \: Kg }}}}}[/tex]
Henceforth :-
The mass of the object is 5 kg.▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
A car traveling at 35.6m/s crashes into a concert barrier and comes to a stop in 0.35 seconds. Calculate the average force applied to the 75kg driver.A 3.2kg steel ball traveling at 4.1m/s strikes a second ball of a mass 2.3kg Initially at rest. Calculate the velocity of the second ball when the first one continues traveling in the same direction with a speed of 1.5m/s2 balls of putty are shot towards one another. Ball 1 has a mass of 4.3kg and is moving at 18.6m/s . Ball 2 has a mass of 5.8kg and is moving at 9.5m/s. They collide and stick together. Calculate their final combine velocity.I really appreciate those attempting the problems. I do know the answers but I’m unaware of the steps to get there. Please include all formulas in your response and steps so I can learn and understand.Check your answer:7629N3.6m/s2.46m/sThank you all!
The force on the driver is 7629 N. The velocity of the second ball is 3.6 m/s. The combined velocity of the balls is 13.37 m/s.
We have to find the acceleration using;
v = u - at
v = final velocity = 0 m/s
u = initial velocity = 35.6m/s
a = acceleration = ?
t = time = 0.35 s
u = at
a = u/t = 35.6m/s / 0.35 s
a = 101.7 ms-2
The force on the driver = 75kg × 101.7 ms-2 = 7629 N
Using the principle of conservation of momentum;
Momentum before collision = momentum after collision
m1u1 +m2u2 = m1v1 + m2v2
Hence
(3.2 × 4.1) + 0 = (3.2 × 1.5) + 2.3v2
13.12 = 4.8 + 2.3v2
13.12 - 4.8 = 2.3v2
v2 = 13.12 - 4.8/2.3
v2 = 3.6 m/s
Using the principle of conservation of linear momentum;
m1u1 + m2u2 = m1v1 + m2v2
(4.3 × 18.6) + (5.8 × 9.5) = (4.3 + 5.8) v
v = 79.98 + 55.1/10.1
v = 13.37 m/s
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