Answer:
1 month
Explanation:
Which phase of matter makes up stars?
O liquid
O gas
O plasma
Answer:
The answer to this question is plasma
Answer:
Plasma
Explanation:
The pressure exerted at the bottom of a column of liquid is 30 kPa. The height of the
column is 3,875 m. What type of liquid is used?
Answer:
For example, the pressure acting on a dam at the bottom of a reservoir is ... pressure = height of column × density of the liquid × gravitational field ... The density of water is 1,000 kg/m 3.
A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positive direction, what is the truck's resultant displacement
Answer:
140m east
Explanation:
If East is positive then lets rephrase the problem into integers
A truck moves +70 m, then moves -120m, and finally moves +90m.
So totally Displacement = +70-120+90= +140m
Since east is positive, the trucks resultant displacement is 140 m east of origin
~~~~NEED HELP ASAP~~~~
Block A slides into block B along a frictionless surface. They are moving in the direction from left o the right.
Block A= 3kg
Block B= 4kg
Block A velocity before collision =30m/s.
Block B velocity before collision = 15 m/s
The velocity of block B after the collision is 20m/s.
a.) What is the velocity of block A after collision?
b.) Is the collision elastic? Show work to explain answer why or why not.
Answer:
Block A velocity is 23.33 m/s and the collission is not elastic.
Explanation:
a) m1v1 + m2v2 = m1v1' + m2v2'
Plug in givens
90+60=3v1'+80
solve for v1'= 23.33m/s
b) Find the initial and final kinetic energy of Block B
Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J
Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J
Since Ki does not equal Kf the collision is not elastic
A uniform ladder of length 24 m and weight w is supported by horizontal floor at A and by a vertical wall at B. It makes an angle 45 degree with the horizontal. The coefficient of friction between ground and ladder is 1/2 and coefficient of friction between ladder and wall is 1/3. If a man whose weight is one-half than the ladder, ascends the ladder, how much length x of the ladder he shall climb before the ladder slips
Answer:
I could not find the answer or do it myself if I did find it I would defenetly share
A stopped organ pipe of length L has a fundamental frequency of 220 Hz. For which of the following organ pipes will there be a resonance if a tuning fork of frequency 660 Hz is sounded next to the pipe?
a. a stopped organ pipe of length L
b. a stopped organ pipe of length 2L
c. an open organ pipe of length L;
d. an open organ pipe of length 2L.
Answer:
Option (a), (d) are correct.
Explanation:
Frequency, f = 220 Hz
Resonant frequency = 660 Hz
The next frequency of stopped organ pipe is
2f, 3 f, 4 f ....
= 2 x 220 , 3 x 220 , 4 x 220 ....
= 440 Hz, 660 Hz, 880 Hz
So, the option (a) is correct.
The next resonant frequency of open organ pipe is
3 f, 5 f,...
= 3 x 220, 5 x 220 , ..
= 660 Hz, 1100 Hz,...
So, option (d) is correct.
In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.
With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.
Part A
What is the most significant conclusion that Thomson was able to draw from his measurements?
He found a different value of q/m for different cathode materials.
He found the same value of q/m for different cathode materials.
From measurements of q/m he was able to calculate the charge of an electron.
From measurements of q/m he was able to calculate the mass of an electron.
Part B
What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.
Part C
Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.
What is the speed v0 of the electrons in this case?
Express your answer in terms of E0 and B0.
Answer:
a) He found the same value of q/m for different cathode materials.
b) y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex] , c) v = [tex]\frac{E_o}{B_o}[/tex]
Explanation:
In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.
b) In the part of the plates the electrons are accelerated by the electric field,
F = ma
- e E = m a
a = - (e/m) E₀
the distance traveled is
X axis
x = v₀ t
the separation of the plates is x = d
t = vo / d
Y axis
y = v_{oy} t + ½ to t²
y = ½ a t²
y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex]
c) In this case there is a magnetic field B₀ and the electrons have no deflection
F = - e E + e v x B
if there is no deviation F = 0
e E = e v B
v = [tex]\frac{E_o}{B_o}[/tex]
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0 m and Ay 450.0 m. Suppose that the coordinate system is rotated counterclockwise by 35.0, but the magnitude (450.0 m) and direction of vector remain unchanged, as in drawing b. What are the scalar components, Ax and Ay, of the vector in the rotated x, y coordinate system
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
Astronauts in space move a toolbox from its initial position ????????→=<15,14,−8>m to its final position ????????→=<17,14,−1>m. The two astronauts each push on the box with a constant force. Astronaut 1 exerts a force ????1→=<18,7,−12>???? and astronaut 2 exerts a force ????2→=<16,−10,16>????.
Required:
What is the total work performed on the toolbox?
If both forces are measured in Newtons, then the net force is
F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N
The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector
d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m
The total work done by the astronauts on the toolbox is then
F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J
The work done by the two astronauts is equal to 96 J.
What is work done?work done?Work done is defined as the product of force applied and the distance moved by the force.
Work done = Force × DistanceThe forces applied = 18+16 N, 7+ -10 N, and -12 + 16N
Forces = 34 N, -3 N, and 4N
Distances = (17 - 15, 14 - 14, -1 - - 8) m
Distances = 2, 0, 7
Work done = 34 × 2 + -3 × 0 + 4 × 7
Work done = 96 J
Therefore, the work done by the two astronauts is equal to 96 J.
Learn more about work done at: https://brainly.com/question/25573309
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What is true when an object floats in water? A. When an object floats, it exceeds the volume of water available. B. When an object floats, it displaces a volume of water equal to its own volume. C. When an object floats, it does not displace its entire volume.
Answer:
C. When an object floats, it does not displace its entire volume.
Explanation:
Buoyancy can be defined as an upward force which is created by the water displaced by an object.
According to Archimede's principle, it is directly proportional to the amount (weight) of water that is being displaced by an object.
Basically, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up. The buoyancy of an object is given by the formula;
[tex] Fb = pgV [/tex]
[tex] But, \; V = Ah [/tex]
[tex] Hence, \; Fb = pgAh [/tex]
Where;
Fb = buoyant force of a liquid acting on an object.
g = acceleration due to gravity.
p = density of the liquid.
v = volume of the liquid displaced.
h = height of liquid (water) displaced by an object.
A = surface area of the floating object.
The unit of measurement for buoyancy is Newton (N).
Additionally, the density of a fluid is directly proportional to the buoyant force acting on it i.e as the density of a liquid decreases, buoyancy decreases and vice-versa.
Furthermore, an object such as a boat, ship, ferry, canoe, etc, are able to float because the volume of water they displace weigh more than their own weight. Thus, if a boat or any physical object weighs more than the volume of water it displaces, it would sink; otherwise, it floats.
In conclusion, the true statement is that when an object floats, it does not displace its entire volume.
What would the radius (in mm) of the Earth have to be in order for the escape speed of the Earth to equal (1/21) times the speed of light (300000000 m/s)? You may ignore all other gravitational interactions for the rocket and assume that the Earth-rocket system is isolated. Hint: the mass of the Earth is 5.94 x 1024kg and G=6.67×10−11Jmkg2G=6.67\times10^{-11}\frac{Jm}{kg^2}G=6.67×10−11kg2Jm
Answer:
The expected radius of the Earth is 3.883 meters.
Explanation:
The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:
[tex]U = K[/tex] (1)
Where:
[tex]U[/tex] - Gravitational potential energy, in joules.
[tex]K[/tex] - Translational kinetic energy, in joules.
Then, we expand the formula by definitions of potential and kinetic energy:
[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the Earth. in kilograms.
[tex]m[/tex] - Mass of the rocket, in kilograms.
[tex]r[/tex] - Radius of the Earth, in meters.
[tex]v[/tex] - Escape velocity, in meters per second.
Then, we derive an expression for the escape velocity by clearing it within (2):
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)
If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]
[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]
[tex]r = 3.883\,m[/tex]
The expected radius of the Earth is 3.883 meters.
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.
Answer:
[tex]\lambda=6.83\times 10^{-5}\ m[/tex]
Explanation:
Given that,
An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 4.39 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
We know that,
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]
So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].
Which indicates the first law of thermodynamics
Answer:
(d)
Explanation:
because dU = Q -W so ,that the option d(D) is correct
Please assist with solving this problem and showing the steps
Answer:
2.21 N
Explanation:
The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.
The fulcrum feels F1 + F2 + 34 * 980
F2 = 141.7 * 980 = 138866
F1 = 50.3 * 980 = 49294
Ruler = 34 * 980= 33320
Total Force = 221480 The units here are dynes
I just saw in the middle of the question that g = 9.80
So the answer becomes 221480 / 1000 = 221.48 because we needed kg
And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980
The total force exerted on the fulcrum is
Please help,it is urgent!)
Answer:
answer is 18.58 because
Answer:
My answer is d.25.1 because
lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.360 s, during which it produces an average 0.690 W from an average 3.00 V. (a) What energy does it dissipate
Energy = (power) x (time)
Energy = (0.69 W) x (0.36 sec)
Energy = 0.25 Joule
Which level of government relies the most on income tax?
OA.
federal
state
OC.
local
Answer:
Its the Federal government
The primary coil in a transformer has 250 turns; the secondary coil has 500. Which is correct?
a. This is a step-down transformer.
b. The voltage in the secondary coil will be higher than in the primary.
c. The power in the secondary coil is greater.
d. The power in the primary coil is greater.
Explanation:
option b is the correct one
A boy throws a ball upward with a velocity of 4.50 m/s at 60.0o. What is the maximum height reached by the ball?
Answer:
3.1m
Explanation:
Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.
use vf²=v²+2a(y)
At the maximum height the velocity is 0 and since the object is in freefall, a=-g
Plug in all values
0=15.1875-2*9.8(y)
solve for y
-15.1875*2/-9.8=y
y=3.1m
Answer:
0.774m
Explanation:
The formula for maximum height is given by:
hmax = ∨₀² ₓ Sin (α)² / 2 × g
where;
∨₀ = initial velocity
Sin (α) = angle of launch
g = gravitational acceleration which is equal to 9.8m/s²
Plugging in our values, we will have:
hmax = (4.50m/s)² × (Sin 60.0)² / 2 × 9.8m/s²
hmax= 20.25m/s × 0.75 / 19.8m/s²
hmax = 15.1875 / 19.8
hmax = 0.774m
When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]
a. Give an example of the conversion of light energy to electrical energy.
b. Give an example of chemical energy converting to heat energy.
c. Give an example of mechanical energy converting to heat energy.
Explanation:
a) photovoltaic cell is a semiconductor device and it converts light energy to electrical energy
b) burning of coal converts chemical energy to heat energy
c) rubbing of both hands against each other converts mechanical to heat energy
Answer:
a. solar cells
b.coal,wood,petroleum
c.rubbing ours palms
Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer section at a distance of 2.0 m from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 1.1 m from the center. The carousel takes 5.8 s to make each complete revolution.
Required:
a. What is Mary's angular speed %u03C9M and tangential speed vM?
b. What is Alex's angular speed %u03C9A and tangential speed vA?
Answer:
you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q
how did kepler discoveries contribute to astronomy
Answer:
They established the laws of planetary motion. They explained how the Sun rises and sets. They made astronomy accessible to people who spoke Italian.
Explanation:
It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?
Answer:
1. 588 N
2. 738 N
3. 588 N
Explanation:
time, t = 4 s
initial velocity, u = 0
final velocity, v = 10 m/s
mass, m= 60 kg
1.
Weight of passenger before starts
W =m g = 60 x 9.8 = 588 N
2.
When the elevator is speeding up
v = u + a t
10 = 0 + a x 4
a = 2.5 m/s2
Now the weight is
W' = m (a + g) = 60 (9.8 + 2.5) = 738 N
3.
When he reaches the cruising speed, the weight is
W = 588 N
Two positive charges, 91 = 5 x 10-'[C] and q2 =1 x 10-9 [C], are
separated by a distance of d=0.05 m. At location 'P' between the
two charges, the net electric field is found to be zero.
b. [10 points] The distance between charge qı and location 'P' is
considered to be 'x'. Find the value of 'x' in [cm]
Answer:
wareffctgggyyggghhhh
Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation results
Answer:
No, it will not affect the results.
Explanation:
For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.
What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.
Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong
Roman ng pagleletra.
Answer:
Explanation:
English alphabets numbered fro 1 to 26
and the numbers 1 to10 so they are written in roman numbers as
1 - I
2 - II
3 - III
4 - IV
5 -V
6 - VI
7 -VII
8 - VIII
9 - IX
10 -X
11 - XI
12 - XII
13 - XIII
14 - XIV
15 - XV
16 - XVI
17 - XVII
18 - XVIII
19 - XIX
20- XX
21 - XXI
22 - XXII
23 - XXIII
24 - XXIV
25 - XXV
26 - XXVI
water contracts on freezing is it incorrect or conrrect
Answer:
hope it helps
much as you can
Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car B is 100 from car A, car A begins to accelerate toward car B with a constant acceleration of 5 m/s/s. Let right be positive.
1) How much time elapses before the two cars meet? 2) How far does car A travel before the two cars meet? 3) What is the velocity of car B when the two cars meet?
4) What is the velocity of car A when the two cars meet?
Answer:
Let's define t = 0s (the initial time) as the moment when Car A starts moving.
Let's find the movement equations of each car.
A:
We know that Car A accelerations with a constant acceleration of 5m/s^2
Then the acceleration equation is:
[tex]A_a(t) = 5m/s^2[/tex]
To get the velocity, we integrate over time:
[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]
Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:
[tex]V_a(t) = (5m/s^2)*t[/tex]
To get the position equation we integrate again over time:
[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]
Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:
[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]
Now let's find the equations for car B.
We know that Car B does not accelerate, then it has a constant velocity given by:
[tex]V_b(t) =20m/s[/tex]
To get the position equation, we can integrate:
[tex]P_b(t) = (20m/s)*t + P_0[/tex]
This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:
[tex]P_b(t) = (20m/s)*t + 100m[/tex]
Now we can answer this:
1) The two cars will meet when their position equations are equal, so we must have:
[tex]P_a(t) = P_b(t)[/tex]
We can solve this for t.
[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]
This is a quadratic equation, the solutions are given by the Bhaskara's formula:
[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]
We only care for the positive solution, which is:
[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]
Car A reaches Car B after 11.48 seconds.
2) How far does car A travel before the two cars meet?
Here we only need to evaluate the position equation for Car A in t = 11.48s:
[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]
3) What is the velocity of car B when the two cars meet?
Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s
4) What is the velocity of car A when the two cars meet?
Here we need to evaluate the velocity equation for Car A at t = 11.48s
[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]
Two balls of known masses hang from the ceiling on massless strings of equal length. They barely touch when both hang at rest. One ball is pulled back until its string is at 45 ∘, then released. It swings down, collides with the second ball, and they stick together.The problem can be divided into three parts: (1) from when the first ball is released and to just before it hits the stationary ball, (2) the two balls collide, and (3) the two balls swing up together just after the collision to their highest point. ..............conserved in parts (1) and (3) as the balls swing like pendulums. During the collision in part (2) ................. conserved as the collision is ................. Explain.Match the words in the left column to the appropriate blanks in the sentences on the rightboth energy and momentum areonly energy is only momentum is.........both energy and momentum are only energy is only momentum iselasticinelastic
Answer:
In parts 1 and 3 the energy
In part 2 moment. inelastic
conserved
Explanation:
In this exercise, we are asked to describe the conservation processes for each part of the exercise.
In parts 1 and 3 the energy is conserved since the bodies do not change
In part 2 the bodies change since they are united therefore the moment is conserved and part of the kinetic energy is converted into potential energy.
Energy
moment .inelastic
conserved
The two balls swing up together just after the collision to their highest point. energy is conserved.
What is the law of conservation of momentum?According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
According to the law of conservation of momentum
Momentum before collision =Momentum after collision
When the first ball is released and just before it hits the stationary ball, The two balls collide, The two balls swing up together just after the collision to their highest point. energy is conserved.
The balls swing like pendulums. During the collision in part (2) energy is conserved as the collision is inelastic.
We are requested to describe the conservation methods for each element of the activity in this exercise.
Because the bodies do not change in sections 1 and 3, energy is conserved.
Because the bodies change in part 2 is joined, the moment is conserved and some of the kinetic energy is transformed into potential energy.
Hence the two balls swing up together just after the collision to their highest point. energy is conserved.
To learn more about the law of conservation of momentum refer to;
https://brainly.com/question/1113396
