Answer:
0.73kg
Explanation:
Potential gravitational energy is equal to [tex]Pe=mgh[/tex], where Pe is potential gravitational energy, m is mass in kilograms, g is gravity in m/s^2 and h is height in meters.
Substitute:
[tex]Pe=mgh\\23=m*9.8*3.2\\23=31.36m\\0.73=m[/tex]
17. How long will it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s? (A) 0.6s (B) 1.2 s (C) 2.4 s (D) 3.6 s (E) 4.8 s
Answer:
(B)
Explanation:
Time = change of velocity ÷ acceleration
= (6-0) ÷ 5
= 1.2
1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.
Hence, the correct option is D.
To calculate the time it takes for an object to change its velocity from 0 to 6 m/s, we can use the formula:
time = change in velocity / acceleration
Given that the change in velocity (Δv) is 6 m/s and the acceleration (a) is 5 m/s², we can plug these values into the formula:
time = 6 m/s / 5 m/s²
time = 1.2 seconds
Therefore, 1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.
Hence, the correct option is D.
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If you go into a store to buy a digital thermometer and there are 8 instruments of the same brand and model, at the same price, but they indicate different temperature values, explain which one you choose and why.
Answer: I would buy the thermometer with the most features because they all cost the same price so why not you getting more than what you are paying for than buying something the same but with less feature and things it can do for the same price.
Hope this helps :) If so brainliest would mean alot thanks!
Explanation:
In a steam engine, steam in a cylinder expands against a piston, exerting 10 atm of external pressure. The volume of the cylinder increases by 10 L and simultaneously the steam cools, losing 3,000 kJ of energy to the surroundings. What is the change in energy of the steam
If steam in a cylinder expands against a piston, exerting 10 atm of external pressure, the change in energy of the steam is mathematically given as
dE= -3010 kJ
What is the change in energy of the steam?Generally, the equation for the workdone is mathematically given as
Work = - (pressure * Volume)
Therefore
w= - (10 x 10)
w= -100 L atm
In conclusion
dEnergy = q + work
dE = -3000 + -10.13
dE= -3010 kJ
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If a lever was a meter long and the fulcrum was placed at 50mm, what would the mechanical advantage be?
Please show work if possible
Answer:
The mechanical advantage would depend on where the load force and the balancing force were placed on the stick
MA = (50 - F) / (L - 50)
Supoose the balancing force was placed at zero and the load at 75 cm
then MA = (50 - 0) / (75 - 50) = 50 / 25 = 2
or the balancing force was applied at 10 cm and the load force was applied at 60 cm then
MA = (50 - 10) / (60 - 50) = 40 / 10 = 4
help me solve the following question below in the attachment
Answer:
Explanation:
The point P is influenced by two electric fields, that are from Q1 and Q2. Because Q1 is a positive point charge, the E-fields is going to the right and the E-fields from Q2 is going to the left because of negative point charge. So:
[tex]E_{1}=\frac{kQ_{1}}{R_{1}^{2}}[/tex]
[tex]E_{2}=\frac{kQ_{2}}{R_{2}^{2}}[/tex]
Where R1 = 2 + 3 = 5 m, and R2 = 3 m. With k = [tex]k=9\times 10^{9}[/tex] in [tex]Nm^{2}/C^{2}[/tex], we can obtain:
[tex]E_{1}=(9\times 10^{9})\frac{500\times10^{-6}}{25}=180\times 10^{3}[/tex] N/C (+)
[tex]E_{2}=(9\times 10^{9})\frac{100\times 10^{-6}}{9}=100\times 10^{3}[/tex] N/C (-)
So: the Electric Field at P : [tex]E_{P}=E_{1}-E_{2}=(180-100)\times 10^{3}=8\times 10^{4}[/tex] N/C in magnitude
If a snowboarder’s initial speed is 4 m/s and comes to rest when making it to the upper level. With a slightly greater initial speed of 5 m/s, the snowboarder is moving to the right on the upper level. His final speed in this case is 3 m/s. Suppose this situation is repeated on planet Epsilon, where gravity is less than gravity on earth.
A. Would the height of the hill on Epsilon cause a reduction in speed from 4 m/s to 0 greater than, less than, or equal to the height of the corresponding hill on earth? Explain.
B. Consider the hill on Epsilon discussed in part A. If the initial speed at the bottom of the hill is 5 m/s, will the final speed at the top of the hill be greater than, less than, or equal to 3 m/s? Explain.
(a) At a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
(b) If the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
Conservation of mechanical energy
The effect of height and gravity on speed on the given planet Epislon is determined by applying the principle of conservation of mechanical energy as shown below;
ΔK.E = ΔP.E
¹/₂m(v²- u²) = mg(hi - hf)
¹/₂(v²- u²) = g(0 - hf)
v² - u² = -2ghf
v² = u² - 2ghf
where;
v is the final velocity at upper levelu is the initial velocityhf is final heightg is acceleration due to gravitywhen u² = 2gh, then v² = 0,
when gravity reduces, u² > 2gh, and v² > 0
Thus, at a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
Final speedv² = u² - 2ghf
where;
u is the initial speed = 5 m/sg is acceleration due to gravity and its less than 9.8 m/s²v is final speedhf is equal heightSince g on Epislon is less than 9.8 m/s² of Earth;
5² - 2ghf > 3 m/s
Thus, if the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
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1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
a. What is surface area being vibrated by the time the sound reaches the listener (you may
have to look up the equation for the area of a sphere).
b. What is the intensity of the sound wave as it reaches the person listening?
c. What is the relative intensity of the sound as heard by the listener?
2. A police car siren has a frequency of 512 Hz. They are chasing a suspect at 65.0 m/s.
a. What is the speed of sound if the air temperature is 15.0o?
b. If the suspect is running away from the police car at 10.0 m/s, what is the frequency of
the sound he hears?
(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².
(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².
(1.c) The relative intensity of the sound as heard by the listener is 103 dB.
(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.
(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.
Surface area being vibratedThe surface area being vibrated by the time the sound reaches the listener is calculated as follows;
A = 4πr²
A = 4π x (20)²
A = 5,026.55 m²
Intensity of the soundThe intensity of the sound is calculated as follows;
I = P/A
I = (100) / (5,026.55)
I = 0.02 W/m²
Relative intensity of the sound[tex]B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB[/tex]
Speed of sound at the given temperature[tex]v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s[/tex]
Frequency of the soundThe frequency of the sound heard is determined by applying Doppler effect.
[tex]f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )[/tex]
where;
-v₀ is velocity of the observer moving away from the source-vs is the velocity of the source moving towards the observerfs is the source frequencyfo is the observed frequencyv is speed of sound[tex]f_0 = f_s(\frac{v-v_0}{v- v_s} )[/tex]
[tex]f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz[/tex]
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The two basic properties of all matter are ______ and ______.
options for first blank
a. length B. volume
options for second blank
a. density B. mass
Answer:
volume , mass
Explanation:
the 3 basic properties of a matter is volume, mass and shape.
a person have a kinetic energy = 32, if the mass increase two times and the velocity decrease four times, What would be the kinetic energy?
Answer:
4 J
Explanation:
Original KE
KE = 1/2mv²⇒ 32 = 1/2mv²Mass increases x 2
KE ∝ mm x 2 ∝ KE x 232 x 2KE' = 64Velocity decreases x 4
KE' ∝ v²KE' x (1/4)² ∝ (1/4)²KE' x 1/16 = 64 x 1/16 = 4 JWhat are the Applications of electronic polarization?
Answer:
it can be used for communication, industrial application and also for instrumentation application
What does the atomic numbers equal? A number of protons in the atomic nucleus B number of protons and neutrons in the atomic nucleus C number of neutrons in the atomic nucleus
number of protons
explanation:the atomic number of an atom is equal to the number of protons
The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e-10 C what is the strength of the other charge ?
Answer:
[tex] \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}[/tex]
Explanation:
Given:
Force between pair of charges= 900 newtons
The distance between the charges = 0.01 meters
Strength of Charge first q1 = 2e-10 Coulomb
To find:
Strength of Charge second q2 = ____ Coulomb?
Solution:
We know that,
Force between two charges separate by distance r is given by the equation,
[tex]|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb[/tex]
[tex] \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}[/tex]
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If the period of oscillation of a simple pendulum is 4s, find its length. If the velocity of the bob
at the mean position is 40cms−1
, find its amplitude. (take gravity = 9.81ms−2
Answer:
Explanation:
Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:
[tex]T=2\pi\sqrt{\frac{L}{g}} \rightarrow 4=2\pi\sqrt{\frac{L}{9.81}} \rightarrow \frac{4}{\pi^{2}}=\frac{L}{9.81} \rightarrow L \approx 3.97 m[/tex]
By using the [tex]y=A\sin(\omega t) \rightarrow v = \frac{dy}{dt}=\omega A \cos\omega t = \omega\sqrt{A^{2}-y^{2}}[/tex]
The mean position is the position when y = 0, so:
[tex]\omega = \frac{2\pi}{T}=\frac{2\pi}{4}=0.5\pi[/tex] rad/s
and [tex]v = \omega A \rightarrow A=\frac{40}{0.5\pi}=\frac{80}{\pi}[/tex] in centimeters (cm).
When you jump, you exert a pushing force against the ground. Gravity pulls you back down. Why can a person jump higher on the moon than on Earth?
Answer:
This is because the force of gravity is much less on the moon than on the earth, therefore the person wont be pulled down much and will jump higher
Answer: This is because the force of gravity is much less on the moon than on the earth, therefore the person won't be pulled down much and will jump higher.
A sound wave has a frequency of 350 Hz and a wavelength of 2.25m. At what speed will the
sound wave move? Show your work.
Answer:
787.5m/s
Explanation:
Speed = wavelength x frequency
speed = 2.25m x 350 Hz
= 787.5m/s
1 If electromagnetic radiation acted like particles in the double-slit experiment, what would be observed?
a The screen would remain dark because no radiation would reach the screen.
b One bright band would appear in the center of the screen.
c A series of light and dark bands would appear on the screen.
d Two bright bands would appear on the screen in line with the slits.
2 Which statement about the interference behavior of electromagnetic radiation seen in the double-slit test experiment is true?
a Waves that make up the radiation collide with each other so that they add together or cancel each other out.
b Particles that make up the radiation collide with each other and scatter randomly.
c Particles that make up the radiation collide with each other so that they add together or cancel each other out.
d Waves that make up the radiation do not interact with each other.
3 Which statement about the observed results of the double-slit experiment is true?
a Waves that are out of phase constructively interfere to create bright bands.
b Waves that are in phase destructively interfere to create bright bands.
c Waves that are out of phase constructively interfere to create bright bands.
d Waves that are in phase constructively interfere to create bright bands.
4 Which statement about the observed results of the double-slit experiment is true?
a Waves that are in phase constructively interfere to form dark bands.
b Waves that are out of phase constructively interfere to form dark bands.
c Waves that are in phase destructively interfere to form dark bands.
d Waves that are out of phase destructively interfere to form dark bands.
5 A scientist decreases the wavelength of the light used in a double-slit experiment and keeps every other aspect the same. What will be true about the new interference pattern seen on the screen compared to the original interference pattern?
a The spacing between the dark fringes will increase.
b The spacing between the bright fringes will increase.
c The spacing between the bright fringes will decrease.
d The spacing between the dark fringes will remain the same.
6 Consider the two-slit interference experiment. Electromagnetic radiation passes through the two slits that are a distance of 0.0170 nm apart. A fourth-order bright fringe forms at an angle of 8.0 degrees relative to the incident beam. What is the wavelength of the light?
a 789 nm
b 420 nm
c 581 nm
d 591 nm
Answer:
1. Two bright bands would appear on the screen in line with the slits.
2. Waves that make up the radiation collide with each other so that they add together or cancel each other out.
3. Waves that are in phase constructively interfere to create bright bands.
4. Waves that are out of phase destructively interfere to form dark bands.
5. The spacing between the bright fringes will decrease.
6. 581 nm
Explanation:
According to Sigmund Freud, memories that easily surface into everyday thought are which part of the mind?
The conscious mind contains all of the thoughts, memories, feelings, and wishes of which we are aware at any given moment. This is the aspect of our mental processing that we can think and talk about rationally. This also includes our memory, which is not always part of consciousness but can be retrieved easily and brought into awareness.
hope it helps...!!!
According to Sigmund Freud, memories that easily surface into everyday thought are a conscious part of the mind.
What is conscious mind?
At any given time, the conscious mind contains all of our thoughts, memories, feelings, and wants.
This is the element of our brain functioning that we can reasonably think about and discuss. This includes our memories, which are not always conscious but can be readily recovered and brought into awareness.
Memories that easily emerge into everyday consciousness, according to Sigmund Freud, are a conscious element of the mind.
Hence, memories that easily surface into everyday thought are a conscious part of the mind.
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When is an object positively charged?
Answer:
when it has a fewer amount of electrons than protons
what does a snow leopard eat?
Answer:
They eat Himalayan blue sheep, boar, deers, wild mountain goats, pikas etc.
Explanation:
QUESTION 9
What determines the evolutionary path a star takes through its life?
O a. Diameter
O b. Who made it
O c. Color
O d. Mass
Answer:
Mass
Explanation:
In the core of the red giant, helium fuses into carbon. All stars evolve the same way up to the red giant phase. The amount of mass a star has determines which of the following life cycle paths it will take from there. The life cycle of a low mass star (left oval) and a high mass star (right oval).May 7, 2015
Calculate the electric field at x = 2 given the electric potential at this point is ϕ=3x3.
Hi there!
Recall the following relationship:
[tex]E = -\frac{dV}{dx}[/tex]
E = Electric field strength (V/m)
V = Potential Difference (V)
Take the derivative of the given electric potential equation using the power rule:
[tex]\frac{d}{dx}x^n = nx^{n-1}[/tex]
[tex]-\frac{dV}{dx} = -(3 \cdot 3 \cdot x^2) = -9x^2[/tex]
Evaluate at x = 2:
[tex]E(2) = -9(2^2) = \boxed{-36 \frac{V}{m}}[/tex]
**The magnitude of the field strength would simply be 36 V/m.
1) How many joules of heat are required to raise the temperature of 550 g of water from
12.7 °C to 28.8°C ? Specific heat of liquid water is 4.184 J/g °C.
Answer:
Explanation:
You can use [tex]Q = mc\Delta T=(550)(4.184)(28.8-12.7)=37049.32[/tex] joule.
Physics Help anyone?
Hi there!
A)
At this point, the pilot is experiencing a DOWNWARD NET centripetal force (ALWAYS points towards the center of the circle) as well as a downward force due to gravity. Additionally, we can say that the normal force exerted by the seat on the pilot is its apparent weight.
[tex]\Sigma F = F_g + F_N[/tex]
**The net force is the centripetal force.
We can assign both to be positive since they work in the same direction. (Assigning down as positive in THIS situation will make things easier).
We know that:
[tex]F_c = \frac{mv^2}{r}[/tex]
Fc = Centripetal force (N)
m = mass (kg)
v = velocity (m/s)
r = radius (m)
[tex]F_g = mg[/tex]
Fg = Force due to gravity (N)
m = mass (kg)
g = acceleration due to gravity (9.81 m/s²)
Solve for mass. (Given weight)
[tex]m = \frac{F_g}{g}\\\\m = \frac{719}{9.8} = 73.367 kg[/tex]
Now, we have everything we need to solve.
[tex]\frac{mv^2}{r} = F_g + F_N \\\\\frac{mv^2}{r} - F_g = F_N\\\\F_N = \frac{(73.367)(225^2)}{3110} - 719 = \boxed{475.278 N}[/tex]
B)
Weightlessness is achieved when the pilot is in free fall. In this instance, FN (Normal force by the seat) = 0 N.
We can use the same setup as above.
[tex]\frac{mv^2}{r} = F_g + 0 (F_N)[/tex]
[tex]\frac{mv^2}{r} = mg[/tex]
Solve for 'v'.
[tex]\frac{v^2}{r} = g\\\\v = \sqrt{gr} = \sqrt{(9.81)(3110)} = \boxed{174.669 \frac{m}{s}}[/tex]
C)
At the BOTTOM, we have the force of gravity working DOWNWARD (Negative in this instance), while the centripetal force (NET FORCE) and normal force are upward (Positive). Do another summation of forces:
[tex]\Sigma F = F_N - F_g\\\\\\frac{mv^2}{r} + F_g = F_N \\\\F_N = \frac{(73.367)(225^2)}{3110} + 719 = \boxed{1913.278 N}[/tex]
The diagram shows a heating curve for water.
Which statement best describes what is happening at
segment W-X?
Heating Curve for Water
O Water is absorbing heat.
O Water is releasing heat.
O Aliquid is changing to a gas.
O A gas is changing to a liquid.
z
110
100
90
x
х
Y
80
Temperature (Cº)
70
60
50
40
w
Heat (J)
Answer:
a water is heating up
Explanation:
the water's temperature is inceasing
A wheel rotates at 24 revolutions every 3 minutes. This is equivalent to
Explain the use of each part of Vernier callipers?
A Vernier caliper has total of four jaws, with two upper jaws used for measuring the internal distances and two lower jaws for measuring the internal distances of objects. The two upper jaws are the smaller jaws of the vernier that are used to measure the internal distances between two parallel sides of an object or an internal diameter.
What is the difference between magnitude of displacement and magnitude of velocity
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figure below. The string goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is k = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the string does not stretch and does not slip on the pulley, and that the mass of the string is negligible. Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown.
The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
Angular Speed of the pulley
The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;
K.E = P.E
[tex]\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\[/tex]
[tex]\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\[/tex]
Substitute the given parameters and solve for the angular speed;
[tex]\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s[/tex]
Linear speed of the blockThe linear speed of the block after travelling 0.7 m;
v = ωR₂
v = 35.39 x 0.03
v = 1.1 m/s
Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
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Two cars collide at an intersection. One car has a mass of 1300 kg and is
moving 12 m/s to the north, while the other has a mass of 1400 kg and is
moving 11 m/s to the south. What is their combined momentum?
O A. 31,000 kg m/s south
O B. 200 kg m/s north
O C. 31,000 kg m/s north
O D. 200 kg m/s south
Answer:B
Explanation:sorry they removed my answer for some reason.
Answer: B
Explanation: Sorry these brainly trolls deleted my first one
In the figure below (Figure 1), the upper ball is released from rest, collides with the stationary lower ball, and sticks to it. The strings are both 50.0 cm long. The upper ball has a mass of 2.20 kg and it is initially 10.0 cm higher than the lower ball, which has a mass of 2.70 kg. Find the frequency of the motion after the collision. Find the maximum angular displacement of the motion after the collision.
(a) The frequency of the motion after the collision is 0.71 Hz.
(b) The maximum angular displacement of the motion after the collision is 16.3⁰.
Speed of the 2.2 kg ball when it collides with 2.7 kg ballThe speed of the 2.2 kg ball which was initially 10 cm higher that 2.7 kg ball is calculated as follows;
K.E = P.E
¹/₂mv² = mgh
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 0.1)
v = 1.4 m/s
Final speed of both balls after collisionThe final speed of both balls after the collision is determined from the principle of conservation of linear momentum.
Pi = Pf
m₁v₁ + m₂v₂ = vf(m₁ + m₂)
2.2(1.4) + 2.7(0) = vf(2.2 + 2.7)
3.08 = 4.9vf
vf = 3.08/4.9
vf = 0.63 m/s
Maximum displacement of the balls after the collisionP.E = K.E
[tex]mgh_f = \frac{1}{2} mv_f^2\\\\h_f = \frac{v_f^2}{2g} \\\\h_f = \frac{(0.63)^2}{2(9.8)} \\\\h_f = 0.02 \ m[/tex]
Maximum angular displacementThe maximum angular displacement of the balls after the collision is calculated as follows;
[tex]cos \theta = \frac{L - h_f}{L} \\\\cos\theta = \frac{0.5 - 0.02}{0.5} \\\\cos\theta = 0.96\\\\\theta = cos^{-1}(0.96)\\\\\theta = 16.3 \ ^0[/tex]
Frequency of the motion[tex]f = \frac{1}{2\pi} \sqrt{\frac{g}{L} } \\\\f = \frac{1}{2\pi } \sqrt{\frac{9.8}{0.5} } \\\\f = 0.71 \ Hz[/tex]
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