Answer:
Q' = 140.859 kJ
Explanation:
Given that, 93.4 g of solid ethanol at − 114.5 °C is converted to gasesous ethanol at 149.8 ° C.
The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol , and its molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .
And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .
The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .
Lets solve this step wise ;
Given 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g
⇒ moles of ethanol given = [tex]\frac{93.4}{46.06}[/tex] = 2.02 moles
step 1: solid ethanol to liquid ethanol at melting point of − 114.5 ° C
⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat
⇒ heat required = 4.60 × 2.02 = 9.292 kJ.
step 2: liquid ethanol at -114 °C to liquid ethanol at 78.4 °C
Q = m×S×ΔT ; Q = heat required
m = mass of the substance
S = specific heat of the substance
ΔT = change in temperature
Here S = S(l);
⇒ Q = 93.4×2.45×(78.4-(-114.5))
= 44.141 kJ
step 3: liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C
1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat
⇒ required heat = 38.56×2.02 = 77.89 kJ
step 4: gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C
Q = m×S×ΔT
Here, S = S(g)
Q = 93.4×1.43×(149.8-78.4)
= 9.536 kJ
⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536
= 140.859 kJ
⇒ Q' = 140.859 kJ
The total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.
Given data:,
93.4 g of solid ethanol at − 114.5 °C is converted to gaseous ethanol at 149.8 ° C.
The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol.
Molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .
And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .
The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .
Since, 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g
⇒ moles of ethanol given is
⇒93.04/46.06 = 2.02 moles
Heat required for conversion of solid ethanol to liquid ethanol at melting point of − 114.5 ° C
⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat
⇒ Q = 4.60 × 2.02 = 9.292 kJ.
Heat required to convert liquid ethanol at -114 °C to liquid ethanol at 78.4 °C
Q' = m×s×ΔT
Here,
m = mass of the substance
s = specific heat of the substance
ΔT = change in temperature
Solving as,
Q' = 93.4×2.45×(78.4-(-114.5))
Q' = 44.141 kJ
Heat required to convert liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C
1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat
Q'' = 38.56×2.02 = 77.89 kJ
Heat required to convert gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C
Q''' = m×S×ΔT
Q''' = 93.4×1.43×(149.8-78.4)
= 9.536 kJ
⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536
= 140.859 kJ
Thus, we can conclude that the total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.
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The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane and slides through a stop sign at 41 mph straight into a 3 ton SUV traveling at 16mph in the other direction. The car and the SUV crumple from the collision, and stick together.
1. What is the final velocity, in MPH (you don't need to enter MPH in your answer) , of the SUV/car entanglement (the positive direction is the direction the car was initially going)
Answer:
final velocity = - 1.75 mph
Explanation:
given data
mass m1 = 1 ton
mass m2 = 3 ton
velocity v = 41 mph
velocity u = 16 mph
to find out
what is final velocity V
solution
we will apply here Conservation of momentum that is express as
mv + Mu = (m + M) × V ...........................1
put here value we get
1 × 41 - 3 × 16 = (1 + 3 ) × V
solve it we get
41 - 48 = 4 V
V = [tex]\frac{-7}{4}[/tex]
final velocity = - 1.75 mph
Final answer:
The final velocity of the SUV/car entanglement is 9.9 m/s.
Explanation:
In order to determine the final velocity of the SUV/car entanglement, we need to first calculate the momentum of each vehicle before the collision.
Momentum is determined by the product of an object's mass and velocity. So, the momentum of the 1 ton car before the collision is 1 ton (or 1000 kg) multiplied by its velocity of 41 mph (which is equivalent to 18.3 m/s). Therefore, the momentum of the car is 1000 kg imes 18.3 m/s = 18300 kg*m/s.
Similarly, the momentum of the 3 ton SUV before the collision is 3 ton (or 3000 kg) multiplied by its velocity of 16 mph (which is equivalent to 7.1 m/s). Therefore, the momentum of the SUV is 3000 kg imes 7.1 m/s = 21300 kg*m/s.
Since momentum is conserved in collisions, the total momentum before the collision is equal to the total momentum after the collision. This means that the final velocity of the SUV/car entanglement can be calculated by dividing the total momentum by the total mass. The total momentum is 18300 kg*m/s + 21300 kg*m/s = 39600 kg*m/s. The total mass is 1000 kg + 3000 kg = 4000 kg. So, the final velocity of the SUV/car entanglement is 39600 kg*m/s / 4000 kg = 9.9 m/s.
While skiing in Jackson, Wyoming, your friend Ben (of mass 69.8 kg) started his descent down the bunny run, 16.8 m above the bottom of the run. If he started at rest and converted all of his gravitational potential energy into kinetic energy, what is Ben’s kinetic energy at the bottom of the bunny run? Use g = 9.8 m/s 2 Answer in units of J.
Answer:
11484.18 J
Explanation:
Applying conservation of energy we get
potential energy = kinetic energy
mgh = 0.5mv^2
gh= 0.5v^2
[tex]v= \sqrt{2gh}[/tex]
h= 16.8 m
g= 9.8 m/s^2
[tex]v= \sqrt{2\times16.8\times9.8}[/tex]
v=18.14 m/s at the bottom of the bunny run
therefore, Kinetic energy at the bottom of the bunny run
K= [tex]\frac{1}{2} mv^2=\frac{1}{2}69.8\times18.14^2[/tex]
therefore K= 11484.18 J
= 11.48 KJ
Answer: What is his final velocity?
Answer in units of m/s.
Explanation: i need help
How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)
a. 0.252b. 3.97c. 252d. 397
Explanation:
We need to find how many calories is 1 BTU.
Given
1 BTU = 1054 J
1 calorie = 4.186 J
So we have
1 BTU = 4.186 x 251.79 J
1 BTU =251.79 calorie
1 BTU = 252 calorie.
Option C is the correct answer.
A hypodermic needle can be thought of as a large area pipe leading to a small pipe that opens into the vein. The diameter of the plunger is 1.2 cm and the needle diameter is 0.15 cm. If the gauge pressure in the vein is 25 mmHg, what minimum is needed on the plunger in order for a fluid flow into the vein to occur
According to the statement and the data presented, it is presumed that the variable to look for is force. To solve this problem it is necessary to apply the concepts related to the Force according to the pressure and the Area.
Mathematically the Force can be expressed as
[tex]F = PA[/tex]
Where,
P = Pressure (At this case, the plunger pressure)
A = Cross-sectional Area ( At this case the plunger area), defined for a circle.
Our values are given as,
[tex]r = 0.006m[/tex]
[tex]P = 25mmHg[/tex]
Replacing we have that
[tex]F = (25mmHg \big [\frac{133N/m^2}{1mm\cdot Hg} \big ])(\pi 0.006^2)[/tex]
[tex]F = 0.376N[/tex]
Therefore the minimum force needed on the plunger inorder for a fluid flow into the vein to occur is 0.376
This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broadcasts (or aliens find us from ours). The closest stars are 4 light years away from us. How far away must you be from a 781 kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume that the photons spread out spherically. The area of a sphere is 4????????2.
Answer:
[tex]2.77287\times 10^{15}\ m[/tex]
Explanation:
P = Power = 50 kW
n = Number of photons per second
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
[tex]\nu[/tex] = Frequency = 781 kHz
r = Distance at which the photon intensity is i = 1 photon/m²
Power is given by
[tex]P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s[/tex]
Photon intensity is given by
[tex]i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m[/tex]
The distance is [tex]2.77287\times 10^{15}\ m[/tex]
You must stay at a distance of [tex]2.77287*10^1^5m[/tex]
How can we arrive at this result?First, we have to find the number of protons per second. This will be done using the equation: [tex]n= \frac{P}{h*v}[/tex]In this equation, the "h" represents Planck's constant and will take on the value of [tex]6.626*10^-^3^4m^2\frac{Kg}{s}[/tex]
The "r" will be equal to 1 photon/m² and the "P' will be equal to 50 kW.
Therefore, we will solve the equation as follows:
[tex]n= \frac{50000}{(6.626*10^-^3^4*781000)}= 9.66201*10^3^1 \frac{protons}{s}[/tex]
From this value, we can calculate the appropriate distance for you to position yourself. For this, we will use the equation:[tex]r^2=\frac{n}{4*\pi} \\r= \sqrt{\frac{9.6621*10^3^1}{4*\pi } } = 2.77287*10^1^5m[/tex]
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There is a skier at the top of a ski slope. The skier has potential energy. What gives the skier his potential energy? A. his speed B. friction C. gravity D. his efficiency
Answer:C
Explanation:
Skier at the top of a ski has Potential Energy due to gravity.
Potential Energy is the Energy Possessed by an object when it attains a height concerning some zero level Position.
During the process of attaining the height, some work has to be done against gravity and this energy stored within the object after attaining some height w.r.t relative zero position.
The answer is C: gravity
After coming down a steep hill at a constant speed of 43 m/s, a car travels along the circumference of a vertical circle of radius 618 m until it begins to climb another hill. r x What is the magnitude of the net force on the 34 kg driver of the car at the lowest point on this circular path? Answer in units of kN.
Answer:
F=0.101 kN
Explanation:
Newton's 2nd law, F = ma, but this is circular path, the acceleration (a) is the centripetal acceleration.
a = (v²) / r
F = (m×v²) /r
F=(34 kg)×(43 m/s)² / 618 m
F=101.72 N
To convert Newtons into kilo-Newtons divide it 1000.
F=0.101 kN
The deflection of air masses to the right or left (depending on latitude) as they move from one latitude to another is called the:
a. Coriolis effect.
b. Hadley cell.
c. Saffir-Simpson scale.
d. Cyclonic effect.
e. Ekman spiral.
The coriolis effect is the force produced by the rotation of the Earth in space, which tends to deflect the trajectory of objects that move on the surface of the earth; to the right in the northern hemisphere and to the left, in the south. Said 'object' for this particular case is the mass of air. Therefore the correct answer is A: Coriolis effect.
Explain in detail why it may take longer for you to fight off the flu virus the first time this season than it would if you caught the same strain a second time.
Answer:
"when an infection has cleared, a small number of B and T cells are present in the blood with memory of the virus, allowing them to activate and destroy viruses more quick and fast next time when they enter the body."which is known as immune system."
Explanation:
Our immune system is not only designed to destroy disease-causing codes but remembers codes it's encountered so they are able to fight against them with when they return. when we all come across these types of invaders our layer of protection stops it from entering our body. When body senses a virus or other infections, like bacteria, the body tends to try and destroy the foreign invaders. The first step is the macrophages is activated. Macrophages destroy the foreign invaders. If the viruses penetrates deep into the body it result into infection. Where the T cells and B cells are activated to fight against these viruses.T & B cells contains antibodies that attach to the virus and label it as foreign for other cells to destroy.
A 100 kg box is hanging from two strings. String #1 pulls up and left, making an angle of 80o with the horizontal on the left, and string #2 pulls up and to the right, making an angle of 65o with the horizontal on the right. FInd the tension in string #2 (on the right side)
Answer:296.76 N
Explanation:
Given
mass of box [tex]m=100 kg[/tex]
Let [tex]T_1[/tex] be the Tension in left side and [tex]T_2[/tex] be the Tension in the right side
From diagram
[tex]T_1\cos 80=T_2\cos 65[/tex]
[tex]T_1=T_2\cdot \frac{\cos 65}{\cos 80}[/tex]
and
[tex]T_1\sin 80+T_2\sin 65=100\cdot g[/tex]
[tex]T_2\left [ \tan 80\cdot \cos 65+\sin 65\right ]=100\cdot g[/tex]
[tex]T_2=\frac{100\cdot g}{\left [ \tan 80\cdot \cos 65+\sin 65\right ]}[/tex]
[tex]T_2=\frac{980}{3.3023}=296.76 N[/tex]
Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2, and oxygen gas, O2 to produce the following combustion reaction: 2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g) Imagine that you have a 6.50 L gas tank and a 4.00 L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 115 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
Answer:
Pressure in acetylene gas tank will be 74.8atm
Explanation:
Step 1: Using the ideal gas equation, determine the number of moles of oxygen
[tex]n=\frac{PV}{RT}[/tex]
[tex]P_O=115atm[/tex]
[tex]V_O=6.50L[/tex]
[tex]n_O=\frac{P_OV_O}{RT}[/tex]
[tex]n_O=\frac{115*6.50}{RT}[/tex]
[tex]n_O=\frac{747.5}{RT}[/tex]
As temperature is unknown and assumed to be the same for both gases, and the ideal gas constant will be the same for both cases, these values are left as constants 'T' and 'R'.
Step 2: Determine the proportionate number of moles of acetylene required based on the chemical equation
2 moles of acetylene require 5 moles of oxygen for complete combustion
Thus, 0.4 moles of acetylene are required per mole of oxygen
[tex]n_A=0.4n_O[/tex]
[tex]n_A=0.4*\frac{747.5}{RT}[/tex]
[tex]n_A=\frac{299}{RT}[/tex]
Step 3: Determine the pressure of acetylene tank required in 4.00L tank
[tex]n_A=\frac{P_AV_A}{RT}[/tex]
[tex]\frac{299}{RT}=\frac{P_A*4.00}{RT}[/tex]
[tex]P_A=\frac{299}{4.00}[/tex]
[tex]P_A=74.75 atm[/tex]
Assumptions:
Temperature is the same for both gases and constant
The ideal gas constant is the same for both gases
The combustion reaction is complete and there are no limiting factors
A row of five 1.2 N wooden blocks is being pushed across a tabletop at a constant speed by a toy tractor that exerts a force of 1.7 N on the row. What is the coefficient of kinetic friction between the wooden blocks and the tabletop?
To solve this problem we will apply the concepts related to the Kinetic Friction Force for which we define as
[tex]F = \mu_k N[/tex]
Where,
N = Normal Force (Mass for gravity)
[tex]\mu_k =[/tex] Kinetic frictional coefficient
The total force applied is 1.7N and the Force from the (normal) weight is equivalent to five times 1.2N, therefore:
[tex]F = 5(N)\mu_k[/tex]
[tex]1.7 = 5(1.2)(\mu_k)[/tex]
[tex]\mu_k = \frac{1.7}{5*1.2}[/tex]
[tex]\mu_k =0.283[/tex]
Therefore the coefficient of kinetic friction between the wooden blocks and the tabletop is 0.283
At high noon, the sun delivers 1 000 W to each square meter of a blacktop road.
What is the equilibrium temperature of the hot asphalt, assuming its emissivity e = 1? (σ = 5.67 × 10−8W/m2⋅K4) .
a. 75°Cb. 84°Cc. 91°Cd. 99°C
Answer:
The correct answer is c, T = 91.3°C
Explanation:
For this exercise let's use Stefan's equation on the emission of a black body
P = σ A e T⁴
Where σ is the Stefan-Boltzmann constant, A the area, and 'e' emissivity and T the absolute temperature
In this case give the absorbed power is 1000W per square meter, let's clear the temperature equation
T⁴ = (P / A) 1/σ e
Let's calculate
T⁴ = 1000 1 / (5.67 10⁻⁸ 1)
T⁴ = 176.37 10⁸
T =[tex]\sqrt[4]{176.37 10^8}[/tex]
T = 3.6442 10² K
Let's reduce to degrees Celsius
T = 364.42 -273.15
T = 91.3 ° C
The correct answer is c
The equilibrium temperature of the hot asphalt, assuming it behaves as a perfect blackbody with emissivity of 1, can be calculated using the Stefan-Boltzmann law, resulting in approximately 91°C, making choice (c) the correct answer.
To determine the equilibrium temperature of the hot asphalt, we can use the concept of blackbody radiation.
Since the emissivity (e) is 1, the asphalt behaves as a perfect blackbody, which means it absorbs and emits radiation efficiently.
The power per unit area absorbed by the asphalt is:
P = 1000 W/m²According to the Stefan-Boltzmann law, the power radiated per unit area by a blackbody is given by:
P = σeT⁴where
σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²·K⁴), and T is the equilibrium temperature in KelvinGiven that e = 1 for a perfect blackbody, we can set the absorbed power equal to the emitted power:
1000 = 5.67 × 10⁻⁸ T⁴Solving for T:
T⁴ = 1000 / 5.67 × 10⁻⁸T⁴ ≈ 1.76 × 10¹⁰T ≈ (1.76 x 10¹⁰)∧1/4T ≈ 278.8 KConverting to Celsius:
T ≈ 278.8 K - 273.15 ≈ 5.65°CThis result does not match any answer choices, which suggests a potential issue. The correct calculations should yield a higher temperature due to an error in an earlier assumption or value misunderstanding. Revisiting the calculations correctly:
Solving again for higher accuracy:
T ≈ (1000 / 5.67 x 10⁻⁸)∧1/4 ≈ 364 K ≈ 91°CTherefore, the correct equilibrium temperature is 91°C, making the correct choice: (c) 91°C
A concrete pump of 120-cy/hr max. output is used to place 250 cy of concrete. It is supplied by 8 cy transit-mix trucks arriving to the site every 15 mins. Barring interruptions how many hours will the operation require, considering 80% efficiency factor.
Answer:
Completing the operation will require 9.765625 hours.
Explanation:
First, let's check which of the stages will be the process limitation:
Regarding the pump, it can perform up to 120 cy/h. The trucks will deliver 8 cy every 15 min, i.e., 32 cy/h, then this is the smaller capacity within the whole process.
Now, let's establish a simple relation:
Placement speed x Efficiency = Placed quantity / time,
solving for time. we have
[tex]t = \frac{q}{\epsilon s} = \frac{250 cy}{0.8\times 32\frac{cy}{h}} = \mathbf{9.765625 h}.[/tex]
The operation will require approximately 2.6 hours, taking into account the efficiency factor.
Explanation:To determine how many hours the operation will require, we need to calculate the total time it takes for all the trucks to arrive at the site and unload the concrete. Since each truck delivers 8 cy of concrete every 15 mins, we can calculate the time it takes for all the trucks to deliver 250 cy of concrete: 250 cy / (8 cy/truck * 15 mins/truck) = 250 cy / (120 mins) = 2.08 hours.
However, we need to account for the 80% efficiency factor, which means the pump is only operating at 80% of its maximum output. Therefore, the total time required would be 2.08 hours / 0.8 = 2.6 hours.
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Our favorite program runs in 10 seconds on computer A, which has a 4 GHz. clock. We are trying to help a computer designer build a new machine B, that will run this program in 6 seconds. The designer can use new (or perhaps more expensive) technology to substantially increase the clock rate, but has informed us that this increase will affect the rest of the CPU design, causing machine B to require 1.2 times as many clock cycles as machine A for the same program. What clock rate should we tell the designer to target?
Answer:
The rate clock is about
F = 8 GHz
Explanation:
f₁ = 4 G Hz , t₁ = 10 s , t₂ = 6s , f₂ = 1.2 f₁
Can organize to find the rate clock the designer build to the target so
X / 4 Ghz = 10 s , 1.2 X / Y = 6 s
X * Y = 10 s ⇒ F = 10 s
1.2 * 4 G Hz = 6 s
F = 10 * ( 1.2 * 4 G Hz ) / 6
F = 10 * ( 1.2 * 4 x 10 ⁹ Hz ) / 6
F = 8 x 10 ⁹ Hz
F = 8 GHz
Final answer:
To run a program in 6 seconds on Machine B, which needs 1.2 times as many clock cycles as Machine A, the target clock rate should be 8 GHz.
Explanation:
The student asked for the target clock rate needed for computer B to run a program in 6 seconds, given that computer A runs it in 10 seconds with a 4 GHz clock and that machine B needs 1.2 times as many clock cycles as machine A for the same program. To solve this, we know that time (T) is equal to the number of cycles (N) divided by the clock rate (C), or T = N / C. For machine A, TA = NA / CA and for machine B, TB = (1.2 * NA) / CB. If machine A completes the program in 10 seconds, the number of cycles it uses is CA * TA, which is 4 GHz * 10 s, yielding 40 billion cycles.
Machine B needs to run these 40 billion cycles in 6 seconds. Also, machine B requires 1.2 times the number of cycles of machine A; thus we have (1.2 * 40 billion) / 6 s to find CB, the clock rate for machine B. This simplifies to 8 GHz. Thus, for machine B to run the program in 6 seconds, the target clock rate should be 8 GHz.
A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is applied a damping force (in Newtons) of the form F = –0.1v, with v the velocity in m/s. The spring is stretched a distance xm and released. After four complete oscillations, what fraction of the mechanical energy is retained by the system?
Answer:
0.5% per oscillation
Explanation:
The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.
Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).
Where m= Mass, v= velocity.
Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).
Where k= force constant, X= displacement.
Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).
Therefore, we use equation (3) to get the resonance frequency,
W^2= k/m--------------------------------------------------------------------------------------(3)
Slotting values into equation (3).
= 10/2.5.
= ✓4.
= 2 s^-1.
Recall that, F= -kX
F^2= (-0.1)^2
Potential energy,PE= 1/2 ×0.01
Potential energy= 0.05 ×100
= 0.5% per oscillation.
In a damped harmonic oscillator, the fraction of mechanical energy retained by the system after multiple oscillations can be calculated by comparing the initial and final potential energies. The fraction of mechanical energy retained is equal to the ratio of the final amplitude squared to the initial amplitude squared.
Explanation:In a damped harmonic oscillator, the mechanical energy is gradually lost due to the damping force. The fraction of mechanical energy retained by the system after four complete oscillations can be determined by comparing the initial mechanical energy to the final mechanical energy. The initial mechanical energy is the sum of the potential energy and kinetic energy, while the final mechanical energy is only the potential energy.
Using the equation for the potential energy of a spring, U = ½kx², we can calculate the initial and final potential energies. The initial potential energy can be calculated using the initial amplitude, A, and the spring constant, k. The final potential energy can be calculated using the final amplitude, A', and the same spring constant, k.
The fraction of mechanical energy retained by the system is equal to the ratio of the final potential energy to the initial potential energy. This can be calculated using the equation: fraction of mechanical energy retained = (A'/A)².
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The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acceleration. The truck is hauling an empty 460-lb trailer with a uniform 72-lb gate hinged at O and held in the slightly tilted position by two pegs, one on each side of the trailer frame at A. Determine the maximum shearing force developed in each of the two pegs during the acceleration.
Final answer:
To find the maximum shearing force in the pegs during the truck's acceleration, Newton's second law is applied, and the weight and acceleration of the gate are considered. The final shear force for each peg is 7.7021 lbs.
Explanation:
To find the maximum shearing force in each peg during the acceleration of the pickup truck with the trailer, we need to apply Newton's second law.
First, let's convert the speed from miles per hour to feet per second (1 mi/hr = 1.46667 ft/s):
37 mi/hr × 1.46667 ft/s/mi/hr = 54.2667 ft/s
Using the kinematic equation v^2 = u^2 + 2as (where v is final velocity, u is initial velocity, s is distance, and a is acceleration), we can solve for acceleration (a) since the truck starts from rest (u = 0):
54.2667^2 = 0 + 2 × a × 215
a = × 54.2667^2 / (2 × 215) = 6.88776 ft/s^2
The force affecting the gate pegs comes from the horizontal component of the gravitational force and the force due to acceleration. Assume that the mass of the gate is 72 lbs; the gravitational force is:
F_gravity = m × g = 72 lbs (Since g in lbs already accounts for Earth's gravitational acceleration)
To find the force due to acceleration, we use F = m × a:
F_acceleration = 72 lbs × 6.88776 ft/s^2
However, since lbs already include g, we convert lbs to mass in slugs:
m = 72 lbs / 32.2 ft/s^2 = 2.23602 slugs
F_acceleration = 2.23602 slugs × 6.88776 ft/s^2 = 15.4042 lbs
The total force on each peg is thus the shearing force due to the acceleration, but as there are two pegs, we divide the total force by 2, assuming the load is distributed evenly:
Shear force per peg = F_acceleration / 2 = 15.4042 lbs / 2 = 7.7021 lbs (rounded to four significant figures)
Starting from rest, your friend dives from a high cliff into a deep lake below, yelling in excitement at the thrill of free-fall on her way down. You watch her, as you stand on the lake shore, and at a certain instant your keen hearing recognizes that the usual frequency of her yell, which is 919 Hz, is shifted by 55.9 Hz. How long has your friend been in the air when she emits the yell whose frequency shift you hear? Take 342 m/s for the speed of sound in air and 9.80 m/s2 for the acceleration due to gravity.
Answer:
2 seconds
Explanation:
f = Frequency of yell = 919 Hz
[tex]\Delta f[/tex] = Shifted frequency = 55.9 Hz
v = Speed of sound in air = 342 m/s
[tex]v_r[/tex] = Velocity of friend
a = Acceleration due to gravity = 9.81 m/s²
From the Doppler shift formula we have
[tex]\dfrac{f+\Delta f}{f}=\dfrac{v}{v-v_r}\\\Rightarrow v_r=v-\dfrac{vf}{f+\Delta f}\\\Rightarrow v_r=342-\dfrac{342\times 919}{919+55.9}\\\Rightarrow v_r=19.61\ m/s[/tex]
The velocity of the my friend is 19.61 m/s
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{19.61-0}{9.8}\\\Rightarrow t=2\ s[/tex]
The time my friend is in the air is 2 seconds
What is absolute zero? What is the temperature of absolute zero on the Kelvin and Celsius scales?
Answer:
Absolute zero = 0 K or - 273°C
Explanation:
Absolute zero :
When the entropy and enthalpy of the ideal system reach at the minimum value then the temperature at that condition is known as absolute zero condition.
Absolute temperature is the minimum temperature in the temperature scale.The value of absolute zero is 0 K.
We know that
[tex]\dfrac{C-0}{100}=\dfrac{K-273}{100}=\dfrac{F-32}{180}[/tex]
F=Temperature in Fahrenheit scale
K=Temperature in Kelvin scale
C=Temperature in degree Celsius scale
When K = 0
[tex]\dfrac{C-0}{100}=\dfrac{K-273}{100}[/tex]
[tex]\dfrac{C-0}{100}=\dfrac{0-273}{100}[/tex]
C= - 273°C
Absolute zero = 0 K or - 273°C
Absolute zero is the lowest possible temperature, defined as 0 K on the Kelvin scale and -273.15°C on the Celsius scale. At this temperature, particles have minimal vibrational motion.
Explanation:Absolute zero is the lowest possible temperature where nothing could be colder and no heat energy remains in a substance. It is the point at which the fundamental particles of nature have minimal vibrational motion, retaining only quantum mechanical, zero-point energy-induced particle motion.
On the Kelvin scale, absolute zero is defined as 0 K. This is not stated in degrees, as it is an absolute measure. On the Celsius scale, absolute zero is equivalent to -273.15°C.
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the crankshaft of an engine increases its spin from 1,000 to 2,000 rpm in 0.50 s what is its angular acceleration
Answer:
angular acceleration = 209.44 [rad/s^2]
Explanation:
First we have to convert the velocities which are in revolutions per minute to radians on second.
where:
[tex]w_{0} = 1000 [\frac{rev}{min}]*[\frac{2\pi rad }{1rev}] * \frac{1min}{60s} = 104.7[\frac{rad}{s} ]\\w = 2000 [\frac{rev}{min}]*[\frac{2\pi rad }{1rev}] * \frac{1min}{60s} = 209.4[\frac{rad}{s} ][/tex]
Now we can find the angular acceleration:
[tex]w=w_{0} + \alpha *t\\\alpha =\frac{w-w_{0} }{t} \\\alpha =\frac{209.43-104.71}{\0.5 } \\\alpha = 209.44[\frac{rad}{s^{2} } ][/tex]
The angular acceleration of the crankshaft is [tex]\(209.44 \, \text{rad/s}^2\)[/tex].
To find the angular acceleration, we can use the following formula:
[tex]\[ \alpha = \frac{\Delta \omega}{\Delta t} \][/tex]
where:
[tex]\(\alpha\)[/tex] is the angular acceleration
[tex]\(\Delta \omega\)[/tex] is the change in angular velocity
[tex]\(\Delta t\)[/tex] is the change in time
First, we need to convert the angular velocities from revolutions per minute (rpm) to radians per second (rad/s).
1. Initial angular velocity [tex](\(\omega_i\))[/tex]:
[tex]\[ \omega_i = 1000 \, \text{rpm} \][/tex]
2. Final angular velocity [tex](\(\omega_f\))[/tex]:
[tex]\[ \omega_f = 2000 \, \text{rpm} \][/tex]
To convert rpm to rad/s:
[tex]\[ \omega (\text{rad/s}) = \omega (\text{rpm}) \times \frac{2\pi \, \text{rad}}{60 \, \text{s}} \][/tex]
[tex]\[ \omega_i = 1000 \times \frac{2\pi}{60} \, \text{rad/s} = \frac{2000\pi}{60} \, \text{rad/s} \approx 104.72 \, \text{rad/s} \][/tex]
[tex]\[ \omega_f = 2000 \times \frac{2\pi}{60} \, \text{rad/s} = \frac{4000\pi}{60} \, \text{rad/s} \approx 209.44 \, \text{rad/s} \][/tex]
Now, we can find the change in angular velocity:
[tex]\[ \Delta \omega = \omega_f - \omega_i \][/tex]
[tex]\[ \Delta \omega = 209.44 \, \text{rad/s} - 104.72 \, \text{rad/s} = 104.72 \, \text{rad/s} \][/tex]
Given that the change in time [tex](\(\Delta t\))[/tex] is 0.50 s:
[tex]\[ \alpha = \frac{104.72 \, \text{rad/s}}{0.50 \, \text{s}} \][/tex]
[tex]\[ \alpha = 209.44 \, \text{rad/s}^2 \][/tex]
There was an accident at a uranium processing plant in Tokai, Japan, on September 30, 1999. Using your favorite search engine, find out what happened and compare this accident with the ones at Chernobyl and Three Mile Island. How could these accidents have been avoided? Do these incidents prove that nuclear power plants should be phased out? Why or why not?
For the three accidents mentioned there were human factors that caused the destabilization of the reactors and generated catastrophies.
In the case of Tokai, there was an excess of Uranium due to the fact that workers (Not qualified, since such work did not merit it) added to the containers, which generated an excess in the filling generating an emulsion of radiation to all the personnel.
In the case of Chernobyl, it was the experimentation through a series of tests to reduce the power, during which a series of imbalances occurred in the reactor 4 of this nuclear power plant, which led to the uncontrolled overheating of the reactor core nuclear-
In the case of Three Mile Island it was a design error in which the water level was underestimated, believing that the required level was available but in the end it was noted that said water level was not sufficient which caused the melting of a water dipstick.
All this leaves us with reflections on the security protocols followed for the construction or management of these nuclear power plants, and calls into question the human capacity to react to one of these catastrophes. Today, these plants not only put human health at risk but also generate waste that pollutes the planet. Alternatives such as renewable and clean energy already become more popular every day and are about to leave nuclear energy in the past to give rise to a new human stage.
In the experiment to measure specific heat of a metal, you get the following data. What is the specific heat of the metal in J/kg.C?
(answer within +-3% will considered correct) MASSES: water= 259 g, metal=159 g, calorimeter=97 g INITIAL TEMPERATURES: water+calorimeter=21ºC, metal=98ºC Final equilibrium temp=31ºC Calorimeter is made of aluminum.
Answer:
c = 1,100 J/kgºC
Explanation:
Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the unknown metal.
The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:
Q = c * m* Δt
where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.
In our case, we can write the following equality:
(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (cₓ*mₓ*Δtₓ)
Replacing by the givens , and taking cAl = 0.9 J/gºC, we have:
Qg= 0.9 J/gºC*97g*10ºC + 4.186 J/gºC*259g*10ºC = 11,715 J(1)
Ql = cₓ*159g*67ºC (2)
Based on all the previous assumptions, we have:
Qg = Ql
So, we can solve for cx, as follows:
cx = 11,715 J / 159g*67ºC = 1.1 J/gºC (3)
Expressing (3) in J/kgºC:
1.1 J/gºC * (1,000g/1 kg) = 1,100 J/kgºC
Which one of the following statements is not a characteristic of a plane mirror?
A) The image is the same size as the object.
B) The image is always upright.
C) The image is real.
D) The image is reversed left to right compared to the object.
E) The image is the same distance behind the mirror as the object is in front of the mirror.
Answer:
C) false. The image is formed by the prolongation of the rays, so it is VIRTUAL
Explanation:
Let's review each of the statements
A) True. The image is the same size as the object in a flat mirror, m = 1
B) True. The rays are not inverted, so the right images
C) false. The image is formed by the prolongation of the rays, so it is VIRTUAL
D) True. Flat mirrors reverse left and right
E) True. When using trigonometry the angles are equal, therefore two triangles formed have the same leg, and the distance to the object and the image are equal
Answer:
The image is real is not a characteristics of a plane mirror.
Explanation:
A plane mirror is a reflector with a flat reflective covering. For light beams hitting a plane mirror, the angle of reflection matches the angle of incidence. The angle of the incidence is the angle within the incident beam and the surface normal flat mirror.
Reason for Image is not real in the plane mirror:
A real image is that picture that is created when the light beams originating from an object meet each other after reflection or refraction.A real image can be captured on the screen. The real image is constantly inverted. A common example of a real image is the image formed on the cinema screen.Thus we can say, real image is not a characteristic of a plane mirror.
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A cylinder contains 0.300 mol of carbon dioxide (CO2) gas at a temperature of 30.0 ∘C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The cylinder is placed on a hot plate and a 920 J of heat flows into the gas, thereby raising its temperature to 129 ∘C. Assume that the CO2 may be treated as an ideal gas. How much work is done by the gas in this process?
Answer:246.92 kJ
Explanation:
Given
Initial Temperature [tex]T_i=30^{\circ}C[/tex]
Final Temperature [tex]T_f=129^{\circ}C[/tex]
no of moles [tex]n=0.3[/tex]
Pressure is constant [tex]P=1 atm [/tex]
As Pressure is constant therefore work done is given by
[tex]W=P\Delta V[/tex]
[tex]P\Delta V[/tex] can also be written as [tex]P\Delta V=nR\Delta T[/tex]
[tex]W=nR\Delta T[/tex]
[tex]W=0.3\times 8.314\times (129-30)[/tex]
[tex]W=246.92 kJ[/tex]
NASA operates a 2.2-second drop tower at the Glenn Research Centre in Cleveland, Ohio. At this facility, experimental packages are dropped from the top of the tower, on the 8th floor of the building. During their 2.2 seconds of free fall, experiments experience a microgravity environment similar to that of a spacecraft in orbit,
(a) What is the drop distance of a 2.2-s tower?
(b) How fast are the experiments travelling when they hit the air bags at the bottom of the tower?
(c) If the experimental package comes to rest over a distance of 0.75 m upon hitting the air bags, what is the average stopping acceleration?
Answer:
23.7402 m
21.582 m/s
310.521816 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.2^2\\\Rightarrow s=23.7402\ m[/tex]
The drop distance is 23.7402 m
[tex]v=u+at\\\Rightarrow v=0+9.81\times 2.2\\\Rightarrow v=21.582\ m/s[/tex]
When they hit the air bags at the bottom of the tower the speed of the experiments is 21.582 m/s
The final speed of the fall will be the initial velocity of stopping
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-21.582^2}{2\times 0.75}\\\Rightarrow a=-310.521816\ m/s^2[/tex]
The average stopping acceleration is 310.521816 m/s²
Final answer:
Calculations reveal a drop distance of 23.65 meters, a final velocity of 21.56 m/s, and an average stopping acceleration of 311.43 m/s² for the experimental package in the 2.2-second drop tower scenario.
Explanation:
The question encompasses the calculation of drop distance, final velocity, and average stopping acceleration of an experiment package in a microgravity test environment. To solve these, we apply the equations of motion under uniform acceleration.
To find the drop distance (d) for a 2.2-second fall, we use the equation d = 0.5 * g * t, where g is the acceleration due to gravity (9.8 m/s²) and t is time in seconds. This gives us d = 0.5 * 9.8 * 2.22 = 23.65 meters.To calculate the final velocity (v) when the experiments hit the air bags, we employ v = g * t, resulting in v = 9.8 * 2.2 = 21.56 m/s.The average stopping acceleration (a) can be derived from the equation v² = 2 * a * d, rearranged to a = v² / (2*d), where d is the stopping distance of 0.75 m, leading to a = (21.562) / (2*0.75) = 311.43 m/s².This analysis presupposes a vacuum environment in the drop tower to negate air resistance, thereby approximating conditions of free fall and achieving microgravity for the duration of the drop.
A(n) 59 kg astronaut becomes separated from
the shuttle, while on a spacewalk. She finds
herself 63.2 m away from the shuttle and moving with zero speed relative to the shuttle. She
has a(n) 0.695 kg camera in her hand and decides to get back to the shuttle by throwing
the camera at a speed of 12 m/s in the direction away from the shuttle.
How long will it take for her to reach the
shuttle? Answer in minutes.
1. 10.06 min
2. 9.687 min
3. 8.569 min
4. 8.197 min
5. 7.452 min
6. 7.824 min
7. 10.43 min
8. 8.942 min
Answer:
5. 7.452 min
Explanation:
Momentum is conserved, so:
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
0 = (0.695 kg) (-12 m/s) + (59 kg) v
v = 0.1414 m/s
She is 63.2 m away, so the time it takes to reach the shuttle is:
t = 63.2 m / 0.1414 m/s
t = 447.1 s
t = 7.452 min
To find how long it will take for the astronaut to reach the shuttle, the conservation of momentum is used to calculate her velocity after she throws the camera. She will travel at a velocity of 0.141 m/s. It will take her approximately 7.47 minutes to cover the 63.2-meter distance to the shuttle.
Explanation:The given scenario can be solved using the principle of conservation of momentum. The momentum of the system (the astronaut and the camera) before and after the camera is thrown must be equal since there are no external forces acting on the system. The momentum of the astronaut can be determined using the equation p = m × v, where p is momentum, m is mass, and v is velocity.
To find the velocity of the astronaut after throwing the camera, we use the following relationship:
Solving for the velocity of the astronaut, we get:
0 = (59 kg × velocity of astronaut) + (0.695 kg × (-12 m/s))
Velocity of astronaut = (0.695 kg × 12 m/s) / 59 kg = 0.141 m/s
To calculate how long it will take for the astronaut to reach the shuttle, we divide the distance by her velocity:
Time = Distance / Velocity = 63.2 m / 0.141 m/s = 448.227 s
Convert seconds to minutes:
Time in minutes = 448.227 s / 60 = 7.47 minutes
The closest answer from the options provided is 7.452 min, but rounding should be done judiciously, and the calculated value is between 7.452 min and 7.824 min, so it's important to check the context in which the answer will be used, as the rounding might affect the selection of the correct option.
Near the top of the Citigroup Bank building in New York City, there is a 4.00 105 kg mass on springs having adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven — the driving force is transferred to the mass, which oscillates instead of the building.
(a) What effective force constant should the springs have to make the mass oscillate with a period of 3.00 s? N/m
(b) What energy is stored in the springs for a 2.00 m displacement from equilibrium?
Answer:
(a) k = 1.76× 10⁶ N/m
(b) E = 3.52 × 10⁶ J
Explanation:
(a)
The period (T) of a spring = 2π√(m/k)
where m = mass of the spring in kg, k = spring constant.
T = 2π√(m/k)..................... equation 1
making k the subject of the equation,
k = 4π²(m)/T².............................. equation 2
Where m = 4.00 × 10⁵ kg, T = 3.00 s, π = 3.143
Substituting these values into equation 2
k = 4(3.143)²(4.0×10⁵)/3²
k = (1.58 × 10⁷)/9
k = 1.76× 10⁶ N/m
(b)
The energy stored(E) in a spring = 1/2ke²
Where k = spring constant, e = extension.
E = 1/2ke²
k = 1.76× 10⁶ N/m, e= 2.00 m
∴E = 1/2(1.76× 10⁶)(2)²
E = 2 × 1.76 × 10⁶
E = 3.52 × 10⁶ J
An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended into heat for warming water.
How much mechanical energy is required to increase the temperature of 300 g of water (enough for 1 cup of coffee) from 20°C to 95°C?
(1 cal = 4.186 J, the specific heat of water is 4 186 J/kg⋅°C)
a. 94 000 J
b. 22 000 J
c. 5 400 J
d. 14 J
Answer:
[tex]Q=94185\ J[/tex]
Explanation:
Given:
mass of water, [tex]m=0.3\ kg[/tex]initial temperature of water, [tex]T_i=20^{\circ}C[/tex]final temperature of water, [tex]T_f=95^{\circ}C[/tex]specific heat of water, [tex]c=4186\ J.kg^{-1}.K^{-1}[/tex]Now the amount of heat energy required:
[tex]Q=m.c.\Delta T[/tex]
[tex]Q=0.3\times 4186\times (95-20)[/tex]
[tex]Q=94185\ J[/tex]
Since all of the mechanical energy is being converted into heat, therefore the same amount of mechanical energy is required.
An inductor with an inductance of 3.50 H and a resistance of 8.00 Ω is connected to the terminals of a battery with an emf of 4.00 V and negligible internal resistance.a)
Just after the circuit is completed, at what rate is the battery supplying electrical energy to the circuit?Express your answer with the appropriate units.b)When the current has reached its final steady-state value, how much energy is stored in the inductor?Express your answer with the appropriate units.c)What is the rate at which electrical energy is being dissipated in the resistance of the inductor?Express your answer with the appropriate units.d)What is the rate at which the battery is supplying electrical energy to the circuit?
Answer:
a. 0 W
b. 0.4375 J
c. 2 W
d. 2 W
Explanation:
a.
Recall that the power is directly proportional to the current in the circuit. (P=I*V) First, determine the current at t=0. Note that inductor prevents an instantaneous build-up of current. Therefore, initial power is 0 W.
P initial = 0 W
b.
given on the APPC equation sheet (last equation bottom right):
U = (1/2)*L*(I^2)
we know I = V/R
U = (1/2)*3.5*(4/8)^2
U = .4375 J
c.
P = R*I^2
I=V/R
P = 8*(4/8)^2
P=2 W
d. (same as part c.)
P = R*I^2
I=V/R
P = 8*(4/8)^2
P=2 W
a) The battery supplies electrical energy at a certain rate just after the circuit is completed. b) When the current has reached its steady-state value, the energy stored in the inductor can be calculated. c) The rate at which electrical energy is being dissipated in the resistance of the inductor can be found using the power dissipated formula. d) The rate at which the battery is supplying electrical energy to the circuit is the same as the power delivered by it.
Explanation:a) Just after the circuit is completed, at what rate is the battery supplying electrical energy to the circuit?
To determine the rate at which the battery is supplying electrical energy, we need to calculate the power delivered by the battery. Power is given by the formula P = IV, where P is power in watts, I is current in amperes, and V is voltage in volts. Since we know the voltage (4.00 V) and the resistance (8.00 Ω), we can use Ohm's Law (V = IR) to find the current. Once we have the current, we can calculate the power delivered by the battery.
b) When the current has reached its final steady-state value, how much energy is stored in the inductor?
When the current reaches its final steady-state value, the inductor acts like a short circuit and has no effect on the circuit. Therefore, all the energy stored in the inductor is dissipated as heat due to the resistance. Since we know the inductance (3.50 H) and the resistance (8.00 Ω), we can calculate the energy stored in the inductor using the formula E = 0.5LI^2, where E is energy in joules, L is inductance in henries, and I is current in amperes.
c) What is the rate at which electrical energy is being dissipated in the resistance of the inductor?
The rate at which electrical energy is being dissipated in the resistance of the inductor is equal to the power dissipated. Power dissipated is given by the formula P = I^2R, where P is power in watts, I is current in amperes, and R is resistance in ohms. Since we know the resistance (8.00 Ω) and the current, we can calculate the power.
d) What is the rate at which the battery is supplying electrical energy to the circuit?
The rate at which the battery is supplying electrical energy to the circuit is the same as the power delivered by the battery, which we calculated in part a using the formula P = IV.
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A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D along the direction of its motion.
By what multiplicative factor RK does the initial kinetic energy increase, and by what multiplicative factor RWdoes the work done by the force increase (with respect to the case when the particle had a mass M)?
If one of the quantities doubles, for instance, it would increase by a factor of 2. If a quantity stays the same, then the multiplicative factor would be 1.
Answer:
Explanation:
Initial kinetic energy of M = 1/2 M vi²
let final velocity be vf
v² = u² + 2a s
vf² = vi² + 2 (F / M) x D
Kinetic energy
= 1/2 Mvf²
= 1/2 M ( vi² + 2 (F / M) x D
1/2 M vi² + FD
Ratio with initial value
1/2 M vi² + FD) / 1/2 M vi²
RK = 1 + FD / 2 M vi²