Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
The rotor of a helicopter is gaining angular speed with constant angular acceleration. At tt = 0 it is rotating at 1.25 rad/srad/s. From tt = 0 to tt = 2.00 ss, the rotor rotates through 8.00 radrad. What is the angular acceleration of the rotor?
Answer:
[tex]2.75\ rad/s^2[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 1.25 rad/s
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation = 8 rad
t = Time taken = 2 s
[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 8=1.25\times 2+\frac{1}{2}\times \alpha\times 2^2\\\Rightarrow \alpha=\dfrac{2}{2^2}(8-1.25\times 2)\\\Rightarrow \alpha=2.75\ rad/s^2[/tex]
The angular acceleration of the rotor is [tex]2.75\ rad/s^2[/tex]
Every day a certain amount of water evaporates from Earth’s oceans, lakes, and land surface and forms water vapor and clouds in the atmosphere. Every day a certain amount of rain falls back to Earth. Make the reasonable assumption that, on average, the energy absorbed by the evaporation and lifting of the water is equal to the energy released by its condensation and falling back to earth. The evaporation of one mole of water requires approximately 40.6×103 J of heat. An equal amount of heat is released when a mole of water condenses.
QUESTION: The annual volume of rainfall on Earth is approximately 4.9×105 km3 and the average cloud altitude is 8.8 km above Earth’s surface. How much energy, in joules, is required every day to evaporate and lift the water?
Answer:
0.0000000010 joules
Explanation:
Amount : 1540.3 nanojoules (nJ)
Equals : 0.0 joules (J)
The total amount of energy required everyday to evaporate and lift the water is 1.15 × 10¹⁸ J
Since one mole of water requires 40.6 × 10³ J of heat to evaporate, and the volume of annual rainfall is 4.9 × 10⁵ km³ = 4.9 × 10⁸ m³, we need to find the amount of heat required to evaporate this volume of rainfall.
Amount of heat required to evaporate this volume of rainfall.So, E = nH where
n = number of moles of water in rainfall = m/M where m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = 4.9 × 10⁸ m³ and M = molar mass of water = 18 g/mol = 1.8 × 10⁻² kg/mol and H = energy required to evaporate one mole of water = 40.6 × 10³ J/molSo, E = nH
E = mH/M
E = ρVH/M
Substituting the values of the variables into the equation, we have
E = ρVH/M
E = 1000 kg/m³ × 4.9 × 10⁸ m³ × 40.6 × 10³ J/mol/ 1.8 × 10⁻² kg/mol
E = 198.94 × 10¹⁴ J-kg/mol/ 1.8 × 10⁻² kg/mol
E = 110.52 × 10¹⁶ J
E = 1.1052 × 10¹⁸ J
The amount of energy required to raise this volume to an altitude of 8.8 km
Now, the amount of energy required to raise this volume to an altitude of 8.8 km is E' = mgh where
m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = 4.9 × 10⁸ m³, g = acceleration due to gravity = 9.8 m/s² and h = average cloud altiude = 8.8 km = 8.8 × 10³ mSo, E' = mgh
E' = ρVgh
Substituting the values of the variables into the equation, we have
E' = ρVgh
E' = 1000 kg/m³ × 4.9 × 10⁸ m³ × 9.8 m/s² × 8.8 × 10³ m
E' = 422.576 × 10¹⁴ J
E' = 4.22576 × 10¹⁶ J
The total amount of energy requiredSo, the total energy required to evaporate and raise the volume of water is E" = E + E'
E" = 110.52 × 10¹⁶ J + 4.22576 × 10¹⁶ J
E" = 114.74576 × 10¹⁶ J
E" = 1.1474576 × 10¹⁸ J
E" ≅ 1.15 × 10¹⁸ J
The total amount of energy required everyday to evaporate and lift the water is 1.15 × 10¹⁸ J
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A 3.0-µF capacitor charged to 40 V and a 5.0-µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 3.0-µF capacitor?
The final charge on the 3.0-µF capacitor is 90 µC.
Explanation:To find the final charge on the 3.0-µF capacitor, we need to use the concept of charge conservation in a series circuit. In a series circuit, the charge across capacitors connected in series is the same. Therefore, the final charge on the 3.0-µF capacitor will be the same as the charge on the 5.0-µF capacitor.
Using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage, we can calculate the charge on the 5.0-µF capacitor.
Q = (5.0 µF)(18 V) = 90 µC.
Therefore, the final charge on the 3.0-µF capacitor is also 90 µC.
A couple of soccer balls of equal mass are kicked off the ground at the same speed, butat different angles. Soccer ball A is kicked off at an angle slightly above the horizontal,whereas ball B is kicked slightly below the vertical.(a) How does the initial kinetic energy of ball A compare to the initial kinetic energyof ball B?(b) How does the change in gravitational potential energy from the ground to thehighest point for ball A compare to the change in gravitational potential energyfrom the ground to the highest point for ball B?(c) If the energy in part (a) di ers from the energy in part (b), explain why there is adi erence between the two energies.
Answer:
Remember that Kinetic energy is a scalar quantity and it only depends on the speed and not necessary not the angle
Thus,Since the masses and the speed are same for both A and B, the initial kinetic energy of A and B are same.
b]
The difference or variation in gravitational potential energy is again a scalar quantity and so as long as the initial speed is same, the change in gravitational potential energy will also be the same [though they may not occur at the same horizontal position].
therefore, from the ground to the highest point of both A and B, both will have same potential energy.
Also The energy in part (a) differs from part (b),
In part (a) energy mention is kinetic energy that depends on mass and velocity of particle whereas in part (b) energy is potential energy that depends on mass and the position with reference of ground. Potential energy is a state function but kinetic energy is not.
Answer: they are the same
Explanation:
Monochromatic light is incident on a pair of slits that are separated by 0.230 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.) (a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.57 cm, find the wavelength of the incident light
Answer:
The wavelength of incident light is [tex]1.38x10^{-6}m[/tex]
Explanation:
The physicist Thomas Young established, through his double slit experiment, a relation between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.
[tex]\Lambda x = L\frac{\lambda}{d} [/tex] (1)
Where [tex]\Lambda x[/tex] is the distance between two adjacent maxima, L is the distance of the screen from the slits, [tex]\lambda[/tex] is the wavelength and d is the separation between the slits.
The values for this particular case are:
[tex]L = 2.60m[/tex]
[tex]d = 0.230mm[/tex]
[tex]\Lambda x = 1.57cm[/tex]
Then, [tex]\lambda[/tex] can be isolated from equation 1
[tex]\lambda = \frac{d \Lambda x}{L}[/tex] (2)
However, before equation 2 can be used, it is necessary to express [tex]\Lambda x[/tex] and d in units of meters.
[tex]\Lambda x= 1.57cm \cdot \frac{1m}{100cm}[/tex] ⇒ [tex]0.0157m[/tex]
[tex]d = 0.230mm \cdot \frac{1m}{1000mm}[/tex] ⇒ [tex]2.3x10^{-4}m[/tex]
Finally, equation 2 can be used.
[tex]\lambda = \frac{(2.3x10^{-4}m)(0.0157m}{(2.60m)}[/tex]
[tex]\lambda = 1.38x10^{-6}m[/tex]
Hence, the wavelength of incident light is [tex]1.38x10^{-6}m[/tex]
Speculate about some worldwide changes likely to follow the advent of successful fusion reactors. Compare the advantages and disadvantages of electricity coming from a large central power station versus a network of many smaller solar-based stations owned and operated by individuals.
Answer:
Nuclear reaction is the disintegration or integration of nucleus or nuclei of an atom, to produce a different nuclei or nucleus, thereby given out a large amount of heat energy. The energy been extracted from this reaction are very much.
At a successful fusion reaction, the world will have more better opportunity to preserve it's natural resources, as they will be enough energy to produce electricity, cooking, industrial power, and many more. It will also reduce spending, as energy from fusion reaction are not too expensive in obtaining, when compared to energy from natural resources.
ADVANTAGES OF USING CENTRAL POWER STATION THAN SMALLER SOLAR BASED STATION.
1) Central power stations are reliable, because of the abundance of natural resources been used to produce the electricity. Solar based station are not reliable because during winter, when the sun is hardly seen, they will not be enough resource to produce electricity.
2) Central power stations are very efficient because they produce a stable electricity power, while the smaller solar stations are not efficient, because the amount of power produced is dependent on the amount of solar energy it has stored.
DISADVANTAGES OF USING CENTRAL POWER STATION THAN SMALLER SOLAR-BASED STATION.
1) Central power stations causes pollution, and green house effect, because the natural gas been used to produce the electricity causes pollution and green house effect during venting. While smaller solar-based station are pollution free and does not cause green house effect, because it only need sun to store it's charge.
2) Central power station are more dangerous for people to work, as risk of being electrocuted is high. Smaller-based solar stations has less risk, as the power been stored from the sun requires less complex wiring.
3) Central power stations are more expensive to produce and requires much land space. While smaller solar-based station are less expensive, and requires little land space.
Final answer:
Successful fusion reactors could dramatically reshape global politics and economies by reducing reliance on fossil fuels. Centralized fusion power offers consistency, whereas decentralized solar power fosters resilience and local control but is intermittent. Fusion faces technological challenges, while solar technology is readily expanding.
Explanation:
Worldwide Changes with Successful Fusion Reactors
The advent of successful fusion reactors would be a transformative event worldwide. Fusion promises a cleaner, nearly inexhaustible energy source compared to fossil fuels, which could significantly alter the current geopolitical landscape where oil and natural gas are dominant factors in international politics and economics. A shift toward fusion energy could lessen global dependence on fossil fuels, potentially reducing the geopolitical power of oil-rich nations and creating a new paradigm in the world economy. With hydrogen from water as fuel, countries may realign alliances and shift focus towards technological advancements and harnessing fusion energy efficiently.
Central vs. Distributed Power Generation
Electricity from a large central power station, such as a fusion reactor, offers economies of scale and a consistent power supply but requires a significant up-front investment and complex infrastructure. Meanwhile, smaller solar-based stations, which are often owned and operated by individuals, offer distributed generation, empowering local communities and enhancing energy resilience. However, solar energy faces challenges with intermittency and requires storage solutions to provide a reliable power supply.
The disadavantages of fusion, despite its potential, include the complexities and high costs associated with maintaining such a power source, as well as addressing issues related to radioactivity. On the other hand, the technology for solar power is already available today, with deployment expanding rapidly thanks to decreasing costs and improving battery storage technologies.
Challenges in Fusion Technology
Developing fusion reactors for electricity generation requires overcoming significant technical hurdles, such as achieving the necessary high temperatures and creating materials that can contain the fusion reaction without melting. Scientists are optimistic, yet the timeline for fusion reactors to become a commercial reality remains uncertain.
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is low, the penny rides up and down without difficulty. If the frequency is steadily increased, there comes a point at which the penny leaves the surface.A) At what point in the cycle does the penny first lose contact with the piston?midpoint moving uplowest pointhighest pointmidpoint moving downB) What is the maximum frequency for which the penny just barely remains in place for the full cycle?Express your answer with the appropriate units.
Answer:
the penny loses contact at the piston's highest point.
f = 2.5 Hz
Explanation:
Concepts and Principles
1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by
∑F = ma (1)
2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:
a_max = -w^2A (2)
3- The angular frequency w of a wave is related to the frequency f by:
w = 2π f (3)
Given Data
- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.
- The frequency of oscillation of the piston is steadily increased.
Required Data
In part (a), we are asked to determine the point at which the penny first loses contact with the piston.
In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.
Solution
(a)
The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.
figure 1 is attached
Apply Newton's second law from Equation (1) in the vertical direction to the penny:
∑F_y -mg= ma
Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So
0 = m(g + a)
a = —g
Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.
(b)
The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):
a = —w^2A
where a = —g at the highest point. So
g = w^2A
Solve for w:
w =√g/A
Substitute for w from Equation (3):
2πf = √g/A
Solve for f :
f = 1/2π√g/A
Substitute numerical values:
f = 1/2π√9.8 m/s^2/0.04
f = 2.5 Hz
The question is based on Simple Harmonic Motion: The penny first loses contact with the piston at its highest point (option c) in the motion. The maximum frequency for which the penny just barely remains in place for the full cycle is approximately 12.5 Hz.
Explanation:The penny will first lose contact with the piston at its highest point, which is option c) highest point in the cycle.
To determine the maximum frequency for which the penny just barely remains in place for the full cycle, we need to consider the condition for the penny to stay in contact with the piston. At the highest point of the motion, the penny experiences an upward gravitational force and a downward centripetal force due to the circular motion.
When the centripetal force becomes greater than or equal to the gravitational force, the penny will lose contact with the piston. This occurs when
mvmax2/r = mg
Solving for the maximum velocity using vmax = 2πfA, where A is the amplitude, and substituting into the equation above, we find that
fmax = g / (4π2A)
Substituting the given values, with a=4.0 cm = 0.04 m and g=9.8 m/s2:
fmax = 9.8 / (4π2(0.04)) ≈ 12.5 Hz
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A firework shell is launched vertically upward from the ground with an initial speed of 44m/s. when the shell is 65 m high on the way up it explodes into two wequal mass halves, one half is observed to continue to rise straight up to a heigh of 120 m. How high does the other half go?
Answer:
[tex]h = 83.093\,m[/tex]
Explanation:
The speed of the firework shell just before the explosion is:
[tex]v = \sqrt{(44\,\frac{m}{s})^{2}-2\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (65\,m)}[/tex]
[tex]v \approx 25.712\,\frac{m}{s}[/tex]
After the explosion, the initial speed of one of the mass halves is:
[tex]v_{f}^{2} = v_{o}^{2} -2\cdot g \cdot s[/tex]
[tex]v_{o}^{2} = v_{f}^{2} + 2\cdot g \cdot s[/tex]
[tex]v_{o} = \sqrt{v_{f}^{2}+2\cdot g \cdot s}[/tex]
[tex]v_{o} = \sqrt{\left(0\,\frac{m}{s}\right)^{2}+ 2 \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (120\,m-65\,m)}[/tex]
[tex]v_{o} \approx 32.845\,\frac{m}{s}[/tex]
The initial speed of the other mass half is determined from the Principle of Momentum Conservation:
[tex]m \cdot (25.712\,\frac{m}{s} ) = 0.5\cdot m \cdot (32.845\,\frac{m}{s} ) + 0.5\cdot m \cdot v[/tex]
[tex]25.842\,\frac{m}{s} = 16.423\,\frac{m}{s} + 0.5\cdot v[/tex]
[tex]v = 18.838\,\frac{m}{s}[/tex]
The height reached by this half is:
[tex]h = h_{o} -\frac{v_{f}^{2}-v_{o}^{2}}{2\cdot g}[/tex]
[tex]h = 65\,m - \frac{(0\,\frac{m}{s} )^{2}- (18.838\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]
[tex]h = 83.093\,m[/tex]
Answer:
The other half goes 17.4m high
Explanation:
Pls see calculation in the attached file
Electrons are ejected from a metallic surface with speeds of up to 4.60x105 m/s when light with a wave length of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface? (
Answer:
The solution to the question above is explained below:
Explanation:
(a) The work function of the surface is:
The work function of a metal, Φ, it's the minimum amount of energy required to remove electron from the conduction band and remove it to outside the metal. It is typically exhibited in units of eV (electron volts) or J (Joules). Work Function, is the minimum thermodynamic work.
hf = hc /λ = φ + Kmax = φ + 1 /2 mev ²max
where, φ=work function of a metal
eV=electron volts
J= Joules
λ=the Plank constant 6.63 x 10∧-34 J s
f = the frequency of the incident light in hertz
∧= signifies raised to the power in the solution
Kmax= the maximum kinetic energy of the emitted electrons in joules
φ= ( hc /λ -1/2mev ²max= 6.63 × 10∧-34Js × 3.00 × 10∧8 m/s ÷ 625 × 10∧-9) -
(1/2 × 9.11 × 10∧-31 kg × 4.60 × 10∧5 m/s) = 2.21 × 10∧-19 J
ans= 1.38 eV
(b) The cutoff frequency for this surface is:
The cutoff frequency is the minimum frequency that is required for the emission of electrons from a metallic surface at which energy flowing through the metallic surface begins to be reduced rather than passing through.
Light at the cutoff frequency only barely supplies enough energy to overcome the work function. The value of the cutoff frequency is in unit hertz (Hz)
hfcut = φ
fcut = φ /h
ans= 334 THz
When a golfer tees off, the head of her golf club which has a mass of 152 g is traveling 44.8 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 27.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answer:
56.5 m/s²
Explanation:
From the law of conservation of momentum,
mu+m'u' = mv+m'v'........................ Equation 1
Where m = mass of the golf club, u = initial velocity of the golf club, m' = mass of the golf ball, u' = initial velocity of the golf ball, v = final velocity of the golf club, v' = final velocity of the golf ball.
From the question,
The golf ball is at rest, Hence u' = 0 m/s
mu = mv+m'v'
Make v' the subject of the equation
v' = (mu-mv)/m'........................... Equation 2
Given: m = 152 g = 0.152 kg, u = 44.8 m/s, v = 27.7 m/s, m' = 46 g = 0.046 kg.
Substitute into equation 2
v' = (0.152×44.8+0.152×27.7)/0.046
v' = (6.8096-4.2104)/0.046
v' = 2.5992/0.046
v' = 56.5 m/s²
A howler monkey is the loudest land animal and under some circumstances, can be heard up to a distance of 5.0km. Assume the acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0*10^-12 W/m^2. The acoustic power emitted by the howler is clostest to:
A) 0.31mW
B) 1.1mW
C) 3.2mW
D) 11mW
Answer:
Power emitted will be 0.314 mW
So option (A) will be correct option
Explanation:
We have given threshold hearing [tex]I=10^{-12}W/m^2[/tex]
Distance is given r = 5 km =5000 m
We have to find the power emitted
Power emitted is equal to
[tex]P=I\times A[/tex]
[tex]=10^{-12}\times 4\pi r^2[/tex]
[tex]=10^{-12}\times 4\times 3.14\times (5000)^2[/tex]
=[tex]314\times 10^{-6}watt=0.314mW[/tex]
So power emitted will be 0.314 mW
So option (A) will be correct option.
A very long uniform line of charge has charge per unit length λ1 = 4.80 μC/m and lies along the x-axis. A second long uniform line of charge has charge per unit length λ2 = -2.26 μC/m and is parallel to the x-axis at y1 = 0.400 m . (Hint: You will need to know or derive using Gauss law the electric field by a uniform line charge)
Answer:
a) E=228391.8 N/C
b) E=-59345.91N/C
Explanation:
You can use Gauss law to find the net electric field produced by both line of charges.
[tex]\int \vec{E_1}\cdot d\vec{r}=\frac{\lambda_1}{\epsilon_o}\\\\E_1(2\pi r)=\frac{\lambda_1}{\epsilon_o}\\\\E_1=\frac{\lambda_1}{2\pi \epsilon_o r_1}\\\\\int \vec{E_2}\cdot d\vec{r}=\frac{\lambda_2}{\epsilon_o}\\\\E_2=\frac{\lambda_2}{2\pi \epsilon_o r_2}[/tex]
Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.
The net electric field at point r will be:
[tex]E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})[/tex]
a) for y=0.200m, r1=0.200m and r2=0.200m:
[tex]E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C[/tex]
b) for y=0.600m, r1=0.600m, r2=0.200m:
[tex]E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C[/tex]
a) The electric field strength for case 1 will be 228391.8 N/C
b) The electric field strength for case 2 will be 59345.91 N/C
What is gauss law?The total electric flux out of a closed surface is equal to the charge contained divided by the permittivity,
According to Gauss Law. The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane perpendicular to the field.
The given data in the problem is;
λ₁ is the charge per unit length for charge 1 = 4.80 μC/m
λ₂ is the charge per unit length for charge 2 =-2.26 μC/m
y₁ is the distance from charge 1 =0.400 m.
From the gauss law, the net electric field is produced by both lines of charges will be .;
[tex]\rm E= \frac{\lambda_}{2 \pi \epsilon_0 r_2} \\\\ \rm E_1= \frac{\lambda_1}{2 \pi \epsilon_0 r_1} \\\\ \rm E_2= \frac{\lambda_2}{2 \pi \epsilon_0 r_2}[/tex]
The total electric field will be the sum of the electric field for the charge 1 and2;
[tex]\rm E=E_1+E_2 \\\\ \rm E=\frac{\lambda_1}{2 \pi \epsilon_0 r_1}+\frac{\lambda_2}{2 \pi \epsilon_0 r_2} \\\\[/tex]
The net electric field for y=0.200m, r₁=0.200m, and r₂=0.200 m will be;
[tex]\rm E=\frac{4.80 \times 10^{-6}}{2 \times 3.14 8.85 \times 10^{-12} \times 0.200}-\frac{2.26 \times 10^{-6}}{2 \times 3.14 \times 8.85 \times 10^{-12}} \\\\\rm E=228391.8 \ N/C[/tex]
The net electric field for y=0.600m, r₁=0.600m, r₂=0.200 m will be;
[tex]\rm E=\frac{4.80 \times 10^{-6}}{2 \times 3.14 8.85 \times 10^{-12} \times 0.600}-\frac{2.26 \times 10^{-6}}{2 \times 3.14 \times 8.85 \times 10^{-12}} \\\\\rm E=-59345.51 \ N/C[/tex]
Hence the electric field strength for case 1 and case 2 will be 228391.8 N/C and 59345.91 N/C respectively.
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A 50-cm-long spring is suspended from the ceiling. A 330g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 28cm before coming to rest at its lowest point. It then continues to oscillate vertically.
A. What is the spring constant?
B. What is the amplitude of the oscillation?
C. What is the frequency of the oscillation?
Explanation:
Given that,
Length of the spring, l = 50 cm
Mass, m = 330 g = 0.33 kg
(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :
[tex]kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.14}\\\\k=23.1\ N/m[/tex]
(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.
(C) The frequency of the oscillation is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz[/tex]
A coil formed by wrapping 65 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When the magnetic field is increased uniformly from 200 µT to 600 µT in 0.400 s, an emf of magnitude 80.0 mV is induced in the coil. What is the total length of the wire?
Final answer:
To find the total length of the wire in a 65-turn square coil subjected to a changing magnetic field, apply Faraday's law of induction to calculate the magnetic flux change and then determine the side length of the coil. Multiply the coil's perimeter by the number of turns to obtain the total length.
Explanation:
The student's question pertains to electromagnetic induction in a coil exposed to a changing magnetic field. To solve for the total length of the wire, we can use Faraday's law of induction, which states that the induced emf (electromotive force) in a coil is proportional to the rate of change of magnetic flux through the coil. Given an induced emf of 80.0 mV and a change in magnetic field from 200 µT to 600 µT over 0.400 s, the change in magnetic flux φ can be calculated.
Since the coil is square and positioned at a 30.0° angle to the magnetic field, the effective area A for inducing emf can be determined using the cosine of the angle and the side length of the square, assuming all sides are equal. With the number of turns (N) being 65, we can apply Faraday's law to find the magnetic flux and subsequently the side length of the square. The total length of the wire is simply the perimeter of the square (4 times the side length) multiplied by the number of turns (N).
The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional boxer exerts a force of 1.8 × 103 N with an effective perpendicular lever arm of 2.85 cm, producing an angular acceleration of the forearm of 140 rad/s2.
What is the moment of inertia of the boxer's forearm?
Answer:
0.366kgm²
Explanation:
F = 1.8*10³N
r = 2.85cm = 0.0285m
α = 140rad/s²
Torque = applied force * distance
τ = r * F
τ = 0.0285 * 1.8*10³
τ = 51.3N.m
but τ = I * α
I = τ / α
I = 51.3 / 140
I = 0.366kgm²
Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 110 years into the future. According to special relativity, how fast must you travel
Answer:
0.999958c
Explanation:
Remember that the trip involves traveling for six months at a constant velocity thus returning home at same speed . The time interval of this trip will therefore be one year.
Accurate time interval to be measured by the clock on the spaceship Δt0 = 1.0 years
Time interval as advanced on earth observed from the spacecraft Δt = 110 years
the formula for time dilation is
Δt = Δt0 /√(1-v2/c2)
or v = c*√1-(Δt0/Δt)2
=c*√[1-(1year/110years)2]
= 0.999958c
A charged particle moving along the x-axis enters a uniform magnetic field pointing along the z-axis. Because of an electric field along the y-axis, the charge particle does not change velocity. What is the sign of this particle
Answer:
Explanation:
Let the charge particle have charge equal to +q .
force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along - y axis. ( negative of y axis )
force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.
F = q ( v i x B k ) , ( velocity is along x direction and magnetic field is along z axis. )
= (Bqv) - j
= - Bqv j
The force will be along - ve y - direction .
If we take charge as negative or - q
force due to electric field will be along - y axis .
magnetic force = F = -q ( v i x B k )
= + Bqv j
magnetic force will be along + y axis
So it is difficult to find out the nature of charge on the particle from this experiment.
Please select that whether below statements are correct or not?
1. Resistivity is the property that measures a material's ability to provide "obstacles" to the flow of electrons caused by an external electric field. Such a flow of electrons is called an electric current. Resistivity rho is defined as the ratio of the magnitude of the electric field E to the magnitude of the current density J: rho=EJ.
2. Resistance is a measure of an object's ability to provide "obstacles" to electric current. The resistance R of a conductor (often, a metal wire of some sort) is defined as the ratio of the voltage V between the ends of the wire to the current I through the wire: R=VI.
3. Ohm's law is not a fundamental law of physics; it is valid under certain conditions (mostly, metal conductors in a narrow range of temperatures). Still, Ohm's law is a very useful tool, since many circuits operate under these conditions.
Answer:
Statements 1, 2 and 3 are all correct.
A person stands in an elevator
weighing a Cheeseburger with a
kitchen scale « lit could happent
The mass of the cheese burger at
10 ilsoka The scale reads l-ION
Determine the magnitude ana
Lirecteon of the netforce on the
Cheeseburger
Answer:
Noennt ied olod l qeu es ocmo a balazuna de cozcna
in
Explanation:
A new concrete mix is being designed to provide adequate compressive strength for concrete blocks. The specification for a particular application calls for the blocks to have a mean compressive strength µ greater than 1350 kPa. A sample of 100 blocks is produced and tested. Their mean compressive strength is 1356 kPa and their standard deviation is 70 kP
Complete question:
new concrete mix is being designed to provide adequate compressive strength for concrete blocks. The specification for a particular application calls for the blocks to have a mean compressive strength µ greater than 1350 kPa. A sample of 100 blocks is produced and tested. Their mean compressive strength is 1356 kPa and their standard deviation is 70 kP
a) Find the p value.
b) Do you believe it is plausible that the blocks do not meet the specification, or are you convinced that they do? Expain your reasoning.
Answer:
a) p-value= 0.1949
b) It is possible the blocks do not meet the specifications
Explanation:
Given:
n = 100
Sample mean, X' = 1356
Standard deviation, s.d = 70
a) To find p- value.
Null hypothesis:
H0: u ≤ 1350
Alternative hypothesis:
H1 : u > 1350
The test statistic wll be:
[tex] z= \frac{X'-u_o}{s.d/ \sqrt{n}}[/tex]
[tex] = \frac{1356-1350}{70/\sqrt{100}}[/tex]
=0.86
The p value will be:
= P(z>0.86)
= 1-P(z≤0.86)
Using the normal distribution table, we now have:
1 - 0.8051
= 0.1949
P value = 0.1949
b) Since our p-value is 0.1949, we do not reject the null hypothesis, because the p-value, 0.1949, is not small. This means that it is possible the blocks do not meet the specifications.
The probability that the compressive strength is greater than 1350 kPa is 80.51%
Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\Where\ x=raw\ score,\mu=mean, \sigma=standard\ deviation, n=sample\ size[/tex]
Given that n = 100, μ = 1356, σ = 70. For x > 1350 kPa:
[tex]z=\frac{1350-1356}{70/\sqrt{100} } =-0.86[/tex]
From the normal distribution table, P(x > 1350) = P(z > -0.86) = 1 - P(z < -0.86) = 1 - 0.1949 = 80.51%
Hence the probability that the compressive strength is greater than 1350 kPa is 80.51%
Find out more at: https://brainly.com/question/15016913
Why do you suffer more pain than me when I slap your face?
cylinder of mass 6.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 15.0 m/s. (a) Determine the translational kinetic energy of its center of mass. J (b) Determine the rotational kinetic energy about its center of mass. J (c) Determine its total energy.
Answer:
a). 675J
b). 337.5J
c). 1012.5J
Explanation:
M = 6.0kg
V = 15.0m/s
a). Translational energy
E = ½ *mv²
E = ½ * 6 * 15²
E = 675J
b). Rotational kinetic energy K.E(rot) = Iw²
But moment of inertia of a cylinder (I) = ½Mr²
I = ½mr²
V = wr, r = v / w
K.E(rot) = ¼ mv²
K.E(rot) = ¼* 6 * 15²
K.E(rot) = 337.5J
Total energy of the system = K.E(rot) + Translational energy = 337.5 + 675
T.E = 1012.5J
Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.40. The thickness t of the air layer can be varied. Light with wavelength λ in air is at normal incidence onto the top of the air film. There is constructive interference between the light reflected at the top and bottom surfaces of the air film when its thickness is 700 nm. For the same wavelength of light the next larger thickness for which there is constructive interference is 980 nm.
a. What is the wavelength λ of the light when it is traveling in air?
b. What is the smallest thickness t of the air film for which there is constructive interference for this wavelength of light?
Answer:
the wavelength λ of the light when it is traveling in air = 560 nm
the smallest thickness t of the air film = 140 nm
Explanation:
From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)
Now for constructive interference;
Δx= [tex](m+ \frac{1}{2} \lambda)[/tex]
replacing ;
Δx = 2t ; we have:
2t = [tex](m+ \frac{1}{2} \lambda)[/tex]
Given that thickness t = 700 nm
Then
2× 700 = [tex](m+ \frac{1}{2} \lambda)[/tex] --- equation (1)
For thickness t = 980 nm that is next to constructive interference
2× 980 = [tex](m+ \frac{1}{2} \lambda)[/tex] ----- equation (2)
Equating the difference of equation (2) and equation (1); we have:'
λ = (2 × 980) - ( 2× 700 )
λ = 1960 - 1400
λ = 560 nm
Thus; the wavelength λ of the light when it is traveling in air = 560 nm
b)
For the smallest thickness [tex]t_{min} ; \ \ \ m =0[/tex]
∴ [tex]2t_{min} =\frac{\lambda}{2}[/tex]
[tex]t_{min} =\frac{\lambda}{4}[/tex]
[tex]t_{min} =\frac{560}{4}[/tex]
[tex]t_{min} =140 \ \ nm[/tex]
Thus, the smallest thickness t of the air film = 140 nm
Answer:
1.4x10^7m & 98nm
Explanation:
Pls the calculation is in the attached file
In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way in which this is done is to seat them in a chair of known mass attached to a spring of known force constant and measure the period of the oscillations of this system. The 36.4 kg chair alone oscillates with a period of 1.00 s, and the period with the astronaut sitting in the chair is 2.20 s.
Find the force constant of the spring.
Answer:
Approximately [tex]1.44\times 10^3 \; \rm N \cdot m^{-1}[/tex] assuming that the spring has zero mass.
Explanation:
Without any external force, a piece of mass connected to an ideal spring (like the chair in this question) will undergo simple harmonic oscillation.
On the other hand, the force constant of a spring (i.e., its stiffness) can be found using Hooke's Law. If the spring exerts a restoring force [tex]\mathbf{F}[/tex] when its displacement is [tex]\mathbf{x}[/tex], then its force constant would be:
[tex]\displaystyle k = -\frac{\mathbf{F}}{\mathbf{x}}[/tex].
The goal here is to find the expressions for [tex]F[/tex] and for [tex]x[/tex]. By Hooke's Law, the spring constant would be ratio of these two expressions.
Let [tex]T[/tex] represent the time period of this oscillation. With the chair alone, the period of oscillation is [tex]T = 1.00\; \rm s[/tex].
For a simple harmonic oscillation, the angular frequency [tex]\omega[/tex] can be found from the period:
[tex]\displaystyle \omega = \frac{2\pi}{T}[/tex].
Let [tex]A[/tex] stands for the amplitude of this oscillation. In a simple harmonic oscillation, both [tex]\mathbf{F}[/tex] and [tex]\mathbf{x}[/tex] are proportional to [tex]A[/tex]. Keep in mind that the spring constant [tex]k[/tex] is simply the opposite of the ratio between [tex]\mathbf{F}[/tex] and [tex]\mathbf{x}[/tex]. Therefore, the exact value of [tex]A[/tex] shouldn't really affect the value of the spring constant.
In a simple harmonic motion (one that starts with maximum displacement and zero velocity,) the displacement (from equilibrium position) at time [tex]t[/tex] would be:
[tex]\displaystyle \mathbf{x}(t) = A \cos(\omega \cdot t)[/tex].
The restoring velocity at time [tex]t[/tex] would be:
[tex]\displaystyle \mathbf{v}(t) = \mathbf{x}^\prime(t) = -A\, \omega \sin(\omega\cdot t)[/tex].
The restoring acceleration at time [tex]t[/tex] would be:
[tex]\displaystyle \mathbf{a}(t) = \mathbf{v}^\prime(t) = -A\, \omega^2 \cos(\omega\cdot t)[/tex].
Assume that the spring has zero mass. By Newton's Second Law of motion, the restoring force at time [tex]t[/tex] would be:
[tex]\begin{aligned}& \mathbf{F}(t) \\ &= m(\text{chair}) \cdot \mathbf{a}(t) \\&= -m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)\end{aligned}[/tex].
Apply Hooke's Law to find the spring constant, [tex]k[/tex]:
[tex]\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= -\left(\frac{-m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)}{A\cos(\omega \cdot t)}\right) \\ &= \omega^2 \cdot m(\text{chair}) \end{aligned}[/tex].
Again, [tex]\omega[/tex] stands for the angular frequency of this oscillation, where
[tex]\displaystyle \omega = \frac{2\pi}{T}[/tex].
Before proceeding, note how [tex]A[/tex] was eliminated from the ratio (as expected.) Additionally, [tex]t[/tex] is also eliminated from the ratio. In other words, the spring constant is "constant" at all time. That agrees with the assumption that this spring is indeed ideal. Back to [tex]k[/tex]:
[tex]\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= \cdots \\ &= \omega^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{T}\right)^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{1.00\; \rm s}\right)^2 \times 36.4\; \rm kg\end{aligned}[/tex].
Side note on the unit of [tex]k[/tex]:
[tex]\begin{aligned} & 1\; \rm kg \cdot s^{-2} \\ &= 1\rm \; \left(kg \cdot m \cdot s^{-2}\right) \cdot m^{-1} \\ &= 1\; \rm N \cdot m^{-1}\end{aligned}[/tex].
Which of the following statements are true concerning the reflection of light?
Check all that apply.
a. The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.
b. The reflection of light from a smooth surface is called specular reflection.
c. The reflection of light from a rough surface is called diffuse reflection.
d. For diffuse reflection, the angle of incidence is greater than the angle of reflection.
e. For specular reflection, the angle of incidence is less than the angle of reflection.
Answer:
b. The reflection of light from a smooth surface is called specular reflection.
c. The reflection of light from a rough surface is called diffuse reflection.
Explanation:
a. The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.
This is wrong: Based on law of reflection "The angle of incidence is equal to the angle of reflection when light strikes any plane surface" examples plane mirrors, still waters, plane tables, etc
b. The reflection of light from a smooth surface is called specular reflection.
This is correct
c. The reflection of light from a rough surface is called diffuse reflection.
This is correct
d. For diffuse reflection, the angle of incidence is greater than the angle of reflection.
This is wrong: the angle of incident is equal to angle of reflection. The only difference between this type of reflection and specular reflection, is that the normal for diffuse reflection is not parallel to each due to the rough surface in which the light incidents.
For specular reflection, the angle of incidence is less than the angle of reflection.
This is wrong: the angle of incident is equal to angle of reflection
The correct statements are statement 2 and statement 3.The true statements are: the reflection of light from a smooth surface is called specular reflection, and the reflection of light from a rough surface is called diffuse reflection. The others are incorrect as the angle of incidence always equals the angle of reflection.
To answer the question about the reflection of light, let's analyze each statement:
The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.In summary, the true statements concerning the reflection of light are:
The reflection of light from a smooth surface is called specular reflection.The reflection of light from a rough surface is called diffuse reflection.
A turntable is off and is not spinning. A 0.8 g ant is on the disc and is 9 cm away from the center. The turntable is turned on and 0.8 s later it has an angular speed of 45 rpm. Assume the angular acceleration is constant and determine the quantities for the ant 0.4 s after the turntable has been turned on. Express all quantities using appropriate mks units.
Answer:
I] alpha = wf/t = (45*2pi/60)/1 = 4.712 rad/s
ii] W = alpha*t = 2.356 rad/s
iii] v = Wr =2.356*0.10 = 0.2356 m/s
iv] atan = alpha*r = 0.4712 m/s^2
v] arad= v^2/r = 0.2356^2/0.10 = 0.555 m/s^2
vi] a = sqrt(0.4712^2+0.555^2) = 0.728 m/s^2
vii] Fnet = ma = 1.6e-3*0.728 = 0.00116 N
Explanation:
Two children are playing with a roll of paper towels. One child holds the roll between the index fingers of her hands so that it is free to rotate, and the second child pulls at constant speed on the free end of the paper towels. As the child pulls the paper towels, the radius of the roll of remaining towels decreases.
a)How does the torque on the roll change with time?
increases
decreases
stays the same
b)How does the angular speed of the roll change in time?
increases
decreases
stays the same
Answer:
a )Decreases
b ) Increases
c.) It's likely to break
Explanation:
a )The torque is as a result of friction, and also the magnitude is dependent on the normal force that exist between the surface in contact. The torque decrease because there is decrease in frictional force as the weight of the roll decrease
b.)The angular velocity will increase as radius o f the roll decrease because it is operating at constant speed.
c)if the child suddenly hit the roll, the papper will break, because it's angular acceleration might not be able to move along with the papper. Situation like this occur when there is large radius
A cosmic-ray proton in interstellar space has an energy of 19.5 MeV and executes a circular orbit having a radius equal to that of Mars' orbit around the Sun (2.28 1011 m). What is the magnetic field in that region of space?
Answer:
B = (2.80 × 10⁻¹²) T
Explanation:
First of, we convert the 19.5 MeV to Joules
19.5 MeV = 19.5 × 10⁶ × 1.602 × 10⁻¹⁹
= (3.124 × 10⁻¹²) J
The velocity of the cosmic-ray proton can be calculated from the kinetic energy formula
E = (1/2) mv²
m = mass of a proton = (1.67 × 10⁻²⁷) kg
(3.124 × 10⁻¹²) = (1/2)(1.67 × 10⁻²⁷)v²
v = (6.117 × 10⁷) m/s
And since the magnetic force keeps the cosmic ray proton in uniform circular motion,
Magnetic force = Centripetal force keeping the proton in circular motion.
qvB = (mv²/r)
q = charge on a proton = (1.602 × 10⁻¹⁹) C
v = velocity of the proton = (6.117 × 10⁷) m/s
B = magnetic field = ?
r = radius of circular motion = (2.28 × 10¹¹) m
B = (mv/qr)
B = (1.67 × 10⁻²⁷ × 6.117 × 10⁷) ÷ (1.602 × 10⁻¹⁹ × 2.28 × 10¹¹)
B = (2.797 × 10⁻¹²) T
Hope this Helps!!!
Answer:
The magnetic field strength in that region of space is 2.7983 x 10⁻¹² T
Explanation:
Given;
Energy of the cosmic-ray proton, E = 19.5 MeV = 19.5 x 10⁶ x 1.6 x 10⁻¹⁹
E = 3.12 x 10⁻¹² J
Radius of the circular orbit, r = 2.28 x 10¹¹ m
Step 1:
determine the speed of cosmic-ray proton
Kinetic energy of the cosmic-ray proton;
K.E = ¹/₂mv²
[tex]v = \sqrt{\frac{2K.E}{m} }[/tex]
where;
m is mass of proton, m = 1.67 x 10⁻²⁷ kg
[tex]v = \sqrt{\frac{2*3.12*10^{-12}}{1.67*10^{-27}} } \\\\v = 6.1127*10^7 \ m/s[/tex]
Step 2:
determine the magnetic field strength in that region of space
magnetic force = centripetal force
[tex]qvB = \frac{mv^2}{r} \\\\B = \frac{mv}{rq}\\\\[/tex]
Where;
q is charge of proton, q = 1.6 x 10⁻¹⁹ C
[tex]B = \frac{mv}{rq} = \frac{(1.67*10^{-27})(6.1127*10^7)}{(2.28*10^{11})(1.6*10^{-19})}\\\\B =2.7983 *10^{-12} \ T[/tex]
Therefore, the magnetic field strength in that region of space is 2.7983 x 10⁻¹² T
The earth’s magnetic field points toward (magnetic) north. For simplicity, assume that the field has no vertical component (as is the case near the earth’s equator). (a) If you hold a metal rod in your hand and walk toward the east, how should you orient the rod to get the maximum motional emf between its ends? (i) East-west; (ii) north-south; (iii) up-down; (iv) you get the same motional emf with all of these orientations. (b) How should you hold it to get zero emf as you walk toward the east? (i) East-west; (ii) north-south; (iii) up-down; (iv) none of these. (c) In which direction should you travel so that the motional emf across the rod is zero no matter how the rod is oriented? (i) West; (ii) north; (iii) south; (iv) straight up; (v) straight down. ❙
Answer:
A. Up-down
B. East-west & north -south
C. North or south
Explanation:
See attached handwritten document for more details
A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is given an additional 6-in displacement in the positive direction and then released. The mass is in a medium that exerts a viscous resistance of 6 lb when the mass has a velocity of 3 ft/s. Under the assumptions discussed in this section, formulate the initial value problem that governs the motion of the mass.
Answer:
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
[tex]my'' + \zeta y' + ky=0[/tex]
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
[tex]w - kL=0[/tex]
[tex]mg - kL=0[/tex]
[tex]4 - k\frac{1}{6}=0[/tex]
[tex]k = 4\times 6 =24[/tex]
The damping constant can be found by
[tex]F = -\zeta y'[/tex]
[tex]6 = 3\zeta[/tex]
[tex]\zeta = \frac{6}{3} = 2[/tex]
Finally, the mass m can be found by
[tex]w = 4[/tex]
[tex]mg=4[/tex]
[tex]m = \frac{4}{g}[/tex]
Where g is approximately 32 ft/s²
[tex]m = \frac{4}{32} = \frac{1}{8}[/tex]
Therefore, the required differential equation is
[tex]my'' + \zeta y' + ky=0[/tex]
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
The initial position is
[tex]y(0) = \frac{1}{2}[/tex]
The initial velocity is
[tex]y'(0) = 0[/tex]
We formulated the initial value problem for a damped spring-mass system:
1/8 * d²x/dt² + 2 * dx/dt + 24 * x = 0
with initial conditions x(0) = 0.5 ft and dx/dt(0) = 0 ft/s.
Let's break down the problem with the given data:
The mass weighs 4 lb.The spring stretches 2 inches under this weight.Additional displacement given is 6 inches in the positive direction.Viscous resistance is 6 lb when the velocity is 3 ft/s.First, find the spring constant k:
The weight of the mass stretches the spring by 2 inches (0.1667 feet).
The force exerted by the weight = 4 lb = mg
The spring force F = kx
So,
k = F/x k= 4 lb / 0.1667 ft k ≈ 24 lb/ftThe general form of the second-order differential equation governing the motion of the spring-mass-damper system is:
m*d²x/dt² + c*dx/dt + k*x = 0The viscous resistance given is 6 lb at 3 ft/s. Therefore, the damping coefficient c:
c = 6 lb / 3 ft/s c = 2 lb·s/ftThe initial conditions are displacement 6 inches (0.5 feet) and initial velocity 0:
x(0) = 0.5 ftCombining these elements, the initial value problem is:
1/8 * d²x/dt² + 2 * dx/dt + 24 * x = 0