Answer:
The new angular speed is 2/3omega
Explanation:
See attached handwritten document for more details
Answer:
2ω/3
Explanation:
The moments of inertia of the 1st disk is:
[tex]I_1 = (2MR^2)/2 = MR^2[/tex]
The moments of inertia of the 2nd disk is:
[tex]I_2 = MR^2/2[/tex]
So the total moments of inertia of the system of 2 disks after the drop is:
[tex]I = I_1 + I_2 = MR^2 + MR^2/2 = 1.5MR^2[/tex]
Using law of angular momentum conservation we have the following equation for before and after the drop
[tex]I_1\omega = I\omega_2[/tex]
[tex]\omega_2 = \omega\frac{I_1}{I} = \omega\frac{MR^2}{1.5MR^2} = 2\omega/3[/tex]
2. Turn off the Parallel line and turn on the Line through focal point. Move the light bulb around. What do you notice about the reflected ray in this situation? __________________________________________________
Answer:
The group of light rays is reflected back towards the focal point thereby producing a magnifying effect.
Explanation:
Organs are made up of vast numbers of cells that perform various tasks. 33 poi
When cells die within an organ, homeostasis is interrupted. What will
most likely happen so that homeostasis can be maintained? *
O
The organ will continue to work, but with decreased function.
O
The organ will immediately shut down so that no further damage will occur.
The remaining cells will be unable to divide and the organ will quit working.
0
The remaining cells will reproduce in order to replace the ones that have died.
Answer:
Option D
Explanation:
In order to maintain homeostasis, a function is to be performed that leads to functional normality of the body and its organs.
Here, the death of cells is affecting the homeostasis hence in order to restore homeostasis the remaining cells must divide at higher pace so that they can replenish the lost cells and their functions.
Hence, option D is correct
Final answer:
To maintain homeostasis after cell death in an organ, the remaining cells will typically reproduce to replace the ones that have died, restoring organ function and balance within the body.
Explanation:
When cells within an organ die, homeostasis may be disrupted. Homeostasis is the complex process by which organisms maintain a stable internal environment despite external changes. To maintain homeostasis after cell death, the remaining cells will attempt to compensate for the loss. Typically, these cells will reproduce and divide to replace the dead cells, thereby restoring the organ's function and maintaining the necessary balance within the body. This is a form of homeostatic regulation, which involves continuous adjustments in cellular activity to sustain a set point or a normal level of function.
What is the moment of inertia of a cube with mass M=0.500kg and side lengths s=0.030m about an axis which is both normal (perpendicular) to and through the center of a face of the cube? Watch your units. *Note that moment of inertia is a purely geometric property of a rigid body, and in lab we will call this the static moment of inertia. We will also experimentally determine a dynamic moment of inertia which will be measured from the rigid body's angular acceleration.
The moment of inertia of the given cube, which is the rotational equivalent of mass in linear motion, is 0.000075 kg.m^2. This is calculated by applying the formula I = (M*s^2)/6 with given parameters. The parallel axis theorem helps in finding the moment of inertia about any axis parallel to and a distance away from an object's center of mass.
Explanation:The moment of inertia of a cube about an axis normal to and through the center of one of its faces is calculated using the formula I = (M*s^2)/6, where M is mass, and s is the side length of the cube. In this case, M = 0.500 kg and s = 0.030 m. Therefore, the moment of inertia I = (0.500 kg * (0.030 m)^2)/6 = 0.000075 kg.m^2. Moment of inertia can be thought of as the rotational equivalent of mass for linear motion.
The moment of inertia of a cube, or any object, about any axis through its center of mass can be calculated using the parallel axis theorem. The theorem links the moment of inertia about an axis passing through the object's center of mass to the moment of inertia about any other parallel axis. It states that the moment of inertia about any axis parallel to and a distance d away from the axis through the center of mass is equal to the moment of inertia of the object about the axis through the center of mass plus the mass of the object times the square of the distance between the axes (I=I_cm + md^2). Therefore, the moment of inertia is also related to the distribution of mass within an object.
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In the 2016 Olympics in Rio, after the 50 m freestyle competition, a problem with the pool was found. In lane 1 there was a gentle 1.2 cm/s current flowing in the direction that the swimmers were going, while in lane 8 there was a current of the same speed but directed opposite to the swimmers' direction. Suppose a swimmer could swim the 50.0 m in 25.0 s in the absence of any current.
Part A: How would the time it took the swimmer to swim 50.0 m change in lane 1?
Part B: How would the time it took the swimmer to swim 50.0 m change in lane 8?
The time it would take for the swimmer to swim 50.0m would increase in lane 1, where there is a current flowing in the same direction of the swimmers. In lane 8, where there is a current flowing in the opposite direction, the time would decrease.
Explanation:In lane 1, where there is a 1.2 cm/s current flowing in the direction of the swimmers, the time it would take for the swimmer to swim 50.0 m would increase. This is because the current would push the swimmer backwards, making it harder for them to reach the finish line.
In lane 8, where there is a 1.2 cm/s current flowing in the opposite direction of the swimmers, the time it would take for the swimmer to swim 50.0 m would decrease. This is because the current would push the swimmer forward, helping them reach the finish line faster.
Two students suggest different functions for the battery plays a flashlight.
Student A: "The battery supplies the charge which allows the light bulb to light. When the battery no longer works, it is because it has run out of charge."
Student B: "The battery pumps the charge around the circuit. The charge itself comes from the conducting elements of the circuit; the battery supplies the energy to pump this charge. When the battery no longer works, it is because the energy-producing chemicals have been consumed."
(a) Which student (A or B) do you agree with?
Answer:
Student b
Explanation:
Because, Battery pumps the charges around the circuit. Here, charges are the free electrons. And this charges are already present in the conductor. When we connect this conductor with the battery - It set up an electric field and The diffrence between the potential between the ends of the wire pushes this free charges to move in perticular direction. (This is called electric energy)
This electric energy is converted by chemical reactions in the battery. When the battery is dead - It means that the chemical energy producing chemicals has been consumed. thereafter there will be no more chemical reactions takes place to produce electric energy. Hence we say that the battery is dead.
Answer:
Student B is correct
Explanation:
The battery acts like a pump to drive the electric charge. The electric charge comes from the conducting material but the battery provides a potential difference or a gradient of flow for this charge which is necessary for the movement of the charge along the conductor.
A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experimenter then briefly touches the electroscope with a finger. The finger is removed, followed by the removal of the positively charged object. What happens to the leaves of the electroscope when a negative charge is now brought near but not in contact with the top of the electroscope?
Answer:
The leaves of the electroscope move further apart.
Explanation:
This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.
Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.
So, the leaves move further apart.
The electroscope initially becomes positively charged, but is neutralized when touched. Once a negatively charged object is brought near, the leaves become negatively charged and repel each other.
Explanation:When a positively charged object is brought near an uncharged gold leaf electroscope, the free electrons in the electroscope are attracted towards the positive charge, leaving the leaves positively charged and causing them to repel each other.
When you touch the electroscope, you provided a grounding path which allows electrons to flow from the ground to the positively charged electroscope to neutralize it again, causing the leaves to fall back together.
After the positively charged object is removed, the electroscope remains neutral until a negatively charged object is brought near it. The electrons in the electroscope will be repelled towards the leaves, giving them a negative charge and causing them to repel again.
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the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes from the 5.49 MeV alpha particle emitted by Rn-222, how much energy is deposited in your body each year from radon. Approximately how many decays does this represent.
Answer:
The approximate number of decays this represent is [tex]N= 23*10^{10}[/tex]
Explanation:
From the question we are told that
The amount of Radiation received by an average american is [tex]I_a = 2.28 \ mSv[/tex]
The source of the radiation is [tex]S = 5.49 MeV \ alpha \ particle[/tex]
Generally
[tex]1 \ J/kg = 1000 mSv[/tex]
Therefore [tex]2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg[/tex]
Also [tex]1eV = 1.602 *10^{-19}J[/tex]
Therefore [tex]2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg[/tex]
An Average american weighs 88.7 kg
The total energy received is mathematically evaluated as
[tex]1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x[/tex]
Cross-multiplying and making x the subject
[tex]x = 88.7 * 1.423*10^{16} eV[/tex]
[tex]x = 126.2*10^{16}eV[/tex]
Therefore the total energy deposited is [tex]x = 126.2*10^{16}eV[/tex]
The approximate number of decays this represent is mathematically evaluated as
N = [tex]\frac{x}{S}[/tex]
Where n is the approximate number of decay
Substituting values
[tex]N = \frac{126 .2*10^{16}}{5.49*10^6}[/tex]
[tex]N= 23*10^{10}[/tex]
The energy deposited in the body from radon exposure and the number of decays it represents. It also discusses the significance of radon-222 as a source of radiation exposure.
The energy deposited in your body each year from radon can be calculated by converting the dose equivalent to energy. Given that the average American receives 2.28 mSv annually and all of it comes from a 5.49 MeV alpha particle emitted by Rn-222, the energy deposited would be 5.49 MeV.
To determine how many decays this represents, you divide the total energy deposited by the energy per decay. In this case, 5.49 MeV divided by 5.49 MeV per decay gives you 1 decay annually.
Radon-222 is a significant source of radiation exposure due to its radioactivity and ability to seep into homes from the ground. It highlights the importance of understanding the impact of radioactive elements on our environment.
In a live fire exercise, an Army artillery team fires an artillery shell from a howitzer. The barrel of the howitzer makes a 60.0° angle above horizontal, and the speed of the shell upon exiting the barrel is 380 m/s. The shell hits a target on the side of a mountain 32.0 s after firing. Assuming the point where the shell exits the barrel to be the origin, and assuming as usual that the x-axis is horizontal and the y-axis is vertical, find the x and y coordinates, in meters, of the target.
Answer:
Explanation:
speed of shell =380 m /s
x - component = 380 cos 60
= 190 m /s
y- component = 380 sin 60
= 329 .1 m /s
time taken to hit target = 32 s
horizontal distance covered = horizontal velocity x time
= 190 x 32 = 6080 m
vertical distance covered
h = ut - 1/2 gt²
= 329.1 x 32 - .5 x 9.8 x 32²
= 10531.2 - 5017.6
= 5513.6 m .
x coordinate = 6080 m
y-coordinate = 5513.6 m
M84, M87, and NGC 4258 all have accretion disks around their central black holes for which the rotational velocities have been measured in HST spectra. In M84, the disk extends 8 pc from the center and exhibits Doppler velocities as large as ±400 km/s with respect to the galaxy's overall radial velocity. In M87, the corresponding figures are 20 pc and 500 km/s. In NGC 4258, the figures are 0.5 ly and 1000 km/s. Calculate the black hole masses to two significant figures. Comment also on the assumptions under which you did your calculation (e.g., orbital plane viewed edge-on; note the appearance of the galaxies and their disks in the notes) and the effect this may have on the accuracy of the answers.
Answer:
For M84:
M = 590.7 * 10³⁶ kg
For M87:
M = 2307.46 * 10³⁶ kg
Explanation:
1 parsec, pc = 3.08 * 10¹⁶ m
The equation of the orbit speed can be used to calculate the doppler velocity:
[tex]v = \sqrt{\frac{GM}{r} }[/tex]
making m the subject of the formula in the equation above to calculate the mass of the black hole:
[tex]M = \frac{v^{2} r}{G}[/tex].............(1)
For M84:
r = 8 pc = 8 * 3.08 * 10¹⁶
r = 24.64 * 10¹⁶ m
v = 400 km/s = 4 * 10⁵ m/s
G = 6.674 * 10⁻¹¹ m³/kgs²
Substituting these values into equation (1)
[tex]M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }[/tex]
M = 590.7 * 10³⁶ kg
For M87:
r = 20 pc = 20 * 3.08 * 10¹⁶
r = 61.6* 10¹⁶ m
v = 500 km/s = 5 * 10⁵ m/s
G = 6.674 * 10⁻¹¹ m³/kgs²
Substituting these values into equation (1)
[tex]M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }[/tex]
M = 2307.46 * 10³⁶ kg
The mass of the black hole in the galaxies is measured using the doppler shift.
The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.
Due to a technical malfunction a space explorer had to crash land on Planet-X. She manages to fix her space ship and now she is preparing for launch. However she needs to know the gravitational acceleration on the surface of the planet in order to take off successfully. She builds a mathematical pendulum out of a piece of string and a left over steel bolt. The bolt has a mass of 35.5 g, and the string is 133 cm long. She attaches the pendulum to a fixed point and she lets it swing. She counts 12 complete oscillations in a time period of 70.7 seconds. What is the gravitational acceleration on the surface of Planet-X
Answer:
The gravitational acceleration on the surface of Planet-X is 1.5 [tex]ms^{-2}[/tex]
Explanation:
Firstly, we are to list out the parameters given:
Mass (m) = 35.5 g, Length of string (L) = 133 cm = 1.33 m, t = 70.7 seconds for 12 oscillations, T = 70.7 ÷ 12 = 5.89 seconds
The formula for calculating the period of a simple pendulum (assuming the angle of deflection is lesser than 15º) is given by:
[tex]T=2\pi\sqrt{\frac{L}{g}}\\[/tex]
We want to calculate for the gravitational acceleration (g), hence, we have to make g the subject of the formula
[tex]g=4\pi^{2}\frac{L}{T^{2}}\\[/tex]
Substitute the parameters into the equation, we have:
g = [tex]4\pi^{2}[/tex] * 1.33 ÷ [tex]5.89^{2}[/tex] = 1.51
g = 1.5 [tex]ms^{-2}[/tex]
The gravitational acceleration on the surface of Planet-X is 1.5 [tex]ms^{-2}[/tex]
Answer:
the gravitational acceleration of the Xplanet is 1.344m/s^2
Explanation:
You can use the formula for the calculation of the frequency of a pendulum, in order to find an expression for the gravitational constant:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{l}}\\\\g=4\pi^2 f^2l[/tex]
Where you can notice that mass ob the object does not influence of the gravitatiolan acceleration. By the information of the question, you have the values of f and l. By replacing these values (with units of meter and seconds) you obtain:
[tex]f=\frac{12}{70.7}=0.16s^{-1}\\\\g=4\pi^2(0.16s^{-1})^2(1.33m)=1.344\frac{m}{s^2}[/tex]
Hence, the gravitational acceleration of the Xplante is 1.344m/s^2
Two ice skaters, Skater X and Skater Y, are at rest on a horizontal surface made of ice. The friction between the ice and the skaters is negligible. Skater X, who has a smaller mass than Skater Y, pushes Skater Ysuch that Skater Y travels with a speed of 2vo to the right. Which of the following indicates the direction of the velocity of Skater X and the direction of the velocity of the center of mass of the two-skater system after the push?
a. Skater X Center of Mass
To the left To the right
b. Skater X Center of Mass
To the left Zero
c. Skater X Center of Mass
Zero To the right
d. Skater X Center of Mass
Zero Zero
Answer:
Explanation:
Both skaters are initially at rest
Then,
Ux = Uy = 0m/s
It is assumed that it is a frictionless surface
Mass of skater X is less than Y
Mx < My
My > Mx
This shows that,
My = a•Mx,
Such that a > 1
Skater X pushes Y such that it moves with velocity.
Vy = 2Vo to the right, I.e positive x axis
Vy = 2Vo •i
We want to find direction of the velocity of skater X
Using conservation of momentum
Initial momentum =Final momentum
Mx•Ux + My•Uy = Mx•Vx + My• Vy
Ux = Uy = 0,
also My = aMx. And Vy = 2Vo •i
0 + 0 = Mx•Vx + a•Mx•2Vo •i
0 = Mx•Vx + 2a•Mx•Vo •i
Mx•Vx = —2a•Mx•Vo •i
Divide through by Mx
Vx = —2a Vo •i
Therefore, since a is always positive
Then, Vx is in the negative direction or opposite direction to Vy
Now, center of Mass
The center of mass calculated using
Vcm = 1 / M Σ Mi•Vi
Where
M is sum of all the masses
M = Mx + My = Mx + aMx = (a+1) Mx
Then,
Vcm = (Σ Mi•Vi ) / M
Vcm = (MxVx+MyVy) / (a+1)Mx
Vcm=Mx•(-2aVo) +aMx(2Vo)/(a+1)Mx
Vcm = -2a•Mx•Vo + 2a•MxVo / (a+1)Mx
Vcm = 0 / (a+1) Mx
Vcm = 0
So, conclusion
Skater X direction is to the left and the centre of mass is 0.
This options are not written well
Check attachment for correct option
A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length λ.
1. Calculate the electric field (a) at any point between the cylinders a distance r from the axis and (b) at any point outside the outer cylinder.
Answer:
(a) E=λ/(2\pi e0 r)
(b) E = 0
Explanation:
(a) We can use the Gaussian's Law to calculate the electric field at any distance r from the axis. By using a cylindrical Gaussian surface we have:
[tex]\int \vec{E}\cdot d\vec{r}=\frac{\lambda}{\epsilon_o}[/tex]
where λ is the total charge per unit length inside the Gaussian surface. In this case we have that the Electric field vector is perpendicular to the r vector. Hence:
[tex]E\int dr=E2\pi r=\frac{\lambda}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r \epsilon_o}[/tex]
(b) outside of the outer cylinder there is no net charge inside the Gaussian surface, because charge of the inner radius cancel out with the inner surface of the cylindrical conductor.
Hence, we have that E is zero.
hope this helps!!
Answer:
1a. E(r) = lambda/ 2πrEo
1b. Electric field outside the outside cylinder = lambda/ 2πrEo
Explanation:
Problem: A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is mounted on the end of a low-mass rigid rod of length b. This apparatus is started in such a way that while the rod rotates clockwise with angular speed ω1, the barbell rotates clockwise about its center with an angular speed ω2. What is the total angular momentum of this system about point B?
Answer:
mass of ball 1=m1
mass of ball 2=m2
velocity of ball=r1w1
velocity of ball 2=r2w2
Total angular momentum=m1*v1+m2*v2
but
v1=r1*w1
v2=r2*w2
Substitute values in above equation
Total angular momentum of the system=m1*r1*w1+m2*r2*w2
The total angular momentum of the system at point B will be [tex]\rm L=m_1 r_1\omega_1 +m_2 r_2 \omega_2[/tex].
What is angular momentum?The rotating counterpart of linear momentum is angular momentum also known as moment of momentum or rotational momentum.
The given data in the problem is;
m₁ is the mass of ball 1
m₂ mass of ball 2=m2
v₁ is the velocity of ball=r₁ω₁
v₂ is the velocity of ball 2=r₂ω₂
The total angular momentum of a system at point B is given as;
[tex]\rm V_{total}= r_1\omega_1 + r_2 \omega_2 \\\\ \rm L=m_1 r_1\omega_1 +m_2 r_2 \omega_2[/tex]
Hence the total angular momentum of the system at point B will be [tex]\rm L=m_1 r_1\omega_1 +m_2 r_2 \omega_2[/tex].
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Suppose a collection of quantum harmonic oscillators occupies the lowest 4 energy levels, and the spacing between levels is 0.04 eV. What is the complete emission spectrum for this system? That is, what photon energies will appear in the emissions? Include all energies, whether or not they fall in the visible region of the electromagnetic spectrum. Enter the photon energies in order of increasing energy.
a. Smallest photon energy = ____eV
b. Next larger photon energy = ____eV
c. Largest photon energy = ____eV
Answer:
a. Smallest photon energy = 0.04 eV
b. Next larger photon energy = 0.08 eV
c. Largest photon energy = 0.12 eV
Explanation:
Since the spacing between the levels is 0.04 even
The smallest photon energy, E= 0.04 eV
The next larger photon energy = 2E = 2×0.04 = 0.08 eV
Largest photon energy = 3E = 3×0.04 = 0.12
The relationships for the atomic transitions allowed to find all the transitions in the four-level system are:
The minimum energy is: E = 0.04 eV. Tthe following is: E = 0.08 eV. Tthe maximum energy is: E = 0.12 eV.
Atomic Transitions are the energy deference between two specific atomic levels.
[tex]\Delta E = E_f - E_i[/tex]
They indicate that we have 4 equally spaced states with an energy difference of 0.04 eV between each one.
In the attachment we can see a diagram of the system with an arrow indicating the possible transitions.
From level n = 2 to level n = 1 the energy is E₂₁ = 0.04 eV.
from level n = 3 to n = 1 the energy is E₃₁ = 0.08 eV.
all other transitions in the table.
Initial state final state Energy (eV)
4 1 0.12
3 1 0.08
2 1 0.04
4 2 0.08
3 2 0.04
4 3 0.04
These are all the possible transitions, we can see that some energies have several possible states, these currents that have several possibilities are called degenerate.
the energies are:
the minimum is: E = 0.04 eV.
the following is: E = 0.08 eV.
the maximum is E: = 0.12 eV.
In conclusion using the relationships for the atomic transitions we can find all the transitions in the four-level system are:
The minimum energy is: E = 0.04 eV The following is: E = 0.08 eV Tthe maximum energy is: E = 0.12 eV
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Boltzmann’s constant is 1.38066×10^−23 J/K,
and the universal gas constant is
8.31451 J/K · mol.
If 2.9 mol of a gas is confined to a 6.6 L
vessel at a pressure of 7.1 atm, what is the average kinetic energy of a gas molecule?
Answer in units of J.
Answer:
1.18266×10⁻²⁰ J
Explanation:
Applying,
E = (3/2)kT.................... Equation 1
Where E = kinetic energy of the gas molecule, k = Boltzmann's constant, T = Temperature
But,
PV = nRT.............. Equation 2
Where P = pressure, V = Volume, n = number of moles, R = Universal gas constant.
make T the subject of the equation
T = PV/nR............. Equation 3
Substitute equation 3 into equation 1
E = (3/2)k(PV/nR)
E = 3kPV/2nR............... Equation 4
Given: k = 1.38066×10⁻²³ J/K, V = 6.6 L = 0.0066 m³, P = 7.1 atm = (101325×7.1) N/m² = 719407.5 N/m², n = 2.9 mol, R = 8.31451 J/K.mol
Substitute into equation 4
E = (3×1.38066×10⁻²³×0.0066×719407.5)/(2×8.31451)
E = 1182.66×10⁻²³ J
E = 1.18266×10⁻²⁰ J
The loudness level of a sound, D, in decibels, is given by the formula Upper D equals 10 log (10 Superscript 12 Baseline Upper I )D=10log1012I, where I is the intensity of the sound, in watts per metersquared2. Decibel levels range from 0, a barely audible sound, to 160, a sound resulting in a ruptured eardrum. The sound of a certain animal can be heard 500 miles away, reaching an intensity of 2.62.6times×10 Superscript 6106 watts per meter squared .meter2. Determine the decibel level of this sound. At close range, can the sound of this animal rupture the human eardrum?
complete question
attached
Answer:
188.0 decibels
Explanation:
Substitute I = 6.3 x 10^6 into the given formula
D(6.3 x 10^6) = 10log(10^12 x 6.3 x 10^6) = 188.0 to one decimal place.
This is greater than 160 decibels, so the sound could rupture the human eardrum.
Substitute I = 6.3 x 10^6 into the given formula
The sound produced by the animal has a decibel level of 174 dB, which is intense enough to rupture a human eardrum at close range. However, the actual harm it might cause depends on other factors such as the frequency of the sound.
Explanation:The intensity of the animal's sound is given as 2.6 × 10^6 watts/meter squared. We can calculate the decibel level by substituting it into the given formula D = 10 log (10^12* I). Solving this equation gives us D = 174 dB.
Decibel levels of 160 and above can rupture the human eardrum. Since the sound produced by this animal is 174 dB, it's intense enough at close range to rupture the human eardrum.
However, it's worth noting that loudness is not only determined by intensity, but by frequency as well, which can affect how we perceive the loudness of a sound. Therefore, frequency might play a role in whether or not this sound would actually cause harm to a human ear.
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A beam of light is shined on a thin (sub-millimeter thick) single crystal wafer of material. The light source is special since it can be tuned to provide any wavelength of visible light on demand. The specimen is illuminated such that the wavelength of light is increased over time while the transmitted intensity of the light is measured. If the sample becomes transparent when the wavelength is greater than 650 nanometers, what is the band gap of the material, in eV
Answer:
The band gap of the material is 1.9113 eV
Explanation:
Given data:
λ = wavelength = 650 nm = 650x10⁻⁹m
Question: What is the band gap of the material, E = ?
[tex]E=\frac{hc}{\lambda }[/tex]
Here
h = Planck's constant = 6.626x10⁻³⁴J s
c = speed of light = 3x10⁸m/s
[tex]E=\frac{6.626x10^{-34}*3x10^{8} }{650x10^{-9} } =3.058x10^{-19} J=1.9113eV[/tex]
Two hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 1.20 × 10−3 m2. The two ends of the wire are connected to a resistor. The total resistance in the circuit is 18.0 . If an externally applied uniform longitudinal magnetic filed in the core changes from 1.65 T in one direction to 1.65 T in the opposite direction, how much charge flows through a point in the circuit during the change?
Answer:
The charge flows through a point in the circuit during the change is 0.044 C.
Explanation:
Given that,
Number of turns in the copper wire, N = 200
Area of cross section, [tex]A=1.2\times 10^{-3}\ m^2[/tex]
Resistance of the circuit, R = 118 ohms
If an externally applied uniform longitudinal magnetic filed in the core changes from 1.65 T in one direction to 1.65 T in the opposite direction.
We need to find the charge flows through a point in the circuit during the change. Due to change in magnetic field an emf is induced in it. It is given by :
[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]
Using Ohm's law :
[tex]\epsilon=IR[/tex]
[tex]IR=-\dfrac{d\phi}{dt}\\\\I=-\dfrac{1}{R}\dfrac{d\phi}{dt}[/tex]
Electric current is equal to the rate of change of electric charge. So,
[tex]dq=\dfrac{NA(B(0)-B(t))}{R}\\\\dq=\dfrac{200\times 1.2\times 10^{-3}(1.65+1.65)}{18}\\\\dq=0.044\ C[/tex]
So, the charge flows through a point in the circuit during the change is 0.044 C.
A(n) 77.5 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 37.3 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.94 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. Answer in units of min
Answer:
Explanation:
The speed of the astronaut can be found with the help of law of conservation of momentum .
mv = MV , M is mass of astronaut , m is mass of object thrown , v is velocity of object thrown and V is velocity of astronaut.
Putting the values
77.5 x V = .94 x 12
V = .14554 m /s
This will be the uniform velocity of astronaut.
Distance to be covered = 37.3 m
time taken = distance / velocity
= 37.3 / .14554
= 256.28 s
= 4.27 minutes.
Benzene gas (C6H6) enters a well-insulated reactor operating at steady state at 77oF, 1 atm, and burns completely with dry air entering as a separated stream at 300oF, 1 atm. The combustion products exit the reactor at 1600oF. Kinetic and potential energy effects are negligible. Determine the percent of theoretical dry air provided.
Answer:
35.71%
Explanation:
See attached file for calculation
Learning Goal: To learn the definition and applications of angular momentum including its relationship to torque. By now, you should be familiar with the concept of momentum, defined as the product of an object's mass and its velocity: p⃗ =mv⃗ . You may have noticed that nearly every translational concept or equation seems to have an analogous rotational one. So, what might be the rotational analogue of momentum? Just as the rotational analogue of force F⃗ , called the torque τ⃗ , is defined by the formula τ⃗ =r⃗ ×F⃗ , the rotational analogue of momentum p⃗ , called the angular momentum L⃗ , is given by the formula L⃗ =r⃗ ×p⃗ , for a single particle. For an extended body you must add up the angular momenta of all of the pieces. There is another formula for angular momentum that makes the analogy to momentum particularly clear. For a rigid body rotating about an axis of symmetry, which will be true for all parts in this problem, the measure of inertia is given not by the mass m but by the rotational inertia (i.e., the moment of inertia) I. Similarly, the rate of rotation is given by the body's angular speed, ω. The product Iω⃗ gives the angular momentum L⃗ of a rigid body rotating about an axis of symmetry. (Note that if the body is not rotating about an axis of symmetry, then the angular momentum and the angular velocity may not be parallel.)
Answer:
Let's analyse the definition and applications of angular momentum, and its relation with torque.
First of all, it's important to consider that the angular momentum is a property of rotational dynamics. Also, it's the analogue of the linear momentum.
Mathematically, the angular momentum is defined as
[tex]L=r \times p[/tex]
Where [tex]L[/tex] represents the angular momentum vector, [tex]r[/tex] represents is the position vector and [tex]p[/tex] is the linear momentum vector.
Notice that the angular momentum is also a vector, which is the cross product of two vectorial magnitudes. In other words, the direction of the resulting vector (linear momentum) follows the right hand rule, which means that the resulting direction is according to the rotation direction, also means that the cross product is not commutative, which is a common assumption students make.
Now, the realtion between angular momentum and torque is that the change of the angular momentum with respect to time is equivalent to its torque:
[tex]\frac{d}{dt}[L]=\frac{d}{dt}[r \times p] =\tau[/tex]
Remember that torque is defined as [tex]\sum \tau = r \times \sum F[/tex], and the derivative of the cross product is
[tex]\frac{d}{dt}[L]=\frac{d}{dt}[r \times p] = \frac{dr}{dt} \times p + r \times \frac{dp}{dt}[/tex]
Then,
[tex]\frac{d}{dt}[L] =(v \times mv)+r \times \frac{dp}{dt}[/tex]
But, [tex]v \times mv = 0[/tex], because those vector are parallel.
So, [tex]\frac{d}{dt}[L] = r\times \sum F = \tau[/tex]
At this point, we demonstrate it the relation between torque and rotational momentum.
In words, the net torque on a particle is equal to the rate of change of the angular momentum with respect to time.
Now, the application of angular momentum can be seen in skating spins, notice that when the skater puts his arms closer to its body, he'll rotate faster. The reason of this phenomenon is because arms represents rotating mass and the axis is the body, so the postion of this arm mass changes to zero distance to the rotational axis, that will increase the angular momentum, making higher. If the angular momentum is higher, the torque will be also higher, that's way the skater increses its rotational velocity.
Angular momentum is the rotational equivalent of linear momentum, defined as L = Iω for extended bodies. It is conserved when net external torque is zero, similar to how linear momentum is conserved in the absence of external forces. Torque and angular momentum are related by net τ = ΔL/Δt.
In physics, the concept of angular momentum is the rotational equivalent of linear momentum. It is defined as L = r × p for a single particle, where r is the position vector and p is the linear momentum vector. For an extended body rotating about an axis of symmetry, angular momentum (L) is given by the product of the moment of inertia (I) and angular velocity (ω): L = Iω.
Just as linear momentum is conserved in the absence of external forces, angular momentum is conserved when the net external torque is zero. This principle is crucial in various applications, such as planetary motion and rotating machinery.
The relationship between torque (τ) and angular momentum is given by the equation: net τ = ΔL/Δt, meaning the net torque acting on a system is equal to the rate of change of angular momentum.
Angular momentum plays a significant role in rotational dynamics, similar to how linear momentum is essential for translational motion.In systems where no external torque acts, the initial angular momentum is equal to the final angular momentum, explaining phenomena like gyroscopic motion.Puck 1 (1 kg) travels with velocity 20 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After the collision, puck 1 moves with a velocity of 5 m/s. Assume that no external forces are present and therefore the momentum for the system of pucks is conserved. What is the final velocity (in m/s) of puck 2 after the collision
Answer:
Explanation:
Parameters given:
Mass of Puck 1, m = 1 kg
Mass of Puck 2, M = 1 kg
Initial velocity of Puck 1, u = 20 m/s
Initial velocity of Puck 2, U = 0 m/s
Final velocity of Puck 1, v = 5 m/s
Since we are told that momentum is conserved, we apply the principle of conservation of momentum:
Total initial momentum of the system = Total final momentum of the system
mu + MU = mv + MV
(1 * 20) + (1 * 0) = (1 * 5) + (1 * V)
20 = 5 + V
V = 20 - 5 = 15 m/s
Puck 2 moves with a velocity of 15 m/s
A current of 3.2 Amps flows through a 7.0 Ω resistor in a circuit. Calculate the voltage.
You must show your work.
Answer:voltage=22.4volts
Explanation:
current=3.2A
Resistance=7ohms
Voltage =current x resistance
Voltage=3.2 x 7
Voltage=22.4volts
A goose with a mass of 2.0 kg strikes a commercial airliner with a mass of 160,000 kg head-on. Before the collision, the goose was flying with a speed of 60 km/hr and the aeroplane’s speed was 870 km/hour. Take the length of the goose to be 1.0 m long. (a) What is the change in momentum of the goose during this interaction?
Answer:
The change in momentum of the goose during this interaction is 33.334 m/s
Explanation:
Given;
mass of goose, m₁ = 2.0 kg
mass of commercial airliner, m₂ = 160,000 kg
initial velocity of the bird, u₁ = 60 km/hr = 16.667 m/s
initial velocity of the airliner, u₂ = 870 km/hr = 241.667 m/s
Change in momentum is given as;
ΔP = mv - mu
where;
u is the initial velocity of the bird
v is the final velocity of the bird
Apply the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the final velocity of bird and airliner after collision;
(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)
38,666,753.334 = 160,002v
v = 38,666,753.334 / 160,002
v = 241.664 m/s
Thus, the final velocity of the bird is negligible compared to final velocity of the airliner.
ΔP = mv - mu
ΔP = m(v - u)
ΔP = 2(0 - 16.667)
ΔP = -33.334 m/s
The negative sign implies a deceleration of the bird after the impact.
Therefore, the change in momentum of the goose during this interaction is 33.334 m/s
A solid block of copper, which is a good conductor, has a cavity in its interior. Within the cavity, insulated from the conductor, is a point charge of 3.00 nCnC. The block of copper itself has an excess charge of -8.00 nCnC. How much charge is on the outer surface of the block and the surface of the cavity
Answer:
The charge on the outer surface of the block = -5.00 nC
The charge on the surface of the cavity (on the inner surface of the block) = -3.00 nC
Explanation:
The point charge within the cavity will induce a charge equal in magnitude and opposite in sign on the inside cavities of the copper block.
Charge of the point charge = 3.00 nC
Charge induced on the inner surface of the Copper block's cavity = -3.00 nC
Since the charge on a conductor should usually be neutral, the charge on the inner surface causes a charge equal in magnitude and also opposite in sign on the outer surface of the block; that is, 3.00 nC.
But this block already has an excess charge of -8.00 nC (which resides on the surface because excess charge for conductors reside on the surface of the conductors)
So, net charge on the outer surface of the Copper block = -8.00 + 3.00 = -5.00 nC.
Hope this Helps!!!
5 points total) The latent heat of vaporization (boiling) of helium at a pressure of 1 atm and a temperature of 4.2 K is 21.8 kJ kg-1 . The densities at 4.2 K of the liquid and vapor are 125 kg m-3 and 19 kg m-3 respectively. a. (2 points) What proportion of latent heat is involved in work against the interatomic attraction? b. (2 points) Estimate the depth e of the potential well which results from the forces between two helium atoms. Give your answer in Joules, electron-volts (eV) and kelvin (K), where the answer in kelvin corresponds to an energy kT. (Take the average number of nearest neighbors to be z = 10 and note that the atomic number of helium is 4.) c. (1 point) Why does helium turn into a gas at such low temperatures?
Answer:
a) proportion of latent heat involved in work against the interatomic attraction = 0.794
b)Depth of the well in Joules = 23 * 10^-24 J
ii) In eV, E = 0.000144 eV
III) in Kelvin, E = 1.67 K
C) Check the explanation section for C
Explanation:
a) Latent heat, Q = 21.8 kJ/kg
Vapor density, Vd = 19 kg/m^3
Liquid density, Ld = 125 kg/ m^3
Pressure, P = 1 atm = 1 * 10^5 Pa
Volume change from liquid to vapor = (1/Vd) - (1/ Ld)
Volume change = (1/19) - (1/125)
Volume change = 0.045 m^3
Work done in converting from liquid to vapor, W = P * (Volume change)
W = 1 *0.045 * 10^5
W = 4.5 kJ
Let the proportion of latent heat involved in work against the interatomic attraction be Pr
Pr = (Q - W)/Q
Pr = (21.8 - 4.5)/21.8
Pr = 0.794
b) To calculate the depth of the potential well :
I) In joules
n(L-W) = 0.5 z Na E
z = 10
Where E = depth of the well
4(21.8-4.5) = 0.5 * 10 * 6.02 * 10^23 * E
E = 23 * 10^-24 J
ii) In eV
E = ( 23 * 10^-24)/(1.6 * 10^-19)
E = 0.000144 eV
III) In Kelvin
E = ( 23 * 10^-24)/(1.38 * 10^-23)
E = 1.67 K
C) Helium turns to a gas at that low temperature because of the large workdone (4.5 kJ) against the interatomic attraction.
Final answer:
a. About 83% of the latent heat is involved in work against the interatomic attraction. b. The depth of the potential well is approximately 4.62 x 10^-23 J, 2.88 x 10^14 eV, and 33.6 K. c. Helium turns into a gas at low temperatures due to its weak interatomic attraction.
Explanation:
a. To calculate the proportion of latent heat involved in work against interatomic attraction, we need to first calculate the work done using the formula W = P(V2 - V1), where P is the pressure and V2 and V1 are the volumes of the vapor and liquid respectively. The difference in volume is V2 - V1 = 1 - 19 = -18 m^3. Since the work done is negative (work is done against the interatomic attraction), the proportion of latent heat involved in work is given by the ratio |W|/Q, where Q is the latent heat of vaporization. Therefore, the proportion is |(-1 atm)(-18 m^3)/(21.8 kJ/kg)| = 0.8292, or about 83%.
b. The depth e of the potential well can be estimated using the formula e = kT/2, where k is Boltzmann's constant and T is the temperature. The average number of nearest neighbors z is 10, and the atomic number of helium is 4. Therefore, the depth e can be calculated as e = (4 * (1.38 x 10^-23 J/K) * 4.2 K)/(2 * 10) = 4.62 x 10^-23 J. This is equivalent to approximately 2.88 x 10^14 eV and 33.6 K.
c. Helium turns into a gas at such low temperatures because its interatomic attraction is weak. The low mass and low atomic number of helium result in weak intermolecular forces, making it easier for helium atoms to overcome the attractive forces and transition from a liquid to a gas state at low temperatures.
A diagram of a closed circuit with power source on the left labeled 30 V, a resistor on the top labeled 2 Ohms, a resistor on the right labeled 3 Ohms and a resistor on the bottom labeled R Subscript 3 baseline.
If the current in this circuit is 3 A, what must be the value of R3?
Answer:
5
Explanation:
i dont have one
Answer:
5
Explanation:
good luck :)
A cable passes over a pulley. Because the cable grips the pulley and the pulley has nonzero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 137 N, and the force on the other side is 43 N. Assuming that the pulley is a uniform disk of mass 1.21 kg and radius 0.723 m, find the magnitude of its angular acceleration. [For a uniform disk, I = (1/2)mr2 .] Answer in units of rad/s 2
Answer:
[tex]\alpha=214.8 rad/s^2[/tex]
Explanation:
We are given that
[tex]F_1=137 N[/tex]
[tex]F_2=43 N[/tex]
Net force=F=[tex]F_1-F_2=137-43=94 N[/tex]
Mass,m=1.21 kg
Radius,r=0.723 m
We have to find the magnitude of its angular acceleration.
Moment of inertia ,[tex]I=\frac{1}{2}mr^2[/tex]
Substitute the values
Torque ,[tex]\tau=I\alpha[/tex]
[tex]F_{net}\times r=\frac{1}{2}mr^2\alpha[/tex]
[tex]\alpha=\frac{2F_{net}}{mr}[/tex]
[tex]\alpha=\frac{2\times 94}{1.21\times 0.723}[/tex]
[tex]\alpha=214.8 rad/s^2[/tex]
The 10.2-ft steel beam is being hoisted from its horizontal position by the two cables attached at A and B. If the initial angular accelerations are α1 = 0.52 rad/sec2 and α2 = 0.35 rad/sec2 , determine the initial values of
(a) the angular acceleration α (positive if counterclockwise, negative if clockwise) of the beam,
(b) the acceleration of point C (positive if up, negative if down), and
(c) the distance d (positive if to the right, negative if to the left) from A to the point on the centerline of the beam which has zero acceleration.
Is Fahrenheit an SI unit?
Answer:
No
Explanation:
Because the people put in charge of defining SI units said so