Answer:
w = 7.89 10⁻² rad/s
Explanation:
We will solve this exercise with the conservation of the annular moment, let's write it in two moments
Initial. With the insect in the center
L₀ = I w₀
End with the bug on the edge
[tex]L_{f}[/tex]= I w + [tex]I_{bug}[/tex] w
The moments of inertia are
For a rod
I = 1/3 M L²
For the insect, taken as a particle
I = m L²
The system is formed by the rod and the insect, this is isolated, therefore the external torque is zero and the angular momentum is conserved
L₀ = [tex]L_{f}[/tex]
I w₀ = I w + [tex]I_{bug}[/tex] w
w = I / (I + [tex]I_{bug}[/tex]) w₀
w = I / (I + m L²) w₀
Let's calculate
w = 1.43 10⁻³ / (1.43 10⁻³ + 5 10⁻³ 0.620²)² 0.185
w = 1.43 10⁻³ / 3.352 10³ 0.185
w = 7.89 10⁻² rad/s
When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is 278 nm. What is the vacuum wavelength of the light in nm?
Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.
From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as
[tex]2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}[/tex]
We are given the second smallest nonzero thickness at which destructive interference occurs.
This corresponds to, m = 2, therefore
[tex]2t = 2\lambda_{film}[/tex]
[tex]t = \lambda_{film}[/tex]
The index of refraction of soap is given, then
[tex]\lambda_{film} = \frac{\lambda_{vacuum}}{n}[/tex]
Combining the results of all steps we get
[tex]t = \frac{\lambda_{vacuum}}{n}[/tex]
Rearranging, we find
[tex]\lambda_{vacuum} = tn[/tex]
[tex]\lambda_{vacuum} = (278)(1.33)[/tex]
[tex]\lambda_{vacuum} = 369.74nm[/tex]
If the surface air pressure is 1000 mb and the pressure at the top of the atmosphere (75 km) is 0 mb, at what altitude would I find half of the atmosphere air pressure?
Answer: 5.5km
Explanation:
Atmospheric pressure will be 500 mb (that is half of the total 1000mb air pressure).
Pressure decreases with increasing altitude. This is because at At higher altitudes, there are fewer air molecules above a the known or given surface than a similar surface at lower levels.
Pressure decreasing with higher altitudes also means that air pressure decreases rapidly at lowerevels but more slowly at higher levels.
It is also known that more than half of the atmospheric molecules are located below 5.5 km(that is atmospheric pressure decreases within the lowest 5.5 km to about fifty(50) percent( that is 500 millibar).
A ball of mass 0.800 kg is carefully balanced on a shelf that is 3.00 m above the ground. What is its gravitational potential energy?a. 2.08J b. 20.4J c. 7.84J d. 25.5J
Potential energy is the mechanical energy associated with the location of a body within a force field (gravitational, electrostatic, etc.) or the existence of a force field inside a body (elastic energy)
The gravitational potential force is subject to the relationship
[tex]PE = mgh[/tex]
Where,
m = mass
g = Gravitational Energy
h = Height
If our height is 3 meters, the mass is 0.8Kg and the earth has a gravitational acceleration of 9.8m / s, we will have to
PE = (9.8)(3)(0.8)
PE = 23.52J
The closest answer is D.
For a fiber-reinforced composite, the efficiency of reinforcement η is dependent on fiber length l according to where x represents the length of the fiber at each end that does not contribute to the load transfer. What length is required for a 0.62 efficiency of reinforcement, assuming that x is 0.61 mm?
Answer:
l = 3.21 mm
Explanation:
The efficiency of a fiber-reinforced is the following:
[tex] \eta = \frac{l - 2x}{l} [/tex]
Where:
η: is the efficiency
l: is the fiber length
x: is the length of the fiber at each end that doesn't contribute to the load transfer
So the length required for a 0.62 efficiency of reinforcement is:
[tex] l = \frac{2x}{1- \eta} = \frac{2 \cdot 0.61 mm}{1- 0.62} = 3.21 mm [/tex]
I hope it helps you!
The length required for a 0.62 efficiency of reinforcement in a fiber-reinforced composite can be calculated using the equation η = (l-x)/l, where l represents the total length of the fiber and x represents the length at each end that does not contribute to load transfer. By rearranging the equation, we can solve for l.
Explanation:To calculate the length required for a 0.62 efficiency of reinforcement in a fiber-reinforced composite, we can use the equation η = (l-x)/l. Here, l represents the total length of the fiber, and x represents the length at each end that does not contribute to load transfer. We are given that x is 0.61 mm. We can rearrange the equation to solve for l:
η = (l-x)/l
0.62 = (l-0.61)/l
0.62l = l - 0.61
0.62l - l = -0.61
0.62l = -0.61
l = -0.61/0.62
l = -0.983
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A 28 kg child goes down a straight slide inclined 38∘ above horizontal. The child is acted on by his weight, the normal force from the slide, kinetic friction, and a horizontal rope exerting a 30 N. How large is the normal force of the slide on the child? If the child slides with a constant speed, how large is the kinetic frictional force from the slide?
The normal force of the slide on the child is 217.3 N and the kinetic frictional force from the slide is 166.8 N when the child slides down the slide at a constant speed.
Explanation:The normal force on a slope, which is always perpendicular to the surface, is equal to the weight component of the object that is perpendicular to the slope. As the child slides down the slide at a constant speed, the net force on the child is zero. In this scenario, let's denote mass (m) as 28 kg, inclination angle (θ) as 38 degrees, and g as gravitational acceleration which is 9.8 m/s². So, the normal force (N), which is equal to m*g*cosθ, can be calculated as: 28 kg * 9.8 m/s² * cos(38) = 217.3 N.
The frictional force from the slide acts in the opposite direction to the motion. When the sliding speed is constant, this kinetic frictional force equals the component of the child's weight that is parallel to the slope (m*g*sinθ). Hence, the kinetic frictional force would be: 28 kg * 9.8 m/s² * sin(38) = 166.8 N.
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The normal force of the slide on the child can be found by multiplying the child's weight by the cosine of the angle of inclination. The kinetic frictional force from the slide is equal to the horizontal force exerted by the rope.
Explanation:To find the normal force of the slide on the child, we need to determine the component of the child's weight perpendicular to the slide. Since the slide is inclined at 38∘ above horizontal, the normal force is equal in magnitude to the component of the child's weight perpendicular to the slide, which is given by:
Normal force = weight * cos(38∘)
Next, to find the kinetic frictional force from the slide, we need to use the horizontal force exerted by the rope. Since the child slides with a constant speed, the kinetic frictional force must be equal in magnitude to the horizontal force exerted by the rope, which is given as 30 N.
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The distance required for a car to come to a stop will vary depending on how fast the car is moving. Suppose that a certain car traveling down the road at a speed of 10 m / s can come to a complete stop within a distance of 20 m . Assuming the road conditions remain the same, what would be the stopping distance required for the same car if it were moving at speeds of 5 m / s , 20 m / s , or 40 m / s ?
Answer:
5 m
80 m
320 m
Explanation:
[tex]v_{o}[/tex] = Initial speed of the car = 10 ms⁻¹
[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹
[tex]d[/tex] = Stopping distance of the car = 20 m
[tex]a[/tex] = acceleration of the car
On the basis of above data, we can use the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d\\0^{2} = 10^{2} + 2 (20) a\\a = - 2.5 ms^{-2}[/tex]
[tex]v_{o}[/tex] = Initial speed of the car = 5 ms⁻¹
[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹
[tex]d'[/tex] = Stopping distance of the car
[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²
On the basis of above data, we can use the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d'\\0^{2} = 5^{2} + 2 (- 2.5) d'\\d' = 5 m[/tex]
[tex]v_{o}[/tex] = Initial speed of the car = 20 ms⁻¹
[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹
[tex]d''[/tex] = Stopping distance of the car
[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²
On the basis of above data, we can use the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d''\\0^{2} = 20^{2} + 2 (- 2.5) d''\\d'' = 80 m[/tex]
[tex]v_{o}[/tex] = Initial speed of the car = 40 ms⁻¹
[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹
[tex]d'''[/tex] = Stopping distance of the car
[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²
On the basis of above data, we can use the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d'''\\0^{2} = 40^{2} + 2 (- 2.5) d'''\\d''' = 320 m[/tex]
Based on your observations of the six collisions, describe the physical difference between elastic and inelastic collisions.
Answer:
Collisions are basically two types: Elastic, and inelastic collision. Elastic collision is defined as the colliding objects return quickly without undergoing any heat generation. Inelastic collision is defined as the where heat is generated, and colliding objects are distorted.
In elastic collision, the total kinetic energy, momentum are conserved, and there is no wasting of energy occurs. Swinging balls is the good example of elastic collision. In inelastic collision, the energy is not conserved it changes from one form to another for example thermal energy or sound energy. Automobile collision is good example, of inelastic collision.
Final answer:
Elastic collisions conserve both momentum and kinetic energy with objects bouncing off each other, while inelastic collisions conserve momentum but not kinetic energy, often resulting in the objects sticking together or showing permanent deformations. Mathematically analyzing an elastic collision is generally easier due to the conservation of both momentum and kinetic energy.
Explanation:
The primary physical difference between elastic and inelastic collisions lies in how kinetic energy is conserved in the collision process. In an elastic collision, both momentum and kinetic energy are conserved. This means that after the collision, the total kinetic energy of the two objects remains the same as before they collided. The objects bounce off each other and do not stick together. As a helpful trick, remember that 'elastic' implies the ability to return to the original shape, just like objects in an elastic collision bounce off one another and return to their separate ways.
On the other hand, an inelastic collision is one where the kinetic energy is not conserved, although the momentum is still conserved. The objects may stick together after the collision, which can be referred to as a perfectly inelastic collision. Since kinetic energy is lost in inelastic collisions, the objects typically do not separate after colliding; rather, they may move as a single entity or with less kinetic energy than they had initially. An inelastic collision involves a permanent deformation of the colliding bodies or the generation of heat.
To your other questions: Analyzing an elastic collision might be easier, as both momentum and kinetic energy conservation can be used to solve for final velocities, while in an inelastic collision, only momentum is conserved. Verification can be done both mathematically and graphically, but mathematical methods often provide a direct computation of the final velocities. When considering the recoil in two objects thrown by two people, if the objects have the same velocity but different masses, question 15 from 8.3 suggests that the person who threw the heavier object will gain more velocity upon recoil. However, in reality, since momentum is conserved, both individuals would experience the same amount of recoil velocity, opposite in direction to the thrown object.
4) An electron moving along the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, then the magnetic field must have a component
(A) along the +z-axis.
(B) along the -z-axis.
(C) along the -x-axis.
(D) along the +y-axis.
(E) along the -y-axis.
Answer:
B) along the -z-axis.
Explanation:
According to the Fleming's left hand rule we know that when a positive charge moves with velocity v perpendicular to the direction of magnetic field the it experiences a magnetic force perpendicular to the the plane in which the velocity vector and the magnetic field vector lies.As a result all the three vector are mutually perpendicular in such a case.Here as per given condition we have a negative charge moving along x-axis which experiences the deflection in -y direction so the magnetic field will be along the negative z axis i.e. into the screen.Answer:
option (b)
Explanation:
direction of velocity is along + X axis
direction of force is along - Y axis
So, by use of the
[tex]\overrightarrow{F}=q\left ( \overrightarrow{v} \times \overrightarrow{B}\right )[/tex]
Where, B be the magnetic field
As charge of electron is negative
so the direction of magnetic field is along negative z axis .
Fused quartz possesses an exceptionally low coefficient of linear expansion, 5.50 × 10 − 7 ( ∘ C ) − 1 . Suppose a bar of fused quartz of length 3.45 m at 20.0 ∘ C is heated to 235 ∘ C . By how much, in millimeters, would the bar expand?
To solve this problem we will use the concepts related to thermal expansion in a body for which the initial length, the coefficient of thermal expansion and the temperature change are related:
[tex]\Delta L = L0\alpha\Delta T[/tex]
Where,
[tex]\Delta L[/tex] = Change in Length
[tex]\alpha[/tex] = Coefficient of linear expansion
[tex]\Delta T[/tex] = Change in temperature
[tex]L_0[/tex] = Initial Length
Our values are:
[tex]L_0 = 3.45m[/tex]
[tex]\alpha = 5.5*10^{-7} \°C^{-1}[/tex]
[tex]\Delta T = 235-20 = 215\°C[/tex]
Replacing we have,
[tex]\Delta L = (3.49) (5.5*10^{-7}) [(215)[/tex]
[tex]\Delta L = 0.0004126m[/tex]
[tex]\Delta L = 0.4126mm[/tex]
Therefore the change in milimiters was 0.4126mm
According to the videos seen in Modules 8, please write a short paragraph answering the following questions: a. What are some examples of on-the-spot feedback? b. What are some examples of delayed feedback?
Answer:
.
Explanation:
A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.90×10−2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.
Answer:
a. Δx = 2.59 cm
Explanation:
mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m
Using momentum conserved
mb * (0) + mp * vp = ( mb + mp ) * vf
vf = ( mp / mp + mb) * vp
¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²
Solve to Δx '
Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )
Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97 m /s) ² / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]
Δx = 0.02599 m ⇒ 2.59 cm
Suppose that a parallel-plate capacitor has circular plates with radius R = 35.0 mm and a plate separation of 4.1 mm.
Suppose also that a sinusoidal potential difference with a maximum value of 160 V and a frequency of 60 Hz is applied across the plates; that isV=(160.0 V)sin((2.*p)*(60 Hz * t)).Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R.Find B(r = 17.5 mm).Find B(r = 70.0 mm).Find B(r = 105.0 mm).
Answer:
β max = 2.86 x 10 ⁻¹² T
β max = 1.432 x 10 ⁻¹² T
β max = 5.726 x 10 ⁻¹² T
β max = 8.589 x 10 ⁻¹² T
Explanation:
Given:
R = 35.0 mm , d = 4.1 mm , f = 60 Hz , V = 160v
And knowing
μ₀ = 4 π x 10 ⁻⁷ T * m / A , ε₀ = 8.85 x 10⁻¹² C² / N * m²
To find β max can use the equation
β max = [ π * f * μ₀ * ε₀ * r * V ] / d
r = R
β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.035 m * 160 ] / (4.1 x 10⁻³ )
β max = 2.86 x 10 ⁻¹² T
r = 17.5 mm
β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.0175 m * 160 ] / (4.1 x 10⁻³ )
β max = 1.432 x 10 ⁻¹² T
r = 70 mm
β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.070 m * 160 ] / (4.1 x 10⁻³ )
β max = 5.726 x 10 ⁻¹² T
r = 105.0 mm
β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.105 m * 160 ] / (4.1 x 10⁻³ )
β max = 8.589 x 10 ⁻¹² T
A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most of her body as a uniform cylinder. Suppose the skater has a mass of 64 kg . One eighth of that mass is in her arms, which are 60 cm long and 20 cm from the vertical axis about which she rotates. The rest of her mass is approximately in the form of a 20-cm-radius cylinder.A. Estimate the skater's moment of inertia to two significant figures.B. If she were to hold her arms outward, rather than at her sides, would her moment of inertia increase, decrease, or remain unchanged?
Answer:
Explanation:
The body can be split into two parts
1 ) cylinder part of mass 7/8 x 64 = 56 kg of radius 20 cm
2 ) In the form of rod attached with cylinder having length 40 cm and mass
of 1/8 x 64 = 8kg
moment of inertia of cylinder
= 1/2 mr²
= .5 x 56 x ( 20 x 10⁻²)²
= 1.12 kg m²
moment of inertia of rods
= 1/3 ml²
= 1/3 x 8 x ( 40 x 10⁻²)²
= .4267 kg m²
Total MI = 1.5467 Kg m²
A water wave traveling in a straight line on a lake is described by the equation
y(x,t) = (3.30 cm) cos(0.400 cm?1x + 5.05 s?1t)
where y is the displacement perpendicular to the undisturbed surface of the lake.
(a) How much time does it take for one complete wave pattern to go past a fisherman in a boat at anchor?
What horizontal distance does the wave crest travel in that time?
b) What are the wave number and the number of waves per second that pass the fisherman?
(c) How fast does a wave crest travel past the fisherman?
What is the maximum speed of his cork floater as the wave causes it to bob up and down?
Final answer:
Calculations from the provided wave equation involve finding the period (T), horizontal wave travel, wave number and frequency, wave crest speed, and the maximum vertical displacement speed. These reveal wave motion characteristics including time intervals and speeds for complete wave patterns and individual points within a wave.
Explanation:
The student's question about a wave on a lake is one that involves understanding the characteristics and behavior of waves, which is a key concept in physics. Specifically, the student is asked to find the time for a complete wave pattern to pass a stationary observer, the horizontal distance traveled by a wave crest in that time, the wave number and frequency, the wave crest speed, and the maximum speed of a vertical displacement caused by the wave. Here's how we can calculate each of these:
Time for one complete wave pattern: The time taken for one complete wave to pass is the reciprocal of the frequency, known as the period (T). We use the angular frequency (5.05 s-1) provided in the equation y(x,t) = (3.30 cm) cos(0.400 cm-1x + 5.05 s-1t). We find T by taking 2π divided by the angular frequency.
Horizontal distance traveled: To find this, multiply the wave speed by the period (T).
Wave number and frequency: The wave number (k) is already provided as 0.400 cm-1, which can be converted to meters if needed. The frequency (f) is the angular frequency divided by 2π.
Wave crest speed: This can be found by dividing the angular frequency by the wave number.
Maximum speed of vertical displacement (cork floater): This is the amplitude times the angular frequency, which gives the maximum speed of any vertical movement due to the wave.
These calculations help understand the dynamics of wave motion and how waves interact with objects in their path like a fisherman's cork floater.
An adventurous stdent stands on top of a spinning chair holding a Jack-o-lantern in each hand (each of which is 5.0 kg) as she extents her arms horizontally.
She spins around, making one revolution in 2.0 s.
The moment of inertia when her arms are outstretched is 13.0 kgm, and if she brings the Jack-o-lanterns in close to her stomach, the moment of inertia drops to 2.6 kgm?.
What is the final angular velocity?
(A) 2.5 rev/s
(B) 2.5 rad/s
(C) 27 rad/s
(D) 0.5 rev/s
(E) 5.0 rev/s
Answer:
option A
Explanation:
the initial moment of inertia of system , I₀ = 13 kg.m²
the final moment of inertia of the system , I = 2.6 kg.m²
time = t = 2 s
mass = 5 Kg
the initial angular speed ,
[tex]\omega = \dfrac{2\pi}{T}[/tex]
[tex]\omega = \dfrac{2\pi}{2}[/tex]
ω = 3.14 rad/s
[tex]\omega = \dfrac{3.14}{2\pi}[/tex]
[tex]\omega_0= 0.5\ rev/s[/tex]
let the final angular speed be ω₀
using conservation of angular momentum
I₀ x ω₀= I x ω
13 x ω₀= I x ω
13 x 0.5 = 2.6 x ω
ω = 2.5 rev/s
hence, the correct answer is option A
A 2-kg object is initially at the bottom of a long 50° inclined plane, and is beginning to slide up this inclined plane. The initial velocity of the object os 3.0 m/s. The coefficient of kinetic friction between the object and the surface of the inclined plane is µk = 0.3, while the coefficient of static friction is µs = 0.4. Will the object ever slide back down to the initial position? If yes, what will be its speed once it slides back down to the bottom? If no, how high up the incline will it manage to climb?
Answer:
Explanation:
Given
mass of object [tex]m=2 kg[/tex]
inclination [tex]\theta =50^{\circ}[/tex]
[tex]\mu _k=0.3[/tex]
[tex]\mu _s=0.4[/tex]
initial velocity [tex]u=3 m/s[/tex]
acceleration of block during upward motion
[tex]a=g\sin \theta -\mu _kg\cos \theta [/tex]
[tex]a=g(\sin 50-0.3\cos 50)[/tex]
[tex]a=5.617 m/s^2[/tex]
using relation
[tex]v^2-u^2=2a\cdot s[/tex]
where [tex]s=distance\ moved [/tex]
[tex]v=final\ velocity[/tex]
v=0 because block stopped after moving distance s
[tex]0-(3)^2=2\cdot (-5.617)\cdot s[/tex]
[tex]s=\frac{4.5}{5.617}[/tex]
[tex]s=0.801[/tex]
If block stopped after s m then force acting on block is
[tex]F=mg\sin \theta =[/tex]friction force [tex]f_r=\mu mg\cos \theta [/tex]
[tex]F>f_r[/tex] therefore block will slide back down to the bottom
Two long wires hang vertically. Wire 1 carries an upward current of 1.80 A. Wire 2, 20.0 cm to the right of wire 1, carries a downward current of 4.00 A. A third wire, wire 3, is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force.
(a) Is this situation possible?
Is it possible in more than one way?
(b) Describe the position of wire 3.
(c) Describe the magnitude and direction of the current in wire 3.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
(a) Yes, this situation is possible. No, It is not possible in more than one way.
(b) The position of the wire is at a distance of 16.3 cm, in the direction left of the wire
(c) The magnitude of the current in wire 3 (I3) is 3.267 A and in a direction pointing in the downward direction.
(a)
The diagrammatic illustration of the information provided in the question can be seen in the image attached below.
From the image: suppose the forces on the wire as a result of the two other wires are equal and opposite, then the third wire will experience no net force. Hence, we can say Yes, the situation is possible.
Given that:
I₁ = 1.80 AI₂ = 4.00 AIt will not be possible to have more than one way, this is because the force acting on the other two wires will have opposite directions but they would not have the same equal magnitude.
The force per unit length existing in-between two current-carrying parallel wires can be determined by using the formula:
[tex]\mathbf{\dfrac{F}{l} = \dfrac{\mu_o I_1I_2}{2 \pi r} }[/tex]
where:
[tex]\mathbf{I_1,I_2 }[/tex] = current passing via conductorsr = distance in-between the parallel conductorsSInce F1 = F2, Then:
[tex]\mathbf{\dfrac{\mu_o\times I_1\times I_3\times l}{2 \pi r}= \dfrac{\mu_o\times I_2\times I_3\times l}{2 \pi (r+0.2)} }[/tex]
[tex]\mathbf{\dfrac{I_1}{r}= \dfrac{ I_2}{ (r+0.2)} }[/tex]
[tex]\mathbf{\dfrac{1.80}{r_1}= \dfrac{ 4.00}{ (r_1+0.2)} }[/tex]
1.80(r₁ + 0.2) = 4.00r₁
1.80r₁ +0.36 = 4.00 r₁
0.36 = 2.2r₁
r₁ = 0.36/2.2
r₁ = 16.3 cm in the direction left of the wire.
Again, we are to determine the magnitude of the current and its direction in wire 3 (I₃).
From the image, the forces of F21 = F23
∴
[tex]\mathbf{\dfrac{\mu_o \times I_1 \times I_2 \times l }{2 \pi r} = \dfrac{\mu_o \times I_2 \times I_3 \times l }{2 \pi r}}[/tex]
[tex]\mathbf{\dfrac{\mu_o \times 1.8 \times 4.0 \times l }{2 \pi \times 0.2} = \dfrac{\mu_o \times 4 \times I_3 \times l }{2 \pi \times 0.363}}[/tex]
Making I₃ the subject by equating both equations together, we have:
[tex]\mathbf{I_3 = \dfrac{1.8 \times 0.363}{0.2}}[/tex]
I₃ = 3.267 A and the current is pointing in the downward direction
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A 0.180 kg cube of ice (frozen water) is floating in glycerin. The glycerin is in a tall cylinder that has inside radius 5.50 cm. The level of the glycerin is well below the top of the cylinder.
A)If the ice completely melts, by what distance does the height of liquid in the cylinder change?
Express your answer with the appropriate units. Enter positive value if the surface of the water is above the original level of the glycerin before the ice melted and negative value if the surface of the water is below the original level of the glycerin.
Answer:
0.00389 m
Explanation:
Archimedes principle states that the buoyant force when a body is immersed in a liquid equals the weight of the liquid displaced
B = weight of the liquid displaced
density is defined as the mass divided by the volume of the substance and the units is in kg/m³
Density of glycerin = mass / volume
ρg × volume = mass of glycerin displaced
since the object was floating, the upthrust from the liquid equals the weight of the liquid
ρg × volume × g = mg
divide both side by g
ρg × volume = 0.180 where ρg (density of glycerin) = 1260 kg / m³
1260 × volume of glycerin displaced = 0.18 / 1260 = 0.000143 m³
volume of glycerin displaced = πr²h₁ where h₁ = of liquid displaced
πr²h₁ = 0.000143
h₁ = 0.000143 / ( 3.142 × 0.055² ) = 0.01505 m
when the ice completely melted, it will displaces liquid equal to it own volume
density of water = mass of water /volume
1000 = 0.18 / v
v = 0.18 / 1000 = 0.00018 m³
volume = πr²h₂ where h₂ = height of the melted water
πr²h₂ = 0.00018
h₂ = 0.00018 / ( 3.142 × 0.055²) = 0.01894 m
change in height of the liquid = h₂ - h₁ = 0.01894 m - 0.01505 m = 0.00389 m
When the 0.180 kg ice cube melts, it will form water that spreads out in the glycerin-filled cylinder, raising the level by approximately 1.89 cm.
Explanation:To solve this problem, we first need to understand that when the ice cube melts, it will not change the overall volume of liquid in the cylinder. In other words, the total volume before the ice melts (volume of glycerin + volume of ice) is equal to the total volume after the ice melts (volume of glycerin +volume of water).
The volume of the water, resulting from the melted ice, can be calculated using the mass of the ice cube and the density of water. The volume (V) is equal to the mass (m) divided by the density (ρ). For ice, m = 0.180 kg and ρ = 1000 kg/m³ (the density of water), which gives V = 0.00018 m³ or 180 cm³.
This water spreads out in the cylinder, raising the level. The height (h) to which it raises can be found by dividing the volume of the water (V) with the cross-sectional area (A) of the cylinder (h = V/A). The area of the cylinder is πr², where r is the radius, so A = π * (5.5 cm)² = 95.03 cm². Thus, h = 180 cm³ / 95.03 cm² = 1.89 cm. Consequently, when the ice cube melts, the glycerin's height will increase by approximately 1.89 cm.
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A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for this vehicle to maintain a 70 mi/h speed on a 5% upgrade through an air density of 0.002045 slugs/ft3?
Answer:
i put this in the calculator and my answer is 600. hope this helps
Explanation:
A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2, the minimum tractive effort = 220.3 lb.
To find the minimum tractive effort,
vehicle weight is 2500 lb
drag coefficient is 0.35
frontal area 22 ft^2
vehicle speed is 60 mi/hr = 70 ft/sec
gradient - 5%
air density = 0.002045 slugs/ft^3
What is drag force?The force which exerted on a solid body moving in some relation to a fluid by the fluid's movement is known as a drag force.
Drag force is given as,
Fd = [tex]\frac{1}{2}[/tex]CdρAV²
= [tex]\frac{1}{2}[/tex] × 0.35 × 0.002045× 20× 88²
= 55.4 lb.
Force due to vehicle weight,
Fw = 0.01 ( 1+ υ/147)W
= 0.01 ( 1 + 88/147) 2500
= 39.965 lb.
Force due to gradient
Fg = W × g
= 2500 × 0.05
= 125 lb
Minimum tractive effort
F = Fd + Fw +Fg
= 55.4 + 39.965 + 125
= 220.365 lb
Thus, minimum tractive force can be found as 220.365 lb.
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A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?
Answer:
22.1 m
Explanation:
[tex]v_{o}[/tex] = initial speed of ball = 14.3 m/s
[tex]\theta[/tex] = Angle of launch = 27°
Consider the motion of the ball along the vertical direction.
[tex]v_{oy}[/tex] = initial speed of ball = [tex]v_{o} Sin\theta = 14.3 Sin27 = 6.5 ms^{-1}[/tex]
[tex]a_{y}[/tex] = acceleration due to gravity = - 9.8 ms⁻²
[tex]t[/tex] = time of travel
[tex]y[/tex] = vertical displacement = - 3.50 m
Using the kinematics equation that suits the above list of data, we have
[tex]y = v_{oy} t + (0.5) a_{y} t^{2} \\- 3.50 = (6.5) t + (0.5) (- 9.8) t^{2}\\- 3.50 = (6.5) t - 4.9 t^{2}\\t = 1.74 s[/tex]
Consider the motion of the ball along the horizontal direction.
[tex]v_{ox}[/tex] = initial speed of ball = [tex]v_{o} Cos\theta = 14.3 Cos27 = 12.7 ms^{-1}[/tex]
[tex]X[/tex] = Horizontal distance traveled
[tex]t[/tex] = time taken = 1.74 s
Since there is no acceleration along the horizontal direction, we have
[tex]X = v_{ox} t\\X = (12.7)(1.74)\\X = 22.1 m[/tex]
Which of the following phrases best describes the term "magnetic flux"?
A) the direction of the magnetic field relative to a surface
B) the amount of magnetic field that passes through a surface
C) the number of magnetic dipoles moving through a wire
D) the flow of magnetons in space
E) another name for the magnetic field
Answer:
B) the amount of magnetic field that passes through a surface
Explanation:
The magnetic flux is defined as the total magnetic field times the area normal to the magnetic field lines passing through it.
Mathematically:
[tex]\phi_B=\vec{B}.\vec{A}[/tex]
where:
[tex]\vec{A}=[/tex] area vector directed normal to the surface
[tex]\vec{B}=[/tex] magnetic field vector
It is a quantity defined for the convenience of use in Faraday's Law.
Final answer:
The correct phrase for describing 'magnetic flux' is 'the amount of magnetic field that passes through a surface,' represented by the formula Φ = BA cos θ. The unit of magnetic flux is the weber (Wb) or volt second (Vs).
Explanation:
The phrase that best describes the term magnetic flux is 'B) the amount of magnetic field that passes through a surface.' This is due to the fact that magnetic flux is a measure of the total number of magnetic field lines that penetrate a given surface area. The formula for calculating magnetic flux (symbolized by the Greek letter phi, Φ) is Φ = BA cos θ, where B represents the magnetic field strength, A is the area through which the field lines pass, and θ is the angle between the magnetic field and the perpendicular to the surface.
The unit of magnetic flux is the weber (Wb), which can also be expressed in terms of magnetic field per unit area, or tesla per square meter (T/m²), and as a volt second (Vs). Understanding magnetic flux is essential in applying Faraday's Law and in the study of electromagnetic induction, where changes in magnetic flux through a conductor generate an electromotive force (emf).
What is the force of a 500kg car that is accelerating at 20m /sex2?
Answer:10000N
Explanation:formuler for calculating force is given by F=Ma
M(mass)=500kg
a(acceleration)=20m/s^2
Therefore by substitution we have F=500*20
F=10000N
Calculate the number of molecules in a deep breath of air whose volume is 2.35 L at body temperature, 37 ∘C, and a pressure of 735 torr. Express the answer in molecules to three significant figures. NN = nothing molecules Request Answer Part B The adult blue whale has a lung capacity of 5.0×103 L. Calculate the mass of air (assume an average molar mass 28.98 g/mol) contained in an adult blue whale’s lungs at 0.5 ∘C and 1.07 atm, assuming the air behaves ideally. Express the answer in kilograms to two significant figures. mm = nothing kg
Answer:
Part A. 5.36x10²³ molecules of air
Part B. 6.9kg
Explanation:
Part A.
To calculate the number of molecules of the air we need first find the number of moles of air using the equation of ideal gas law:
[tex] PV = nRT [/tex] (1)
where P: is the pressure, V: is the volume, n: is the number of moles of the gas, R: is the gas constant and T: is the temperature
[tex] n = \frac{PV}{RT} = \frac{735torr \cdot 1atm/760torr \cdot 2.35L}{0.082 Latm/Kmol \cdot (37 + 273)K} = 0.089 moles [/tex]
Now by using the Avogadro's number we can find the number of molecules of air:
[tex] number of molecules = \frac{6.022 \dot 10^{23}}{1mol} \cdot 0.089moles = 5.36 \cdot 10^{22} molecules [/tex]
Part B.
Similarly, to calculate the mass of air first we need to detemine the number of moles using equation (1):
[tex] n = \frac{PV}{RT} = \frac{1.07atm \cdot 5.0\cdot 10^{3}L}{0.082 Latm/Kmol \cdot (0.5 + 273)K} = 238.55 moles [/tex]
So, the mass of air is:
[tex] m = moles \cdot M [/tex]
where M: is the average molar mass of air
[tex] m = 238.55moles \cdot 28.98g/mol = 6.9 kg [/tex]
I hope it helps you!
The student needs to use the ideal gas law and Avogadro's number to convert the volume of air to moles and then to molecules. For the blue whale, use the ideal gas law to convert volume to moles, then multiply by the molar mass of air to get mass.
Explanation:To begin with, we first convert the volume into molecules. Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is the number of moles, R is the gas constant and T is temperature. The number of moles, n, can be found by rearranging the equation to n=PV/RT.
Substituting the given values (converting pressure to atm and volume to L, and temperature to Kelvin), we will compute n. After finding the no of moles, the number of molecules is calculated by multiplying the number of moles by Avogadro's number.
In the case of the blue whale, we again use the ideal gas law to find the number of moles of air in the lungs, then multiply by the molar mass of air (28.98 g/mol) to find the mass in grams. Finally, we then convert from grams to kilograms.
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A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from the pivot causing a ccw torque, and a force of 2.60 N is applied at the end of the rod 3.00 m from the pivot. The 2.60-N force is at an angle of 30.0o to the rod and causes a cw torque. What is the net torque about the pivot?
From the definition we have that the Torque corresponds to the Multiplication between the Force (or its respective component) and the radius of distance of the Force to the inertial turning point.
Mathematically this can be expressed,
[tex]\tau = F \times d[/tex]
Where,
F = Perpendicular component of force
d = distance from pivot point
The total sum of the torques would be equivalent to
[tex]\tau_{net} = \tau_1 +\tau_2[/tex]
According to the values given, torque 1 and 2 would be given by
[tex]\tau_1 = 6*1.2 = 7.2N\cdot m (+)[/tex]
[tex]\tau_2 = -5.2sin(30) = -7.8N\cdot m (-)[/tex]
Therefore the net Torque is
[tex]\tau_{net} = \tau_1+\tau_2[/tex]
[tex]\tau_{net} = 7.2-7.8[/tex]
[tex]\tau_{net} = -0.6N\cdot m[/tex]
Therefore the net torque about the pivot is -0.6Nm
In fast-pitch softball, a pitcher swings her arm from straight overhead in a circle, releasing the 0.196 kg ball at the bottom of the swing. If a pitcher's arm has a length of 0.984 m and mass of 11.3 kg, and the ball goes from initially at rest to a tangential speed of 29.8 m/s before release, what average torque must the pitcher apply to the ball? Model the pitcher's arm as a uniform thin rod swung about one end, and make sure you don't forget the mass of the softball (but you can ignore its shape; i.e., treat it like a point mass). Hint: you will need to make use of the equations of circular motion for constant angular acceleration that we derived back in week 2.
Answer:
560.06714 Nm
Explanation:
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation = [tex]\pi[/tex] (Half rotation)
v = Velocity of bat = 29.8 m/s
M = Mass of bat = 11.3 kg
m = Mass of ball = 0.196 kg
R = Radius of swing = 0.984 m
[tex]\omega_f=\dfrac{v}{r}\\\Rightarrow \omega_f=\dfrac{29.8}{0.984}\\\Rightarrow \omega_f=30.28455\ rad/s[/tex]
From equation of rotatational motion
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{30.28455^2-0^2}{2\times \pi}\\\Rightarrow \alpha=145.96958\ rad/s^2[/tex]
Moment of inertia is given by
[tex]I=\dfrac{1}{3}MR^2+mR^2\\\Rightarrow I=\dfrac{1}{3}11.3\times 0.984^2+0.196\times 0.984^2\\\Rightarrow I=3.83687577\ kgm^2[/tex]
Torque is given by
[tex]\tau=I\alpha\\\Rightarrow \tau=3.836875776\times 145.96958\\\Rightarrow \tau=560.06714\ Nm[/tex]
The torque the pitcher applies is 560.06714 Nm
The average torque this pitcher must apply to the softball is 560.64 Newton.
Given the following data:
Mass of ball = 0.196 kg.Mass of pitcher's arm = 11.3 kg.Length of pitcher's arm (radius) = 0.984 m.Initial speed = 0 m/s (since it starts from rest).Tangential speed = 29.8 m/s.To determine the average torque this pitcher must apply to the ball:
How to calculate the average torque.First of all, we would determine the final angular speed of the ball by using this formula:
[tex]\omega_f =\frac{v}{r} \\\\\omega_f =\frac{29.8}{0.984} \\\\\omega_f = 30.29\;rad/s[/tex]
Next, we would determine the constant angular acceleration by using the equation of circular motion:
[tex]\alpha =\frac{\omega^2_f - \omega^2_i}{2\theta} \\\\\alpha =\frac{30.29^2 - 0^2}{2\times \pi} \\\\\alpha =\frac{917.48}{2\times 3.142} \\\\\alpha =\frac{917.48}{6.284}\\\\\alpha = 146\;rad/s^2[/tex]
For the moment of inertia:
For a point mass, moment of inertia is given by this formula:
[tex]I =\frac{1}{3} Mr^2 + mr^2\\\\I =\frac{1}{3} \times 11.3 \times 0.984^2 + 0.196 \times 0.984^2\\\\I = 3.84 \;Kgm^2[/tex]
Now, we can determine the average torque:
[tex]Torque = I\alpha \\\\Torque =3.84 \times 146[/tex]
Torque = 560.64 Newton.
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An 800-kHz radio signal is detected at a point 8.5 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 0.90 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the average electromagnetic energy density at that point? (c = 3.0 x 108 m/s, μ0 = 4π × 10-7 T ∙ m/A, ε0 = 8.85 × 10-12 C2/N ∙ m2)
A. 7.2 pJ/m3
B. 10 pJ/m3
C. 3.6 pJ/m3
D. 14 pJ/m3
E. 5.1 pJ/m3
To solve this problem we need to apply the concepts related to the average electromagnetic energy density. Which is given as
[tex]U = \frac{1}{2}\epsilon_0 E^2[/tex]
Where,
\epsilon_0 = Permettivity of free space constant
E = Electric Field amplitude
Since the average electromagnetic energy density is directly proportional to the amplitude of the magnetic field then we have to
[tex]E = \frac{1}{2} (8.85*10^{-12}C^2/N\cdot m^2)(0.9V/m)^2[/tex]
[tex]E = 3.6*10^{-12}J/m^3[/tex]
[tex]E = 3.6pJ/m^3[/tex]
Therefore the correct answer is C.
Assume that the home construction industry is perfectly competitive and in long-run competitive equilibrium. It follows that: A. marginal cost exceeds long-run average total cost.B. marginal cost equals long-run average total cost.C. there will be incentive for new firms to enter the industry.D. firms in the industry enjoy economic profits.
Answer:
B. Marginal cost equals long-run average total cost.
Explanation:
The zero profit condition implies that entry continues until all firms are producing at minimum long run average total cost. Since the marginal cost curve cuts the long run average total cost curve at its minimum point, marginal cost and long run average total cost must be equal in long run equilibrium.
The deepest point in any ocean is in the Mariana Trench, which is about 11 km deep, in the Pacific. The pressure at this depth is huge, about 1.13 108 N/m2. (Take the bulk modulus of seawater to be 2.34 109 N/m2) (a) Calculate the change in volume of 0.9 m3 of seawater carried from the surface to this deepest point. (b) The density of seawater at the surface is 1.03 103 kg/m3. Find its density at the bottom
Answer:
a) ΔV = - 4.346 10⁻² , b) ρ’= 1.082 10³ kg / m³
Explanation:
The volume module is defined as the ratio of the pressure and the unit deformation, with a negative sign, for the module to be positive
B = - P / (ΔV/V)
a) The ΔV volume change
ΔV/V = -P / B
ΔV = - P V / B
ΔV = - 1.13 10⁸ 0.9 /2.34 10⁹
ΔV = - 4.346 10⁻²
b) Density at the bottom of the sea
On the surface
ρ = m / V
m = ρ V
m = 1.03 10³ 0.9
m = 0.927 10³ kg
Body mass does not change with depth
Deep down
ρ’= m / V’
ΔV = 4.346 10⁻²
[tex]V_{f}[/tex]- V₀ = 4,346 10⁻²
[tex]V_{f}[/tex] = 0.0436 + Vo
[tex]V_{f}[/tex]= -0.04346 + 0.9
[tex]V_{f}[/tex] = 0.85654 m³
ρ’= 0.927 10³ / 0.85654
ρ’= 1.082 10³ kg / m³
Using the formula for volume change under pressure and density calculations, it can be determined that the change in volume of 0.9 m³ seawater when taken to the Mariana Trench is -0.043 m³ and its density at the bottom is approximately 1072 kg/m³.
Explanation:The pressure in the ocean increases with depth due to the weight of the overlying water. This high pressure can compress the volume of the water at such depths, altering its density. As the question provides, the bulk modulus of seawater is given as 2.34 x 10⁹ N/m² and the pressure in the Mariana Trench is 1.13 x 10⁸ N/m².
(a) To find the change in volume, we can use the formula ΔV = -(PΔV/B), where P is the pressure, ΔV is the change in volume, and B is the bulk modulus. Inserting the given values, we get the change in volume to be -0.043 m³.
(b) The density of a substance is its mass divided by its volume. At the bottom of the Mariana Trench, the volume of water has decreased due to the high pressure, but its mass remains the same. Therefore, as the volume decreases, the density increases. The new density, ρ', is calculated using the formula ρ' = ρ/(1-ΔV/V), where ρ is the initial density and V is the initial volume. Substituting into this equation, we get ρ' = 1.03 x 10³ kg/m³ / (1 -(-0.043), which gives us a density at the bottom of the Mariana Trench of approximately 1072 kg/m³.
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A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0.54 m below the surface of the water. The water surface drops very slowly and its speed is approximately zero. Keep 2 decimal places in all answers. (a) Find the speed v (in m/s) at which water squirts out of the hole.
Speed can be found through the application of concepts related to potential energy and kinetic energy, for which you have
[tex]KE = PE[/tex]
[tex]\frac{1}{2}mv^2 = mgh[/tex]
Where,
m = mass
v = Velocity
g = Gravitational acceleration
h = Height
Re-arrange to find the velocity we have,
[tex]v^2 = 2gh[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2(9.8)(0.54)}[/tex]
[tex]v = 3.253m/s[/tex]
Therefore the speed at which water squirts out of the hole is .3253m/s
Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω , and the reactance of the capacitor is 473 Ω . The voltage amplitude across the capacitor is 364 V . Part A What is the current amplitude in the circuit? II = nothing A Request Answer Part B What is the impedance? ZZ = nothing Ω Request Answer Part C What two values can the reactance of the inductor have? Enter your answers in ascending order separated by a comma. XLX L = nothing Ω Request Answer Part D For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? the less value of XL the larger value of XL Request Answer Provide Feedback
Answer:
a. I = 0.76 A
b. Z = 150.74
c. RL₁ = 34.41 , RL₂ = 602.58
d. RL₂ = 602.58
Explanation:
V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω
a.
Using law of Ohm
V = I * R
I = Vc / Rc = 364 V / 473 Ω
I = 0.76 A
b.
The impedance of the circuit in this case the resistance, capacitance and inductor
V = I * Z
Z = V / I
Z = 116 v / 0.76 A
Z = 150.74
c.
The reactance of the inductor can be find using
Z² = R² + (RL² - Rc²)
Solve to RL'
RL = Rc (+ / -) √ ( Z² - R²)
RL = 473 (+ / -) √ 150.74² 77.0²
RL = 473 (+ / -) (129.58)
RL₁ = 34.41 , RL₂ = 602.58
d.
The higher value have the less angular frequency
RL₂ = 602.58
ω = 1 / √L*C
ω = 1 / √ 602.58 * 473
f = 285.02 Hz
The current amplitude in the circuit is 1.51A. The impedance of the circuit is 168Ω. The reactance of the inductor can have two values, which can be calculated using the equation XL = 2&pivL.
Explanation:The current amplitude in the circuit can be calculated using the equation:
II = Vo/R = 116V / 77.0Ω = 1.51A.
The impedance of the circuit can be calculated using the equation:
ZZ = √(R2 + (XL - XC)2) = √(77.0Ω2 + (473Ω - 364Ω)2) = 168Ω.
The reactance of the inductor can have two values. From the equation XL = 2&pivL, we can set XL = XC = 473Ω and solve for L, which gives L = XC / (2&piv) = 473Ω / (2πf) where f is the frequency of the AC voltage source.
To find the values for which the angular frequency is less than the resonance angular frequency, we need to compare the values of XL obtained for different frequencies. If the frequency is such that XL > XC, then the angular frequency is less than the resonance angular frequency.
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