Answer:
option C.
Explanation:
The correct answer is option C.
Vernier Calipers is a precision instrument that is used to measure the internal or outer dimensions or the depth of the material.
Vernier Caliper consists of two jaws, the main scale, and the vernier scale.
Graduation is present on both the main scale and vernier.
To measure the width of any material it should be placed in between the jaws and reading is taken with the help of the main scale and vernier scale.
The thickness of the candle can be measured using vernier calipers precisely.
Celsius
The world's most common temperature scale is Celsius. Abbreviated C, it is virtually the same as the old centigrade scale and therefore has 100 degrees between the melting point and boiling point of water, taken to occur at 0 and 100 degrees, respectively.
Kelvin
Temperature is a measure of the thermal energy of a system. Thus cooling can proceed only to the point at which all of the thermal energy is removed from the system, and this process defines the temperature of absolute zero. The Kelvin scale, also called the absolute temerature scale, takes its zero to be absolute zero. It uses units of kelvins (abbreviated K), which are the same size as the degrees on the Celsius scale.
Fahrenheit This anachronistic temperature scale, used primarily in the United States, has zero defined as the lowest temperature that can be reached with ice and salt, and 100 degrees as the hottest daytime temperature observed in Italy by Torricelli.
Required:
A. In the equation of state for the perfect gas, pV = nRT , which of the following three temperature scales must be used?
O Celsius
O Kelvin
O Fahrenheit
Answer:
In this equation the scale used must be Kelvin
Explanation:
The absolute temperature is the value of the measured temperature relative to a scale starting at Absolute Zero (0 K or -273.15 °C). It is one of the main parameters used in thermodynamics and statistical mechanics. In the international system of units it is expressed in kelvin, whose symbol is K
The absolute temperature should always be used in the ideal gas state equation as this can only be in kelvin grade scale.
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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An alternating current is set up in an LRC circuit.
For which of the following circuit elements are the current and voltage in phase?
A) inductor only
B) resistor only
C) capacitor only
D) resistor and capacitor only
E) inductor, resistor, and capacitor
Answer:
(B) Resistor only
Explanation:
Alternating Current: These are currents that changes periodically with time.
An LRC Ac circuit is an AC circuit that contains a Resistor, a capacitor and an inductor, connected in series.
In a purely resistive circuit, current and voltage are in phase.
In a purely capacitive circuit, the current leads the voltage by π/2
In a purely inductive circuit, the current lags the voltage by π/2.
Therefore when a alternating current is set up in LRC circuit, in the resistor, the current and the voltage are in phase.
The right option is (B) Resistor only.
The circuit elements are the current and voltage in phase in:
B) Resistor only
What is Alternating current?These are flows that changes occasionally with time. A LRC AC circuit is an AC circuit that contains a Resistor, a capacitor and an inductor, associated in series. In an absolutely resistive circuit, current and voltage are in stage. In an absolutely capacitive circuit, the current leads the voltage by π/2. In an absolutely inductive circuit, the current slacks the voltage by π/2.
Thus, when an alternating current is set up in LRC circuit, in the resistor, the current and the voltage are in phase.
So, correct option is (B).
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Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1 = 12kg and diameter d1 = 60cm is at rest (f1 = 0min−1 ) and disengaged from disk 2 with mass m2 = 8kg and diameter d2 = 40cm that is rotating with a frequency of f2 = 200min−1 . When we engage the clutch both disks become connected and disk 1 is accelerated while disk 2 decelerates due to a portion of its rotational energy being used to accelerate disk 1. In full contact both disks are rotating with the same angular velocity. Calculate this final angular velocity ω and the corresponding frequency f. We are neglecting further loss of energy due to heat as a result of friction. This friction and heat is the reason why real clutches wear out over time. The moment of inertia for a solid disk can be found in the textbook.
Answer:
w = 4,786 rad / s , f = 0.76176 Hz
Explanation:
For this problem let's use the concept of angular momentum
L = I w
The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved
Initial Before sticking
L₀ = 0 + I₂ w₂
Final after coupling
[tex]L_{f}[/tex] = (I₁ + I₂) w
The moments of inertia of a disk with an axis of rotation in its center are
I = ½ M R²
How the moment is preserved
L₀ = [tex]L_{f}[/tex]
I₂ w₂ = (I₁ + I₂) w
w = w₂ I₂ / (I₁ + I₂)
Let's reduce the units to the SI System
d₁ = 60 cm = 0.60 m
d₂ = 40 cm = 0.40 m
f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz
Angular velocity and frequency are related.
w₂ = 2 π f₂
w₂ = 2π 3.33
w₂ = 20.94 rad / s
Let's replace
w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)
w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)
Let's calculate
w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)
w = 20.94 1.28 / 5.6
w = 4,786 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 4.786 / 2π
f = 0.76176 Hz
In grinding a steel knife blade (specific heat = 0.11 cal/g-c),the metal can get as hot as 400C. If the blade has a mass of 80g,what is the minimum amount of water needed at 20C if the water isnot to rise above the boiling point when the hot blade is quenchedin it?
Answer:
33 g.
Explanation:
Assuming no heat transfer can be possible except for heat exchange between water and steel, we can say that the heat lost by the knife, must be equal to the heat gained by the water.
As we have a limit for the maximum temperature of both elements (once reached a final thermal equilibrium), of 100ºC, which means that the maximum allowable change in temperature will be of 300º C for the knife, and of 80º C for the water.
Empirically , it has been showed that for a heat exchange process using only conduction, the heat needed to raise the temperature of a body, is proportional to the mass, being the proportionality constant a factor that depends on the material, called specific heat.
So, we can write the following equation:
cs*mk*Δtk = cw*mw*Δtw
Replacing by the givens of the question, we have:
0.11 cal/gºC * 80 g * 300ºC = 1 cal/gºC*mw*80ºC
Solving for mw = 2,640 cal / 80 cal/g =33 g.
When a battery, a resistor, a switch, and an inductor form a circuit and the switch is closed, the inductor acts to oppose the change in the current.
How is the time constant of the circuit affected by doubling the resistance in the circuit?
Answer:Time constant gets doubled
Explanation:
Given
L-R circuit is given and suppose R and L is the resistance and inductance of the circuit then current is given by
[tex]i=i_0\left [ 1-e^{-\frac{t}{\tau }}\right ][/tex]
where [tex]i_0[/tex] is maximum current
i=current at any time
[tex]\tau =\frac{L}{R}=time\ constant[/tex]
[tex]\tau '=\frac{2L}{R}=2\tau [/tex]
thus if inductance is doubled then time constant also gets doubled or twice to its original value.
A rocket is fired vertically upward. At the instant it reaches an altitude of 2700 m and a speed of 274 m/s, it explodes into three equal fragments. One fragment continues to move upward with a speed of 235 m/s following the explosion. The second fragment has a speed of 484 m/s and is moving east right after the explosion. What is the magnitude of the velocity of the third fragment? Answer in units of m/s.
Answer:
Explanation:
Given
initial velocity of particle u=274 m/s
one Particle moves up with velocity of v=235 m/s
and other moves u=484 m/s towards east
let [tex]v_y[/tex] and [tex]v_x[/tex] be the velocity of third Particle is Y and x direction
conserving momentum in y direction
[tex]m(274)=\frac{m}{3}\times v_y+\frac{m}{3}\times 0+\frac{m}{3}\times 235[/tex]
[tex]v=587 m/s[/tex]
Now conserving momentum in x direction
[tex]m\times 0=\frac{m}{3}\times v_x+\frac{m}{3}\times 0+\frac{m}{3}\times 484[/tex]
[tex]v_x=-484 m/s[/tex]
Net Velocity of third Particle
[tex]v^2=v_x^2+v_y^2[/tex]
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
[tex]v=\sqrt{484^2+587^2}[/tex]
[tex]v=760.80 m/s[/tex]
To find the magnitude of the velocity of the third fragment, we need to consider the conservation of momentum. The magnitude of the velocity of the third fragment is 445 m/s.
Explanation:To find the magnitude of the velocity of the third fragment, we need to consider the conservation of momentum. According to the law of conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion. The momentum of an object is the product of its mass and velocity.
Let's assume the mass of each fragment is m. Since the first fragment continues to move upward with a speed of 235 m/s and the second fragment has a speed of 484 m/s and is moving east, we can write the momentum equation as:
m(235) + m(484) + m(v3) = m(274)
Simplifying the equation, we get:
719m + m(v3) = 274m
445m = - m(v3)
- 445m = m(v3)
Dividing both sides by m, we get:
-445 = v3
Therefore, the magnitude of the velocity of the third fragment is 445 m/s.
A horizontal beam of unpolarized light is incident on a stack of three polarizing filters with their polarization axes oriented, in sequence, 30◦, 60◦ and 90◦ clockwise from the vertical. The intensity of the light emerging from the stack is measured to be 275 W/m2. What is the intensity of the emerging light (in W/m2) if the middle polarizing filter is removed?
Answer:
122.22 W/m²
Explanation:
Let the intensity of unpolarized light is Io.
from first polariser
I' = Io/2
From second polariser
I'' = I' Cos²30 = 3 Io/8
From third polariser
I''' = I'' Cos²30 = 9Io/32
According to the question
9Io/32 = 275
Io = 977.78 watt/m²
Now, from first polariser
I' = Io/2 = 977.78 / 2 = 488.89 W/m²
I'' = 488.89 x cos²60 = 122.22 W/m²
thus, the intensity of light is 122.22 W/m².
Answer:
Explanation:
Given
Intensity of light emerging out is [tex]I=275 W/m^2[/tex]
Polarizer axis are inclined at [tex]30^{\circ}] , [tex]60^{\circ}[/tex] , [tex]90^{\circ}[/tex]
If [tex]I_0[/tex] is the Intensity of Incoming light then
[tex]275=\frac{I_0}{2}\times \cos ^2{30}\times \cos^2 {30}[/tex]
as they are inclined to [tex]30^{\circ}[/tex]to each other
[tex]I_0=\frac{275}{9}\times 32[/tex]
[tex]I_0=977.77 W/m^2[/tex]
If middle Filter is removed then
[tex]I=0.5\cdot I_0\cos ^2{60}[/tex]
[tex]I=0.5\cdot 977.77\cdot \frac{1}{4}[/tex]
[tex]I=122.22 W/m^2[/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 child pulls a 15 kg sled containing a 5.0 kg dog along a straight path on a horizontal surface. He exerts a force of 55 N on the sled at an angle of 20 degrees above the horizontal. The coefficient of friction between the sled and the surface is 0.22.Calculate the word done by the child's pulling force as the system moves a distance of 7.0 m.
The work done by the child pulling the sled and dog, with a pulling force of 55 N, over a distance of 7.0 m, considering both pulling work and the work done against friction, is 59.42 J.
Explanation:The question concerns the work done by a child pulling a sled with a dog. 'Work done' in physics is calculated using the equation, work done = force x distance x cosine of the angle. The force exerted is 55 N, the distance is 7.0 m, and the angle is 20 degrees. Thus, the work done by the child's pulling force as the system moves a distance of 7.0 m, ignoring friction and because cos(20) is approximately 0.94, is calculated as: Work done = 55 N x 7.0 m x cos(20) = 55 N x 7.0 m x 0.94 = 361.3 J.
However, the total work done is reduced due to friction between the sled and the ground. The sled's total weight (15 kg sled + 5.0 kg dog = 20 kg) multiplied by gravity (9.8 m/s²) gives the normal force (20 kg * 9.8 m/s² = 196 N). Multiplying the normal force by the friction coefficient (0.22), gives the frictional force (196 N * 0.22 = 43.12 N). Hence, the work done against friction is: Work done against friction = frictional force x distance = 43.12 N x 7.0 m = 301.88 J. Therefore, the actual work done by the child equals the pull work minus the work against friction, which is 361.3 J - 301.88 J = 59.42 J.
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For years, marine scientist were mystified by sound waves detected by underwater microphones in the Pacific Ocean. These so-called T waves were among the purest sounds in nature. Eventually the researchers traced the source to underwater volcanoes whose rising columns of bubbles resonated like organ pipes. A typical T wave has a frequency of 6.8 Hz. Knowing that the speed of sound in seawaver is 1,530 m/s, determine the wavelength of a T wave.
Answer:
225 m
Explanation:
[tex]f[/tex] = Frequency of the T wave = 6.8 Hz
[tex]v[/tex] = Speed of sound in seawater = 1530 ms⁻¹
[tex]\lambda[/tex] = Wavelength of the T wave
we know that, frequency, speed and wavelength are related as
[tex]wavelength = \frac{speed}{frequency}[/tex]
[tex]\lambda = \frac{v}{f}[/tex]
Inserting the values, we get
[tex]\lambda = \frac{1530}{6.8}\\\lambda = 225 m[/tex]
Final answer:
The wavelength of a T wave with a frequency of 6.8 Hz in seawater, where the speed of sound is 1,530 m/s, is calculated to be approximately 225 meters using the formula for wave speed.
Explanation:
To calculate the wavelength of a T wave with a frequency of 6.8 Hz in the Pacific Ocean, where the speed of sound in seawater is 1,530 m/s, we can use the formula for wave speed: v = f × λ, where v is the speed of sound, f is the frequency, and λ is the wavelength.
By rearranging the formula to solve for the wavelength (λ = v / f), and substituting in the known values, we get λ = 1,530 m/s / 6.8 Hz, which calculates to approximately 225 meters. Therefore, the wavelength of a T wave with a 6.8 Hz frequency in seawater is about 225 meters.
Learning Goal: To practice Problem-Solving Strategy 40.1 for quantum mechanics problems. Suppose a particle of mass m is confined to a one-dimensional box of length L. We can model this as an infinite square well in which the particle's potential energy inside the box is zero and the potential energy outside is infinite. For a particle in its first excited state, what is the probability Prob(center20%) of finding the particle within the center 20% of the box?
Answer
The answer and procedures of the exercise are attached in the following archives.
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.
The probability of finding a particle in its first excited state within the center 20% of an infinite square well of length L is approximately 11.8%. This involves integrating the square of the wavefunction over the specific interval. The key steps are defining the potential energy and solving the Schrödinger equation.
To find the probability of locating a particle in its first excited state within the center 20% of a one-dimensional box of length L, we follow these steps:
Define the Potential Energy, V: Inside the box (0 ≤ x ≤ L), V(x) = 0. Outside the box, V(x) = ∞.
Solve the Schrödinger Equation: The normalized wavefunction for the first excited state (n=2) is ψ2(x) = √(2/L) * sin(2πx/L).
The center 20% of the box is the interval from 0.4L to 0.6L. We calculate the probability of finding the particle in this region by integrating the square of the wavefunction:
Prob(center20%) = ∫0.4L0.6L |ψ2(x)|² dx = ∫0.4L0.6L ">2/L * sin²(2πx/L) dx.
Using integration techniques, the result is:
Prob(center20%) = 2 * [0.1 - (sin(0.4π))/(π)]. This computes to approximately 0.118 or 11.8%.
Mary is an avid game show fan and one of the contestants on a popular game show. She spins the wheel, and after 5.5 revolutions, the wheel comes to rest on a space that has a $1500 value prize. If the initial angular speed of the wheel is 3.50 rad/s, find the angle through which the wheel has turned when the angular speed reaches 2.00 rad/s.
Answer:
Explanation:
Given
after 5.5 revolution wheel comes to stop
i.e. radian turned before stopping
[tex]\theta =2\pi \times 5.5 rad[/tex]
initial angular velocity [tex]\omega _0=3.5 rad/s[/tex]
[tex]\omega ^2-\omega _0^2=2\cdot \alpha \cdot \theta [/tex]
where [tex]\alpha =angular\ acceleration\ or\ deceleration[/tex]
[tex]0-(3.5)^2=2(\alpha )(2\pi \cdot 5.5)[/tex]
[tex]\alpha =-0.1772 rad/s^2[/tex]
angle turned when final angular velocity is [tex]2 rad/s[/tex]
[tex]2^2-3.5^2=2\cdot (-0.1772)(\theta )[/tex]
[tex]\theta =23.27\ radians[/tex]
For a block to move down an inclined plane what force has to be the greatest?
A. Compression
B. Normal
C. tension
D. Gravity
E. Shear
Answer:
D) True. This is what creates the body weight
Explanation:
Let's write Newton's second law for this case. For inclined planes the reference system takes one axis parallel to the plane (x axis) and the other perpendicular to the plane (y axis)
X axis
Wx -fr = ma
Y Axis
N - Wy = 0
With trigonometry we can find the components of weight
sin θ = Wₓ / W
cos θ = [tex]W_{y}[/tex] / W
Wₓ = W sin θ
[tex]W_{y}[/tex] = W cos θ
W sin θ - fr = ma
From this expression as it indicates that the body is descending the force greater is the gravity that create the weight of the body
Let's examine the answers
A False This force does not apply because it is not a spring
B) False. It is balanced at all times with the component (Wy) of the weight
C) False. For there to be a rope, if it exists you should be less than the weight component for the block to lower
D) True. This is what creates the body weight
E) False. The cutting force occurs for force applied at a single point and gravity is applied at all points
For a block to move down an inclined plane, the force of gravity must be greater than the other forces. This includes the normal force and any friction that may be present. Other forces such as compression, tension, and shear are not directly involved in this process.
Explanation:For a block to move down an inclined plane, the greatest force must be gravity. The force of gravity acts downward and causes the block to slide down the slope. This force must be stronger than the others such as the normal force (the force exerted by the plane on the block) and any friction forces that may be present.
The other forces listed (compression, tension, and shear) are not directly related to the movement of the block down the inclined plane. Compression and tension are forces that act in opposite directions either pushing (compression) or pulling (tension) an object. Shear is a force that causes materials to slide past each other and is not directly applicable to this scenario
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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 coil of wire 0.12 m long and having 340 turns carries a current of 13 A. (a) What is the magnitude of the magnetic field strength H (in A/m)? (b) Calculate the flux density B (in tesla) if the coil is in a vacuum. (c) Calculate the flux density (in tesla) inside a bar of metal positioned within the coil that has a magnetic susceptibility of 1.90 x 10-4. (d) Calculate the magnitude of the magnetization M (in A/m).
To solve this problem it is necessary to apply the concepts related to the magnetic field
the flux density and the magnitude of the magnetization.
Each of these will be tackled as the exercise is carried out, for example for the first part we have to:
Part A) Magnitude of the magnetic field
[tex]H = \frac{NI}{L}[/tex]
Where,
N = Number of loops
I = Current
L = Length
If we replace the given values the value of the magnitude of the magnetic field would be:
[tex]H= \frac{340*13}{0.12}[/tex]
[tex]H = 36833 A\cdot turns/m[/tex]
For the second and third part we will apply the concepts of density both in vacuum and positioned at a point, like this:
PARTE B) Flux density in a vacuum
[tex]B = \mu_0 H[/tex]
Where,
[tex]\mu_0 =[/tex] Permeability constant
[tex]B = (4\pi*10^{-7})(36833)[/tex]
[tex]B = 0.04628T[/tex]
PART C) To find the Flux density inside a bar of metal but the magnetic susceptibility is given
[tex]X_m = 1.9*10^{-4}[/tex]
[tex]\mu_R = 1+X_m[/tex]
[tex]\mu_R = 1.00019[/tex]
Then the flux density would be
[tex]B = \mu_0 \mu_R H[/tex]
[tex]B = (4\pi*10^{-7})(1.00019)(36833)[/tex]
[tex]B = 0.04629T[/tex]
PART D) Finally, the magnetization describes the amount of current per meter, and is given by the magnetic susceptibility, that is:
[tex]M = X_m H[/tex]
[tex]M = 1.9*10^{-4}*36833[/tex]
[tex]M = 6.996A/m[/tex]
This solution calculates the magnetic field strength, flux density and magnetization in a wire coil, as well as the flux density inside a metal bar within the coil, using Ampere's law and the formulas for magnetic field and flux densities.
Explanation:The calculations are derived from Ampere's law and the formulas for magnetic field strength and flux density. For part (a), to find the magnitude of the magnetic field strength H (in A/m), we consider the given length of the coil and the number of turns. H= nI, where n is the number of turns per unit length and I is the current. In the given case, n= 340/0.12 = 2833.33, so H = 2833.33 turns/m * 13 A = 36833 A/m.
For part (b), we need to calculate the flux density B (in T) inside the coil which is in a vacuum. Using the formula B = μH, where μ = 4π x 10-7 T. m/A is the permeability of vacuum, we find B= 4π x 10^-7 T.m/A * 36833 A/m = 4.6 x 10^-2 T.
For part (c), we need to calculate the flux density B inside a bar of metal with susceptibility χ_m = 1.90 x 10-4. In this case, B = μ(H + M), where M = χ_m H is the magnetization. We find M= 1.90 x 10^-4 * 36833 A/m = 7 A/m, so B in metal = 4π x 10^-7 T.m/A * (36833 A/m + 7 A/m) = 4.6 x 10^-2 T.
For part (d), we've already calculated the magnitude of the magnetization M to be 7 A/m.
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The buildup of plaque on the walls of an artery may decrease its diameter from 1.1 cm to 0.90 cm. The speed of blood flow was 17 cm/s before reaching the region of plaque buildup.
A). Find the speed of blood flow within the plaque region. This was 25 cm/s.
B.) Find the pressure drop within the plaque region. I got 17.808 but says I'm wrong. Am I missing a step?
Answer
Initial radius of the artery is (1.1 cm) / 2 = 0.55 cm
final radius of the artery is (0.90 cm) / 2 = 0.45 cm
initial velocity of the blood is 17 cm/s
Using equation of continuity is
A₁v₁=A₂v₂
π r₁² x v₁ = π r₂² x v₂
r₁² x v₁ = r₂² x v₂
0.55² x 17 =0.45² x v₂
v₂=25.39 cm/s
Bernoulli's equation is
[tex]P_1 - P_2 = \dfrac{1}{2}\rho (v_2^2-v_1^2)[/tex]
rho is the density of blood = 1060 kg/m^3
[tex]P_1 - P_2 = \dfrac{1}{2}\times 1060 \times (0.254^2-0.17^2)[/tex]
[tex]P_1 - P_2 =18.87\ Pa[/tex]
The student uses the continuity equation to correctly solve the first part, but for pressure drop, they need to apply Bernoulli's principle. Using the equation P1 + 1/2ρv1² = P2 + 1/2ρv2², with ρ being the blood's density and v1 and v2 the speeds, they can find the pressure drop. However, real-life complications due to blood viscosity and turbulence might affect the results.
Explanation:The subject of this question is Physics (specifically, fluid dynamics). The concept being applied here is that of continuity and Bernoulli's principle. The continuity equation for fluid states that the mass flow rate must be constant throughout the length of the pipe. This translates to: Area 1 × Speed 1 = Area 2 × Speed 2. You correctly set up this equation to find Speed 2, which is the speed of the blood flow within the plaque region.
For the pressure drop, we need to use Bernoulli's principle, which describes that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume is constant in a non-viscous, steady-flowing system. Applying Bernoulli's Equation: P1 + 1/2ρv1² = P2 + 1/2ρv2². Solving this equation gives us the pressure difference within the plaque region.
Keep in mind that you might have to use the change in blood speed, and also the density of blood within the equations to find the accurate pressure drop. Also, Bernoulli's principle applies to ideal situations and there could be changes in real-life situations due to the viscosity of blood and turbulence caused by plaque.
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In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12 cm2 is rotated in 0.040 s from a position where its plane is perpendicular to the earth’s magnetic field to a position where its plane is parallel to the field. The earth’s magnetic field at the lab location is 6.0×10−5 T. (a) What is the magnetic flux through each turn of the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?
To solve this problem it is necessary to apply the concepts related to the magnetic flow of a coil and take into account the angles for each case.
It is also necessary to delve into part C, the concept of electromotive force (emf) which is defined as the variation of the magnetic flux as a function of time.
By definition the magnetic flux is determined as:
[tex]\phi = NBA cos\theta[/tex]
Where
N = Number of loops [tex]\rightarrow[/tex] We will calculate the value for each of the spins
B = Magnetic Field
A = Cross-sectional Area
[tex]\theta =[/tex] Angle between the perpendicular cross-sectional area and the magnetic field.
PART A) The magnetic flux through the coil after it is rotated is as follows:
[tex]\phi_i = NBA cos\theta[/tex]
[tex]\phi_i = (1turns)(6*10^{-5}T)(12*10^{-4}m^2)cos(0)[/tex]
[tex]\phi_i = 7.2*10^{-8}T\cdot m^2[/tex]
PART B) For the second case the angle formed is perpendicular therefore:
[tex]\phi_f = NBA cos\theta[/tex]
[tex]\phi_f = (1turns)(6*10^{-5}T)(12*10^{-4}m^2)cos(90)[/tex]
[tex]\phi_f = 0[/tex]
PART C) The average induced emf of the coil is as follows:
[tex]\epsilon = - (\frac{\phi_f-\phi_i}{dt})[/tex]
[tex]\epsilon = -(\frac{0-7.2*10^{-8}}{0.04})[/tex]
[tex]\epsilon = 1.8*10^{-6}V[/tex]
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
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.
During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.146 kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 15.3 m/s , shatters the glass as it passes through, and leaves the window at 10.7 m/s with no change of direction. What is the direction of the impulse that the glass imparts to the baseball?
Calculate the magnitude of this impulse (a positive number).
The ball is in contact with the glass for 0.0106 s as it passes through. Find the magnitude of the average force of the glass on the ball (a positive number).
Answer:
Impulse, |J| = 0.6716 kg-m/s
Force, F = 63.35 N
Explanation:
It is given that,
Mass of the baseball, m = 0.146 kg
Initial speed of the ball, u = 15.3 m/s
Final speed of the ball, v = 10.7 m/s
To find,
(a) The magnitude of this impulse.
(b) The magnitude of the average force of the glass on the ball.
Solution,
(a) Impulse of an object is equal to the change in its momentum. It is given by :
[tex]J=m(v-u)[/tex]
[tex]J=0.146\ kg(10.7-15.3)\ m/s[/tex]
J = -0.6716 kg-m/s
or
|J| = 0.6716 kg-m/s
(b) Another definition of impulse is given by the product of force and time of contact.
t = 0.0106 s
[tex]J=F\times \Delta t[/tex]
[tex]F=\dfrac{J}{\Delta t}[/tex]
[tex]F=\dfrac{0.6716\ kg-m/s}{0.0106\ s}[/tex]
F = 63.35 N
Hence, this is the required solution.
The impulse direction is opposite to the baseball's initial direction, and its magnitude is 0.6716 kg·m/s. The average force magnitude exerted by the window on the baseball is 63.36 N.
The direction of the impulse imparted by the window to the baseball is opposite to the baseball's initial direction of motion. This is because impulse is equal to the change in momentum, and since the window slows the ball down, it is applying a force in the opposite direction of the ball's initial velocity.
To calculate the magnitude of the impulse (I), use the formula
I = change in momentum = m(vf - vi)
where m is the mass of the baseball, vf is the final velocity, and vi is the initial velocity. The mass m = 0.146 kg, vi = 15.3 m/s, and vf = 10.7 m/s.
I = (0.146 kg)(10.7 m/s - 15.3 m/s)
I = (0.146 kg)(-4.6 m/s)
I = -0.6716 kg·m/s.
The negative sign indicates the impulse is in the opposite direction of the ball's initial motion, but since the question asks for the magnitude, we take the absolute value: 0.6716 kg·m/s.
The magnitude of the average force (Favg) exerted on the ball can be found using the formula
Favg = I/t
where t is the contact time. For a contact time of 0.0106 s, we have:
Favg = 0.6716 kg·m/s / 0.0106 s
Favg = 63.36 N.
A(n) ____ is a solar system object that enters Earth's atmosphere and becomes very hot due to friction between the object and Earth's atmosphere and does not survive to reach the surface.
a. asteroid
b. meteor
c. comet
d. meteoroid
e. planetesimal
Answer:
d. meteoroid
Explanation:
An asteroid is a small rocky mass of substance that orbits around the sun. They are smaller than a planet in size but larger than a pebble sized mass called meteoroids.Sometimes a meteoroid comes close enough to the Earth and enters the Earth’s atmosphere, due to friction with the atmosphere it vaporizes and turns into a meteor appearing as a streak of light in the sky called meteor.Comets are the mass of ice and dust revolving around the sun. They keep coming closer to the sun in a helical path and when there is enough heat these get vapourised leaving behind a tail of vapour and dust.When one asteroid smashes into another breaking off small pieces which are called meteoroids if they get completely burn into ashes or vapourize when passing through the atmosphere of the earth, but if they fall as a mass of rock they are known as meteorites.A planetesimal is a solid object that arises due to the accumulation of orbiting bodies whose internal strength is dominated by self-gravity and whose orbital dynamics is not significantly affected by gas drag.Four bricks of length L, identical and uniform, are stacked on top of one another in such a way that part of each extends beyond the one beneath. Find, in terms of L, the maximum values of the following, such that the stack is in equilibrium, on the verge of falling.
To find the maximum values that four bricks of length L can be on the verge of falling while stacked on top of each other in equilibrium, we need to consider the moments of each brick and the conditions for equilibrium. The moments of each brick can be calculated by multiplying the mass of the brick by the distance from the axis of rotation, which is half the length of the brick. By setting the sum of the moments equal to zero, we can find the maximum values for the mass of each brick in terms of L.
Explanation:When four identical and uniform bricks are stacked on top of each other such that part of each brick extends beyond the one beneath, the stack is in equilibrium and on the verge of falling. To find the maximum values in terms of L, we need to consider the moment of each brick and the conditions for equilibrium. The moment is the product of the mass of the brick and the distance from the axis of rotation, which is half the length of the brick.
Let's assume the bricks have a length L and that they are stacked vertically with their centers of mass aligned. The distance from the axis of rotation to the center of mass is L/2 for each brick. The moment of the top brick is then (mass of the brick) * (L/2). The moment of the second brick is (mass of the brick) * (3L/2), considering the length of the first brick as the distance from the axis of rotation. Similarly, the moment of the third brick is (mass of the brick) * (5L/2), and the moment of the fourth brick is (mass of the brick) * (7L/2).
For the stack to be in equilibrium, the sum of the moments must be zero. Therefore, we have the equation:
(mass of the first brick) * (L/2) + (mass of the second brick) * (3L/2) + (mass of the third brick) * (5L/2) + (mass of the fourth brick) * (7L/2)
By simplifying this equation and substituting the mass of each brick with a constant value (let's say M), we can find the maximum values for the mass of each brick such that the stack is in equilibrium on the verge of falling, given a certain length L.
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With the assumption of no slipping, determine the mass m of the block which must be placed on the top of the 6.5-kg cart in order that the system period be 0.66 s. What is the minimum coefficient of static friction for which the block will not slip relative to the cart if the cart is displaced 67 mm from the equilibrium position and released?
Answer:
The minimum coefficient of static friction should be 0.62.
Explanation:
Given that,
Mass of block = m
Mass of cart = 6.5 kg
Time period = 0.66 s
Displacement = 67 mm
We need to calculate the mass of block
Using formula of time period
[tex]T=2\pi\times(\dfrac{m}{k})[/tex]
Put the value into the formula
[tex]0.66=2\pi\times(\dfrac{m+6}{600})[/tex]
[tex]m=\dfrac{0.66\times600}{4\pi^2}-6[/tex]
[tex]m=4.03\ kg[/tex]
We need to calculate the maximum acceleration of SHM
Using formula of acceleration
[tex]a_{max}=\omega^2 A[/tex]
Maximum force on mass 'm' is [tex]m\omega^2 A[/tex]
Which is being provided by the force of friction between the mass and the cart.
[tex]\mu_{s}mg \geq m\omega^2 A[/tex]
[tex]\mu_{s}\geq \dfrac{\omega^2 A}{g}[/tex]
[tex]\mu_{s} \geq (\dfrac{2\pi}{T})^2\times\dfrac{A}{g}[/tex]
Put the value into the formula
[tex]\mu_{s} \geq (\dfrac{2\pi}{0.66})^2\times\dfrac{0.067}{9.8}[/tex]
[tex]\mu_{s} \geq 0.62[/tex]
Hence, The minimum coefficient of static friction should be 0.62.
A laser beam of wavelength λ=632.8 nm shines at normal incidence on the reflective side of a compact disc. The tracks of tiny pits in which information is coded onto the CD are 1.60 μm apart.For what nonzero angles of reflection (measured from the normal) will the intensity of light be maximum? Express your answer numerically. If there is more than one answer, enter each answer separated by a comma.
To solve this problem we will apply the concepts given by the principles of superposition, specifically those described by Bragg's law in constructive interference.
Mathematically this relationship is given as
[tex]dsin\theta = n\lambda[/tex]
Where,
d = Distance between slits
[tex]\lambda[/tex] = Wavelength
n = Any integer which represent the number of repetition of the spectrum
[tex]\theta = sin^{-1} (\frac{n\lambda}{d})[/tex]
Calculating the value for n, we have
n = 1
[tex]\theta_1 = sin^{-1} (\frac{\lambda}{d})\\\theta_1 = sin^{-1} (\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_1 = 23.3\°[/tex]
n=2
[tex]\theta_2 = sin^{-1} (\frac{2\lambda}{d})\\\theta_2 = sin^{-1} (2\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_2 = 52.28\°[/tex]
n =3
[tex]\theta_2 = sin^{-1} (\frac{2\lambda}{d})\\\theta_2 = sin^{-1} (3\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_2 = \text{not possible}[/tex]
Therefore the intensity of light be maximum for angles 23.3° and 52.28°
19.55 The 8-kg uniform bar AB is hinged at C and is attached at A to a spring of constant k 5 500 N/m. If end A is given a small displace- ment and released, determine
(a) the frequency of small oscillations,
(b) the smallest value of the spring constant k for which oscillations will occur.
Answer:
Explanation:
C is center of mass of the bar ie middle point. Spring is attached with one end ie A of the bar . When this end is displaced by distance x ( small) a restoring force kx is produced which creates a torque
Torque created = kx . l /2 , which creates angular acceleration α
moment of inertia I = ml²/12
torque = I. x α
kx . L /2 = I. x α
α = L kx / 2I
a = linear acceleration
= α x L/2
= L² kx / 4I
= L²x 12 kx / 4 mL²
a = 3kx/m
This shows that motion is SHM as acceleration is proportional to displacement x .
angular frequency ω² = 3k / m
frequency f = 1/2π √ 3k/m
= (1/6.28) x √ 3x 500/8
= 2.18 Hz
Final answer:
The frequency of small oscillations is approximately 3.98 Hz, and the smallest value of the spring constant for which oscillations will occur is 0 N/m.
Explanation:
(a) To determine the frequency of small oscillations, we can use the formula: f = (1/2π) × (√(k/m)), where f is the frequency, k is the spring constant, and m is the mass of the bar. Plugging in the values, we get:
f = (1/2π) × (√(500/8)) ≈ 3.98 Hz
(b) To find the smallest value of the spring constant k for which oscillations will occur, we need to consider critical damping. Critical damping occurs when the damping force is equal to the force exerted by the spring. The formula for critical damping is c = 2√(mk), where c is the damping coefficient, m is the mass, and k is the spring constant. Since we know that the damping coefficient is 0, we can solve for k to get:
k = (c^2)/(4m). Substituting in the values, we get:
k = (0^2)/(4 x 8) = 0 N/m
A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the speaker? What is the wavelength of this sound wave as it travels through air with a speed of about 341 m/s? What is the wavelength of the same sound wave as it enters some water, where it has a speed of about 1480 m/s?
Answer:
0.003034 s
1.035 m
4.5 m
Explanation:
[tex]f[/tex] = frequency of the tone = 329.6 Hz
[tex]T[/tex] = Time period of the sound wave
we know that, Time period and frequency are related as
[tex]T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s[/tex]
[tex]v[/tex] = speed of the sound in the air = 341 ms⁻¹
wavelength of the sound is given as
[tex]\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m[/tex]
[tex]v[/tex] = speed of the sound in the water = 1480 ms⁻¹
wavelength of the sound in water is given as
[tex]\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m[/tex]
Each milligram of glucose has the same amount of energy available to do work. The series B test tubes produced more bacteria per milligram of glucose than did the series A test tubes. Assuming that each bacterium produced requires a certain amount of energy, which test tube should contain some products of glucose that still contain some "unused" energy?
Answer:
The series A test tube has some left amount of glucose left in it.
Explanation:
Let's assume that a fixed amount of glucose is synthesized, for the fixed quantity the bacteria produced in A and B be x and y respectively,
Therefore, the condition on x and y is, y > x as the no. of bacteria present in B is greater.
As a result B would require a greater amount of energy for its functioning, these energy would be derived from the already fixed amount of glucose present.
A test tube would also require the energy for its x number of bacteria, but it is less than that of B.
Therefore, there would be some unused glucose left in Test Tube Series A which has unused energy.
Assume that the Deschutes River has straight and parallel banks and that the current is 0.75 m/s. Drifting down the river, you fall out of your boat and immediately grab a piling of the Warm Springs Bridge. You hold on for 40 s and then swim after the boat with a speed relative to the water of 0.95 m/s. The distance of the boat downstream from the bridge when you catch it is______________.
Answer:
d = 142.5 m
Explanation:
This is a vector exercise. Let's calculate how much the boat travels in the 40s
d₀ = [tex]v_{b}[/tex] t
d₀ = 0.75 40
d₀ = 30 m
Let's write the kinematic equations
Boat
x = d₀ + [tex]v_{b}[/tex] t
x = 0 + [tex]v_{h}[/tex] t
At the meeting point the coordinate is the same for both
d₀ + [tex]v_{b}[/tex] t = [tex]v_{h}[/tex] t
t ( [tex]v_{h}[/tex] - [tex]v_{b}[/tex]) = d₀
t = d₀ / ( [tex]v_{b}[/tex]- [tex]v_{h}[/tex])
The two go in the same direction therefore the speeds have the same sign
t = 30 / (0.95-0.775)
t = 150 s
The distance traveled by man is
d = [tex]v_{h}[/tex] t
d = 0.95 150
d = 142.5 m
The distance of the boat downstream when you catch it is 60 meters.
Explanation:To find the distance of the boat downstream when you catch it, we can use the equation d = vt, where d is the distance, v is the velocity, and t is the time.
Given that the current of the river is 0.75 m/s and you hold onto the piling for 40 seconds, the distance drift with the current is:
ddrift = (0.75 m/s)(40 s) = 30 mAfter you start swimming with a speed of 0.95 m/s and catch up to the boat, the distance you swim against the current is equal to the distance the boat drifts:
dswim = ddrift = 30 mTherefore, the total distance downstream when you catch the boat is:
d = ddrift + dswim = 30 m + 30 m = 60 mHence, the total distance downstream when you catch the boat is 60 m.
A solid plate, with a thickness of 15 cm and a thermal conductivity of 80 W/m·K, is being cooled at the upper surface by air. The air temperature is 10°C, while the temperatures at the upper and lower surfaces of the plate are 650 50 and 60°C, respectively. Determine the convection heat transfer coefficient of air at the upper surface and discuss whether the value is reasonable or not for force convection of air.
To determine the convection heat transfer coefficient of air at the upper surface, we need to know the surface area of the plate which is not provided in the given question.
Explanation:The convection heat transfer coefficient of air at the upper surface can be determined using Newton's law of cooling. According to Newton's law of cooling, the rate of heat transfer through convection is directly proportional to the temperature difference between the solid surface and the surrounding fluid and the surface area of the solid.
Therefore, the heat transfer rate can be calculated using the formula:
Q = h * A * (Ts - T∞)
Where:
In this case, we are given the following values:
To find the surface area, we need to know the dimensions of the plate. Once we have the surface area, we can solve for the convection heat transfer coefficient using the given formula. However, the surface area is not provided in the question, so we cannot determine the convection heat transfer coefficient without that information.