Answer:
When the midpoint of the magnet is in the plane of the loop, the induced current in the loop viewed from above is essentially zero.
Explanation:
At some point far away from the plane of the loop of the wire, there's no flux and hence, no induced emf, and no current in the loop. As the magnet descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil. The time changing flux induces some EMF and subsequently some induced current.
As the magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves, the time rate of change of total flux increases, so the EMF goes up. Note that the field lines above and below the magnet's midpoint point in the same direction from the the North pole to the south.
As the bar moves through the plane of the coil, the North end first, flux is added by the motion of the magnet and flux is removed by the motion of South end.
At some point, the bar reaches the middle of the coil. At this point, the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half. Therefore, at this point the EMF is zero. And hence, there is no induced current observed in the wire at this point.
Hope this Helps!!!
When viewed from the above, there is no induced current observed in the wire at this point due to zero EMF.
The given problem is based on the concept of electric flux and induced emf. The electric flux is dependent on induced emf, which is produced to current. Since, point far away from the plane of the loop of the wire, there's no flux and hence, no induced emf, and no current in the loop.
As the magnet descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil. The time changing flux induces some EMF and subsequently some induced current.When magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves, the time rate of change of total flux increases, so the EMF goes up.
Note: - The field lines above and below the magnet's midpoint point in the same direction from the the North pole to the south.
As the bar moves through the plane of the coil, the North end first, flux is added by the motion of the magnet and flux is removed by the motion of South end.
When the bar reaches the middle of the coil, then at this point the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half. Therefore, at this point the EMF is zero. And hence, there is no induced current observed in the wire at this point.
Thus, we conclude that when viewed from the above, there is no induced current observed in the wire at this point due to zero EMF.
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Two point charges of +2.50 x 10^-5 C and -2.50 x 10^-5 C are separated by 0.50m. Which of the following describes the force between them?
A): 90 N, repulsive
B): 90 N, attractive
C): 23 N, attractive
D): 45 N, attractive
If one could answer this ASAP, it will be greatly appreciated
Answer:
Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?
Using Coulomb's law:
F = 1/(4πE) x Q1 x Q2/ r^2
Where
k= 1/(4πE) = 9 x 10^9Nm2/C2
Therefore,
F = 9x 10^9 x 2.50 x 10^-5 x2.50 x
10^-5/. ( 0.5)^2
F= 5.625/ 0.25
F= 22.5N approximately
F= 23N.
To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.
Hence To = 23N, attractive. C ans.
Thanks.
Two friends need to displace a sculpture. They place it on a light board 1.5 m long. One friend lifts at one end with a force of 300 N, and the other friend lifts the opposite end of the board with a force of 160 N. The board is perfectly horizontal. Where along the board is the sculpture center of gravity located? (Tip: The horizontal board has no mass).
To solve this problem we will apply the concepts related to the moment. This is defined as the distance of the force to the center of mass (in this case) so the sum of moments must be 0 to maintain the equilibrium condition. The total weight of the sculpture should be
The sculpture must weigh (300 + 160)N = 460 N
From the 300N Force applied, making the summation of moments and fulfilling the condition of static balance, we have to:
[tex]\sum M = 0[/tex]
[tex]160N(1.5m)-460Nx = 0[/tex]
[tex]x = \frac{160(1.5)}{460}[/tex]
x = 0.52m
Therefore the center of gravity is located to 0.52m
Unpolarized light with intensity I0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at 42.0 ∘ to that of the first. Determine the intensity of the beam after it has passed through the second polarizer.
Answer:
[tex]0.276I_0[/tex]
Explanation:
When unpolarized light passes through a polarizer, only the component of the light vibrating in the direction parallel to the axis of the polarizer passes through: therefore, the intensity of light is reduced by half, since only 1 out of 2 components passes through.
So, after the first polarizer, the intensity of light passing through is:
[tex]I_1=\frac{I_0}{2}[/tex]
Where [tex]I_0[/tex] is the initial intensity of the unpolarized light.
Then, the light (which is now polarized) passes through the second polarizer. Here, the intensity of the light passing through the second polarizer is given by Malus Law:
[tex]I_2=I_1 cos^2 \theta[/tex]
where:
[tex]\theta[/tex] is the angle between the axes of the two polarizers
In this problem the angle is
[tex]\theta=42^{\circ}[/tex]
So the intensity after of light the 2nd polarizer is:
[tex]I_2=I_1 (cos 42^{\circ})^2=\frac{I_0}{2}(cos 42^{\circ})^2=0.276I_0[/tex]
The intensity of the beam after it has passed through the second polarizer should be 0.276 I0.
Calculation of the intensity:
Since
after the first polarizer, the intensity of light should be
I1 = I0/2
Here,
Io should be initial intensity of the unpolarized light
Now
The intensity should be
= I0/2(cos 42)
= 0.276 I0
Here theta be 42 degrees
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An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 atm, but inside the pressure is held at 1.0 atm. What net force pushes out on the window
Answer:
Net forces which pushes the window is 30342.78 N.
Explanation:
Given:
Dimension of the office window.
Length of the window = [tex]3.1[/tex] m
Width of the window = [tex]2.1[/tex] m
Area of the window = [tex](3.1\times 2.1) = 6.51\ m^2[/tex]
Difference in air pressure = Inside pressure - Outside pressure
= [tex](1.0-0.954)[/tex] atm = [tex]0.046[/tex] atm
Conversion of the pressure in its SI unit.
⇒ [tex]1[/tex] atm = [tex]101325[/tex] Pa
⇒ [tex]0.046[/tex] atm = [tex]0.046\times 101325 =4660.95[/tex] Pa
We have to find the net force.
We know,
⇒ Pressure = Force/Area
⇒ [tex]Pressure=\frac{Force }{Area}[/tex]
⇒ [tex]Force =Pressure\times Area[/tex]
⇒ Plugging the values.
⇒ [tex]Force =4660.95\times 6.51[/tex]
⇒ [tex]Force=30342.78[/tex] Newton (N)
So,
The net forces which pushes the window is 30342.78 N.
An airplane flying at a distance of 8.3 km from a radio transmitter receives a signal of intensity 15 μW/m2. What is the amplitude of the (a) electric and (b) magnetic component of the signal at the airplane? (c) If the transmitter radiates uniformly over a hemisphere, what is the transmission power?
Answer:
a) = 0.106 V/m
b) = 354.52 pT
c) = 6.5 W
Explanation:
Given,
Intensity, I = 15 μW/m²
Distance, d = 8.3 km
E = √(2Iμc)
where
c = speed of light = 2.99*10^8
μ = permeability of free space = 4π*10^-7
E = √(2 * 15*10^-6 * 4*3.142*10^-7 * 2.99*10^8)
E = √(30*10^-6 * 1.257*10^-6 * 2.99*10^8)
E = √1.13*10^-2
E = 0.106 V/m
B = E/c
B = 0.106 / 2.99*10^8
B = 354.52 pT
P = IA
Where A = 1/2 * 4πr²
A = 0.5 * 4 * 3.142 * (8.3*10^3)²
A = 4.33*10^8
P = I A
P = 15*10^-6 * 4.33*10^-8
P = 6495 W
P = 6.5 kW
Therefore
a) = 0.106 V/m
b) = 354.52 pT
c) = 6.5 W
Answer:
a) Amplitude of the electric field, E = 0.1063 V/m
b) Magnetic field, B = = 3.545 x 10⁻¹⁰ Tesla
c) Transmission power, P = 6492.73 W
Explanation:
Intensity, I = 15 μW/m²
Distance, R = 8.3 km = 8300 m
a) The electric field amplitude
Relationship between the amplitude and intensity of a signal is given by:
[tex]I = \frac{E^{2} }{2c \mu}[/tex]..........(1)
c = 3 * 10⁸ m/s
μ = 4 π * 10⁻⁷
Substitute these values into equation (1)
[tex]15 * 10^{-6} = \frac{E^{2} }{2* 3 * 10^{8} * 4\pi * 10^{-7} }\\15 * 10^{-6} * 240\pi = E^{2}\\ E^{2} = 0.0113097\\E = 0.1063 V/m[/tex]
b)Amplitude of the magnetic field component at the airplane
B = E / c
B = = 0.1063 / (3 x 10⁸ )
B = = 3.545 x 10⁻¹⁰ Tesla
3) The transmission power:
Intensity, [tex]I = \frac{Power}{Area}[/tex]
P = I A
Area of the hemisphere, A = 2πR²
A = 2π * 8300²
A = 432848635.8116 m²
Power, P = 15 * 10⁻⁶ * 432848635.8116
P = 6492.73 W
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of the magnetic field B changes in time according to the equation B ( t ) = ( 3.75 T ) + ( 3.05 T s ) t + ( − 6.95 T s 2 ) t 2 If the radius of the wire loop is 0.220 m , find the magnitude E of the induced emf in the loop when t = 5.71 s .
To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,
[tex]\Phi = BA Cos \theta[/tex]
Here,
[tex]\theta[/tex] = Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,
[tex]\epsilon= -N \frac{d\Phi }{dt}[/tex]
[tex]\epsilon = -N \frac{(BAcos\theta)}{dt}[/tex]
[tex]\epsilon = -NAcos\theta \frac{dB}{dt}[/tex]
[tex]\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)[/tex]
[tex]\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )[/tex]
At time [tex]t = 5.71s[/tex], Induced emf is,
[tex]\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))[/tex]
[tex]\epsilon = 10.9V[/tex]
Therefore the magnitude of the induced emf is 10.9V
10.90V
Explanation:
Faraday's law of induction states that the electromotive force (E) produced in a magnetic field is directly proportional to the change in flux, ΔΦ, inversely proportional to change in time, Δt. i.e
E = - N ΔΦ / Δt ----------------(i)
Where;
N = proportionality constant called the number of coils in the wire.
With a small change in time, equation (i) could be re-written as follows;
E = - N δΦ / δt --------------(ii)
Also, the magnetic flux, Φ, is given as follows;
Φ = BA cos θ --------------------(iii)
Where;
B = magnetic field
A = cross sectional area of the wire
θ = angle between the field and the cross-section of the wire
Substitute equation (iii) into equation (ii) as follows;
E = - N δ(BAcosθ) / δt
E = - NAcosθ δ(B) / δt -----------------------(iv)
From the question;
B(t) = (3.75 T) + (3.05 T/s)t + (− 6.95 T/s²) t² -------------------(v)
Substitute equation (v) into equation (iv) as follows;
E = - NAcosθ δ[(3.75 T) + (3.05 T/s)t + (− 6.95 T/s²) t²] / δt -----(vi)
Solve equation (vi) by taking derivative as follows;
E = - NAcosθ [3.05 − (2)6.95 t]
E = - NAcosθ [3.05 − 13.9t] ----------------(vii)
Solve for A using the following relation;
A = πr² -----------------(viii)
Where;
r = radius of the wire loop = 0.220m
π = 3.142
Substitute these values into equation (viii) as follows;
A = 3.142 x 0.220²
A = 0.152m²
Now substitute A = 0.152m², N = 1 (a single coil), θ = 19.5° and t = 5.71s into equation (vii)
E = - (1) (0.152)cos(19.5)° [3.05 − 13.9(5.71)]
E = - (1) (0.152)(0.94) [3.05 − 13.9(5.71)]
E = 10.90V
Therefore, the induced EMF in the loop is 10.90V
The shape of a planet's orbit is influenced by gravity.
Answer:
True
Explanation:
Gravity is a very important force. Every object in space exerts a gravitational pull on every other, and so gravity influences the paths taken by everything traveling through space. It is the glue that holds together entire galaxies. It keeps planets in orbit.
What is DNA and what is its role in life? List DNA’s four nucleotide bases.
DNA is a molecule that stores and transmits genetic information, and its four nucleotide bases are adenine, thymine, guanine, and cytosine. It plays a crucial role in the growth, development, and functioning of all living organisms, and its understanding has revolutionized various fields of science and medicine.
What is DNA?DNA (Deoxyribonucleic acid) is a molecule that contains genetic information that determines the characteristics and traits of all living organisms. It is a double-stranded helix structure made up of nucleotides, which are composed of a sugar, a phosphate group, and a nitrogenous base.
The role of DNA is to store and transmit genetic information from one generation to the next. It contains the instructions that are necessary for the growth, development, and functioning of all living organisms. DNA is responsible for determining the physical and behavioral characteristics of an individual, including eye color, hair color, height, and susceptibility to diseases.
The four nucleotide bases in DNA are adenine (A), thymine (T), guanine (G), and cytosine (C). These bases pair with each other to form the rungs of the DNA ladder, with A always pairing with T and G always pairing with C.
The sequence of nucleotide bases in DNA is the code that determines the genetic information of an individual. This code is read and translated by cells to synthesize proteins, which are the building blocks of life. DNA is essential for the survival and reproduction of all living organisms.
Therefore, Adenine, thymine, guanine, and cytosine are the four nucleotide bases in DNA, a molecule that stores and transmits genetic information. All living things' growth, development, and functioning depend on it, and our growing understanding of it has revolutionized many disciplines in science and medicine.
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A budding magician holds a 2.24-mW laser pointer, wondering whether he could use it to keep an object floating in the air with the radiation pressure. This might be an idea for a new trick! Assuming the laser pointer has a circular beam 3.00 mm in diameter and the magician rigs up a totally reflecting sail on which to shine the laser, what is the maximum weight the magician could suspend with this technique?
Answer:
W = 1.49 10⁻¹¹ kg
Explanation:
For this problem, let's use Newton's equation of equilibrium
F - W = 0
F = W (1)
Strength can be found from the definition of pressure
P = F / A
F = P A
The radiation pressure for a reflective surface is
P = 2 I / c)
We substitute in equation 1
2 I / c A = W
The intensity is defined by the ratio of the power between the area
I = P / A
P = I A
We substitute
2 P / c = W
W = 2 2.24 10-3 / 3 108
W = 1.49 10⁻¹¹ kg
Workshop:The analysis of a simple pendulum assumed that the mass was a particle, with no size. A realistic pendulum is a small, uniform diskof massMand radiusRat the end of a massless string, withLbeing the distance from the pivot to the center of the sphere.Find an expression for the period of this pendulum.SupposeM
The expression for the period of a pendulum with a small, uniform sphere is [tex]\[ T = 2\pi \sqrt{\frac{2R^2 + 5L^2}{5gL}} \][/tex].
The fractional change in the period with respect to the simple pendulum case is [tex]\[ \frac{\Delta T}{T_0} = \frac{\sqrt{\frac{2R^2 + 5L^2}{5gL}} - \sqrt{\frac{L}{g}}}{\sqrt{\frac{L}{g}}} \][/tex]
Let's analyze the realistic pendulum with a small, uniform sphere of mass M and radius R at the end of a massless string, with L being the distance from the pivot to the center of the sphere.
(a) Expression for the period of the pendulum:
The formula for the period T of a simple pendulum is given by:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
For a solid sphere rotating about an axis through its center (which is the case here), the moment of inertia I is given by:
[tex]\[ I = \frac{2}{5} M R^2 \][/tex]
Using the parallel axis theorem to relate the moment of inertia with respect to the pivot to the moment of inertia of a point mass at the end of the pendulum:
[tex]\[ I_{\text{pivot}} = I_{\text{center}} + M \cdot L^2 \][/tex]
Substitute[tex]\(I_{\text{center}} = \frac{2}{5} M R^2\):[/tex]
[tex]\[ I_{\text{pivot}} = \frac{2}{5} M R^2 + M \cdot L^2 \][/tex]
Now, substitute this into the formula for the period of a simple pendulum:
[tex]\[ T = 2\pi \sqrt{\frac{I_{\text{pivot}}}{MgL}} \][/tex]
[tex]\[ T = 2\pi \sqrt{\frac{\frac{2}{5} M R^2 + M \cdot L^2}{MgL}} \][/tex]
Simplify the expression:
[tex]\[ T = 2\pi \sqrt{\frac{2R^2}{5gL} + \frac{L}{g}} \][/tex]
Combine the terms under the square root:
[tex]\[ T = 2\pi \sqrt{\frac{2R^2 + 5L^2}{5gL}} \][/tex]
So, the expression for the period T of the pendulum with a small, uniform sphere is given by:
[tex]\[ T = 2\pi \sqrt{\frac{2R^2 + 5L^2}{5gL}} \][/tex]
(b) Fractional change in the period with respect to the simple pendulum case:
Let [tex]\(T_0\)[/tex] be the period of the simple pendulum (point mass at the end). The fractional change[tex](\(\frac{\Delta T}{T_0}\))[/tex] is given by:
[tex]\[ \frac{\Delta T}{T_0} = \frac{T - T_0}{T_0} \][/tex]
Substitute the expressions for [tex]\(T\) and \(T_0\):[/tex]
[tex]\[ \frac{\Delta T}{T_0} = \frac{2\pi \sqrt{\frac{2R^2 + 5L^2}{5gL}} - 2\pi \sqrt{\frac{L}{g}}}{2\pi \sqrt{\frac{L}{g}}} \][/tex]
Simplify:
[tex]\[ \frac{\Delta T}{T_0} = \frac{\sqrt{\frac{2R^2 + 5L^2}{5gL}} - \sqrt{\frac{L}{g}}}{\sqrt{\frac{L}{g}}} \][/tex]
This is the fractional change in the period with respect to the simple pendulum case.
Complete question is :
The analysis of a simple pendulum assumed that the mass was a particle, with no size. A realistic pendulum is a small, uniform sphere of mass M and radius R at the end of a massless string, with L being the distance from the pivot to the center of the sphere. (a) Find an expression for the period of this pendulum. (b) What is the fractional change in the period with respect to the simple pendulum case?
To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.
The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit--a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass M. For all parts of this problem, where appropriate, use G for the universal gravitational constant.
Find the orbital speed v for a satellite in a circular orbit of radius R. (Express the orbital speed in terms of G, M, and R),
Answer:
v =√ G m M / r
Explanation:
For this exercise we will use Newton's second law
F = m a
where the acceleration is centripetal
a = v² / r
force is the universal force of attraction
F = G m M / r²
we substitute
G m M / r² = m v² / r
v² = G m M / r
v =√ G m M / r
Final answer:
The orbital speed of a satellite in a circular orbit of radius R around a planet of mass M is given by v = sqrt(G * M / R).
Explanation:
When a satellite is in a circular orbit around a planet, it experiences a centripetal acceleration directed towards the center of the planet. This centripetal acceleration is provided by the gravitational force of the planet. Newton's second law states that the force acting on an object is equal to its mass multiplied by its acceleration. Therefore, we can equate the gravitational force to the mass of the satellite multiplied by its centripetal acceleration to find the relationship between the orbital speed and the radius of the orbit.
The gravitational force acting on the satellite is given by the formula F = G * (m * M) / R^2, where G is the universal gravitational constant, m is the mass of the satellite, M is the mass of the planet, and R is the radius of the orbit. The centripetal acceleration is given by the formula a = v^2 / R, where v is the orbital speed. Equating the force and the mass multiplied by acceleration, we can solve for v:
F = m * a => G * (m * M) / R^2 = m * v^2 / R => v^2 = G * M / R => v = sqrt(G * M / R)
Therefore, the orbital speed of a satellite in a circular orbit of radius R around a planet of mass M is given by v = sqrt(G * M / R).
Beryllium-8 is an unstable isotope and decays into two α particles, which are helium nuclei with mass 6.68×10−27kg . This decay process releases 1.5×10−14J of energy. For this problem, let's assume that the mass of the Beryllium-8 nucleus is just twice the mass of an α particle and that all the energy released in the decay becomes kinetic energy of the α particles.If a Beryllium-8 nucleus is at rest when it decays, what is the speed of the α particles after they are released?
Answer:
1498502.2 m/s
Explanation:
mass of the helium = 6.68×10−27kg
energy released during the decay = 1.5×10−14J
since all the energy released is converted to kinetic energy of the alpha particle then particle will have kinetic energy = 0.5 mh v²
1.5×10−14J = 2 ( 0.5 mh v²) where mh is mass of helium
1.5×10−14J / ( 6.68×10−27kg) = v²
v = 1498502.2 m/s
Answer:
v = 1498502.2 m/s
Explanation:
All energy that was released is been converted to kinetic energy of the alpha particle, particle having a kinetic energy of = 0.5 mh v²
We are given the following as;
mass of the helium = 6.68×10−27kg
decay process release energy of = 1.5×10−14J
Calculating the speed of alpha particle after the release in this equation, we have
1.5×10−14J = 2 ( 0.5 mh v²)
Substituting the value of mh in the equation, we have
v² = 1.5×10−14J / ( 6.68×10−27kg)
To eliminate the square root, we introduce square root to both sides
√v² = √1.5×10−14J / ( 6.68×10−27kg)
v= 1498502.2 m/s
A circular loop of radius 0.10 m is rotating in a uniform magnetic field of 0.20 T. Find the magnetic flux through the loop when the plane of the loop and the magnetic field vector are at an angle of 30 degree with each other.
The magnetic flux through the given rotating circular loop is 0.0054 Weber (Wb) when it is at an angle of 30 degrees with the uniform magnetic field.
Explanation:The magnetic flux through a loop can be calculated by the formula: Φ = B.A.cosθ. In this formula, B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field vector and the normal to the loop. Here B = 0.20 T, the radius of the loop r = 0.10 m (so the area A = πr² = π*0.10² = 0.0314 m²), and the angle θ = 30 degrees. Using these values in the formula gives us Φ = 0.20 T * 0.0314 m² * cos(30) = 0.0054 Wb. Therefore, the magnetic flux through the rotating circular loop is 0.0054 Wb.
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Final answer:
The magnetic flux through a circular loop of radius 0.10 m at a 30-degree angle to a uniform magnetic field of 0.20 T is approximately 0.00544 T•m².
Explanation:
The question involves calculating the magnetic flux through a rotating circular loop in the presence of a magnetic field, utilising concepts from electromagnetism in physics. The magnetic flux (Φ) can be determined using the formula Φ = B • A • cos(θ), where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the plane of the loop and the magnetic field vector.
To find the flux through the loop with a radius of 0.10 m that is at a 30-degree angle to a uniform magnetic field of 0.20 T, first, calculate the area of the loop (A = π • r²), which gives A = π • (0.10 m)² = 0.0314 m². Then, using the angle of 30 degrees, cos(θ) = cos(30°) = √3/2. The magnetic flux is therefore, Φ = (0.20 T) • (0.0314 m²) • (√3/2) ≈ 0.00544 T•m².
Every time your cats paw hits her toy string, it swings away from her. How does this demonstrate Newton’s third law of motion?
Answer:
This demonstrates Newton's Third Law because for every action there is an equal and opposite force.
Explanation:
The cat's paw hitting her toy string would be an action therefore when is swings to and from her it it creating and equal and opposite force. In other words, the cat hits it and when the string comes back to her it's coming back the same force just opposite of the direction she hit.
Jaquan found 2 footballs in his closet to practice throwing long distances. Both footballs are of equal size. However, one has a mass of 1 kg and the other is 2 kg. Which football will travel the farthest distance when thrown and why? Or will they travel the same distance?
Answer:
the football that has a mass of 1 kg will travel the farthest distance when thrown.
Explanation:
Jaquan found 2 footballs in his closet to practice throwing long distances.
The both footballs are of equal size but different mass.
The Newton's Second Law of Motion is used to determine the how far the distance the ball will travel when force is being applied to it.
The second law of motion posits that the rate of change of momentum of a body is directly proportional to the force applied. This implies that the force on the object (football) is equal to the mass of the object multiplied by its acceleration.
Let say the mass of the football with 1 kg = A
the mass of the football wih 2 kg = B
It is obvious and clear that the mass of the football will determine how far the distance of the football will be when force is applied to it by kicking the ball.
Thus; the football that has a mass of 1 kg will travel the farthest distance when thrown.
Two long, straight wires, one above the other, are seperated by a distance d = 1.53 cm and are parallel to the x−axisx−axis. Let the +y−axis+y−axis be in the plane of the wires in the direction from the lower wire to the upper wire. Each wire carries current 34 A in the +x−direction+x−direction. Find the magnetic force on a negative point charge 79 μCμC moving with velocity 3.49×105 m/s in the +y-direction, when the charge is:
Answer:
a) [tex]F_x = F_y = F_x = 0 \ \ N[/tex]
b) [tex]F_x= - 0.05718 \ N ; F_y = 0 \ N ; F_z= 0 \ N[/tex]
Explanation:
Given that:
q = -79 μC
q = - 79×10⁻⁶ C
d = 1.53 cm
d = 0.0153 m
I = 34 A
v = 3.49×10⁵ m/s
[tex]\bar{V} = v \hat{j}[/tex]
The force on charge q is given by
[tex]\bar{ F } = q (\bar {v} * \bar{B})[/tex]
a) At midway (A) , the B will be :
[tex]\bar{B} =\bar{B_1} + \bar{B_2}[/tex]
[tex]B_1[/tex] and [tex]B_2[/tex] will be equally the same in respect to their magnitude but opposite direction at point A.
So; [tex]\bar{|B_1|} = |B_2| = \frac{\mu_oI}{2 \pi d/2}[/tex]
where [tex]|B| =0[/tex]
∴ [tex]\bar {|F|} = 0[/tex]
∴ [tex]F_x = F_y = F_x = 0 \ \ N[/tex]
b)
At a distance d/2 cm above the upper wire:
[tex]\bar{B} =\bar{B_1} + \bar{B_2}[/tex]
where:
[tex]B_1 =B_2 = \frac{\mu_o I }{2 \pi ( d + d/2 )}[/tex] ; upward to the plane of paper
∴ [tex]\frac{\mu_o I}{2 \pi} \ \ [\frac{2}{3/d} +\frac{2}{d}] \ \ \hat{k}[/tex]
B = [tex]\frac{\mu_o I}{ \pi \ d} \ \ [\frac{1}{3} +1] \ \ \hat{k}[/tex]
B = [tex]\frac{\mu_o I}{ \pi \ d} \ [\frac{4}{3} ] \ \hat{k}[/tex]
B = [tex]\frac{7 \pi * 10^{-7} *34}{\pi*0.0153}*\frac{4}{3} \ \hat{k}[/tex]
B = 2.074 × 10⁻³T [tex]\hat {k}[/tex]
∴ [tex]\bar {F} = q ( \bar{v} + \bar {B})[/tex]
[tex]\bar {F} = q ( \bar{v} + B)j*k\\\\\bar {F} = q ( \bar{v} + B) \bar {i}[/tex]
[tex]F = -79*10^{-6}*3.49*10^5*2.074*10^{-3}[/tex]
[tex]F = - 0.05718 \ N \ \hat {i}[/tex]
[tex]F_x= - 0.05718 \ N ; F_y = 0 \ N ; F_z= 0 \ N[/tex]
If you swim the same distance in less time, are you going faster or slower?
Please help me
Answer:
You are swimming faster
Explanation:
Since you are covering the same amount of distance in less time, you must be going faster. If you were going slower, you would cover the same amount of distance in a longer amount of time.
If you swim the same distance in less time, you are going faster.
Speed can be defined as the distance covered by an object per unit time and it is usually measured in meter per seconds (m/s).
Mathematically, speed is given by the formula;
[tex]Speed = \frac{distance}{time}[/tex]
From the above formula, we can deduce that speed is inversely proportional to time and directly proportional to distance. Thus, the faster the speed of an object, the lesser the time to cover a particular distance and vice-versa.
In this context, you are going faster when you swim the same distance in less time.
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The position vector of a particle is directed along the positive y axis. What is the direction of the net force acting on the particle if the net torque is directed along the negative x direction?
a. Posivite z direction
b. Posivite x direction
c. negative x direction
d. negative y direction
e. negative z direction
Answer:
(e) The direction of the net force is negative Z direction.
Explanation:
Let the direction of the particle (positive y) represents direction of current
Let direction of the net torque (negative x) represents direction of the field
Let direction of net force be direction of force
Apply Fleming's right hand rule, which states that when the thumb, the middleave finger and forefinger are held mutually at right angle to each other, with the thumb pointing in the direction of current (particle, positive y ), and the forefinger pointing in the direction of the field (torque, negative x), then the middle finger will be pointing in the direction of the force.
To find direction of the net force:
Y direction is already represented, negative X direction is also represented and finally force will be in Z - direction
Since we know that torque = force x position of particle
Torque is negative, thus force will be negative.
Therefore, the direction of the net force is negative Z direction
In the year 2120, when we have a colony on the moon, an engineer brings an old grandfather clock with her. She knows the clock’s pendulum has a length of 1.0 m and the moon’s gravity is 1.62 m/s^2. If she winds the clock when the time shows 12:00, how many Earth minutes have elapsed when the clock face reads 12:12? Round your answer to 1 decimal place for entry into eCampus. Do not enter units. Example: 12.3
4.9 Earth minutes have elapsed when the clock face on the moon reads 12:12.
To determine how many Earth minutes have elapsed when the grandfather clock on the moon shows 12:12, follow these steps:
Step 1: Calculate the period of the pendulum on the moon
The period (T) of a pendulum is given by the formula:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
where (L) is the length of the pendulum (1.0 m) and (g) is the acceleration due to gravity on the moon (1.62 m/s²).
Plugging in the values:
[tex]\[ T_{\text{moon}} = 2\pi \sqrt{\frac{1.0}{1.62}} \][/tex]
[tex]\[ T_{\text{moon}} = 2\pi \sqrt{0.617} \][/tex]
[tex]\[ T_{\text{moon}} \approx 2\pi \times 0.786 \][/tex]
[tex]\[ T_{\text{moon}} \approx 4.937 \text{ seconds} \][/tex]
Step 2: Calculate the period of the pendulum on Earth
For comparison, the period of the pendulum on Earth with [tex]\(g_{\text{earth}} = 9.81 \text{ m/s}^2\)[/tex] is:
[tex]\[ T_{\text{earth}} = 2\pi \sqrt{\frac{1.0}{9.81}} \][/tex]
[tex]\[ T_{\text{earth}} = 2\pi \sqrt{0.102} \][/tex]
[tex]\[ T_{\text{earth}} \approx 2\pi \times 0.319 \][/tex]
[tex]\[ T_{\text{earth}} \approx 2.006 \text{ seconds} \][/tex]
Step 3: Calculate the number of periods in 12 minutes on the moon
12 minutes on the moon clock is:
[tex]\[ 12 \text{ minutes} \times 60 \text{ seconds/minute} = 720 \text{ seconds} \][/tex]
The number of periods in 720 seconds is:
[tex]\[ \frac{720}{T_{\text{moon}}} = \frac{720}{4.937} \approx 145.9 \text{ periods} \][/tex]
Step 4: Convert the number of periods to Earth seconds
Since each period on Earth takes approximately 2.006 seconds:
[tex]\[ \text{Earth seconds} = 145.9 \times 2.006 \approx 292.6 \text{ seconds} \][/tex]
Step 5: Convert Earth seconds to Earth minutes
[tex]\[ \text{Earth minutes} = \frac{292.6}{60} = 4.9 \text{ minutes} \][/tex]
Conclusion
When the clock face on the moon reads 12:12, approximately 4.9 Earth minutes have elapsed.
One end of a string 4.32 m long is moved up and down with simple harmonic motion at a frequency of 75 Hz . The waves reach the other end of the string in 0.5 s. Find the wavelength of the waves on the string. Answer in units of cm.
To solve this problem, we will apply the concepts related to the kinematic equations of linear motion, which define speed as the distance traveled per unit of time. Subsequently, the wavelength is defined as the speed of a body at the rate of change of its frequency. Our values are given as,
[tex]\text{Length of the string} = L = 4.32 m[/tex]
[tex]\text{Frequency of the wave} = f = 75 Hz[/tex]
[tex]\text{Time taken to reach the other end} = t = 0.5 s[/tex]
Velocity of the wave,
[tex]V = \frac{L}{t}[/tex]
[tex]V = \frac{4.32 m}{0.5s}[/tex]
[tex]V = 8.64m/s[/tex]
Wavelength of the wave,
[tex]\lambda = \frac{V}{f}[/tex]
[tex]\lambda = \frac{8.64m/s}{75Hz}[/tex]
[tex]\lambda = 0.1152m[/tex]
[tex]\lambda = 11.52cm[/tex]
Therefore the wavelength of the waves on the string is 11.53 cm
11.52cm
Explanation:The velocity, v, of a wave undergoing a simple harmonic motion is related to the wavelength, λ, and frequency, f, of the wave as follows;
v = f x λ -----------------(i)
But;
The velocity is also given as the ratio of the length, l, of the body producing the wave to the time taken, t, to undergo the motion. i.e
v = [tex]\frac{l}{t}[/tex] --------------(ii)
Now, substitute the value of v in equation (ii) into equation (i) as follows;
[tex]\frac{l}{t}[/tex] = f x λ ----------------(iii)
From the question,
l = 4.32m
t = 0.5s
f = 75Hz
Substitute these values into equation (iii) as follows;
[tex]\frac{4.32}{0.5}[/tex] = 75 x λ
Make λ subject of the formula;
λ = [tex]\frac{4.32}{0.5*75}[/tex]
λ = 0.1152m
Convert the value to cm by multiplying by 100
λ = 0.1152 x 100cm = 11.52cm
Therefore, the wavelength of the waves on the string is 11.52cm
A rocket takes off from Earth's surface, accelerating straight up at 31.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 85.9 kg, including his space suit. (Assume the rocket's initial motion parallel to the y-direction. Indicate the direction with the sign of your answer.)
Answer:
N = 3522.759 N in the upward direction
Explanation:
acceleration of the rocket, a = 31.2 m/s²
mass of the astronaut, m = 85.9 kg
g = 9.81 m/s²
Since the rocket takes off from Earth's surface, it moves upward and the equation, according to the second law of motion, is given by:
N - mg = ma
N - (85.9*9.81) = (85.9* 31.2)
N - 842.679 = 2680.08
N = 2680.08 + 842.679
N = 3522.759 N in the upward direction
Since the normal force is positive, it is in the upward direction.
To calculate the normal force acting on the astronaut, subtract the net force from the weight of the astronaut. The normal force is directed downwards. Normal Force = -1829.06 N. The negative sign indicates that the normal force is directed downward.
Explanation:The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, because the astronaut is accelerating upwards, the normal force will be less than the weight of the astronaut. We can calculate the normal force using the formula:
Normal Force = Weight - Net Force
where:
Weight = mass × acceleration due to gravity (9.8 m/s²)
Net Force = mass × acceleration (31.2 m/s²)
Plugging in the values, we get:
Weight = 85.9 kg × 9.8 m/s² = 841.42 N
Net Force = 85.9 kg × 31.2 m/s² = 2670.48 N
Therefore, the normal force acting on the astronaut is:
Normal Force = 841.42 N - 2670.48 N = -1829.06 N
The negative sign indicates that the normal force is directed downward.
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A square loop and a circular loop are each made from the same length of wire. Each loop carries the same amount of current. The loops are placed in the same uniform magnetic field. What is the ratio of the maximum torque on the square loop to the maximum torque on the circular loop? Question 2 options: π/4 2/π None of the other answers are correct. 4/π π/2
Explanation:
Let l is the length of the wire. such that, side of square is l/4. Area of square loop is, [tex]A_s=\dfrac{l^2}{16}[/tex]
Radius of the circular loop,
[tex]l=2\pi r\\\\r=\dfrac{l}{2\pi}[/tex]
Area of the circular loop,
[tex]A_a=\pi r^2\\\\A_a=\pi \times (\dfrac{l}{2\pi})^2\\\\A=\dfrac{l^2}{4\pi}[/tex]
Torque in magnetic field is given by :
[tex]\tau=NIAB\sin\theta[/tex]
It is clear that, [tex]\tau\propto A[/tex]
So,
[tex]\dfrac{\tau_s}{\tau_a}=\dfrac{A_s}{A_a}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\dfrac{l^2}{16}}{\dfrac{l^2}{4\pi}}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\pi}{4}[/tex]
So, the ratio of maximum torque on the square loop to the maximum torque on the circular loop is [tex]\pi :4[/tex].
Light travels through a liquid at 2.25e8 m/s. What is the index of refraction of the liquid?
Answer:
Refractive index = 1.33
Explanation:
Refractive index is of a material explains how light passes through a medium. It is the ratio of speed of light in vacuum to the speed of light in the medium.
We know the speed of light in water is [tex] 2.25* 10^8 [/tex] and there is a change iñ speed of light in every medium it passes (except when it passes through vacuum). The change in speed of light is known to be = [tex] 2.99*10^8[/tex]
Given speed of light = 2.25*10^8 m/s
Formula for refractive index:
[tex]n = \frac{c}{v}[/tex]
Where,
change in speed of light, c= [tex] 2.99*10^8[/tex]
speed of light, v = 2.25*10^8 m/s
[tex] Refractive index, n = \frac{2.99*10^8}{2.55*10^8} [/tex]
= 1.328
= 1.33
The liquid is water, and the index of refraction of water is 1.33
As we know,
Speed of light, [tex]v = 2.25\times 10^8 \ m/s[/tex]Change in speed of light, [tex]c = 2.99\times 10^8[/tex]The refractive index will be:
→ [tex]n = \frac{c}{v}[/tex]
By substituting the values,
[tex]= \frac{2.99\times 10^8}{2.55\times 10^8}[/tex]
[tex]= 1.33[/tex]
Thus the approach above is right.
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You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.17 mm and place your screen 8.95 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.57 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength λ expressed in nanometers?
Answer:
laser's wavelength λ = 597.4 nm
Explanation:
Given:
Slit spacing, d = 1.17mm
Tenth bright fringe y = 4.57cm
Distance from slits, D = 8.95m
n = 10
λ = (d * y) / (D * n)
λ = (1.17x10⁻³ * 4.57x10⁻²) / (8.95 x 10)
λ = 5.3469x10⁻⁵ / 8.95x10¹
λ = 0.5974 x 10⁻⁵⁻¹
λ = 0.5974 x 10⁻⁶ m
λ = 597.4 x 10⁻⁹ m
λ = 597.4 nm
A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of refraction in the plastic? A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of refraction in the plastic? The angle of incidence is equal to the angle of refraction. The angle of incidence is greater than the angle of refraction. The angle of incidence is less than the angle of refraction. The two angles cannot be compared without additional information.
Answer:
The angle of incidence is greater than the angle of refraction
Explanation:
Refraction occurs when a light wave passes through the boundary between two mediums.
When a ray of light is refracted, it changes speed and direction, according to Snell's Law:
[tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]
where :
[tex]n_1[/tex] is the index of refraction of the 1st medium
[tex]n_2[/tex] is the index of refraction of the 2nd medium
[tex]\theta_1[/tex] is the angle of incidence (the angle between the incident ray and the normal to the boundary)
[tex]\theta_2[/tex] is the angle of refraction (the angle between the refracted ray and the normal to the boundary)
In this problem, we have a ray of light passing from air into clear plastic. We have:
[tex]n_1=1.00[/tex] (index of refraction of air)
[tex]n_2=1.50[/tex] approx. (index of refraction in clear plastic)
Snell's Law can be rewritten as
[tex]sin \theta_2 =\frac{n_1}{n_2}sin \theta_1[/tex]
And since [tex]n_2>n_1[/tex], we have
[tex]\frac{n_1}{n_2}<1[/tex]
And so
[tex]\theta_2<\theta_1[/tex]
Which means that
The angle of incidence is greater than the angle of refraction
(8c5p80) Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thrust) of 3480 N, the craft descends at constant speed; if the engine provides only 2349 N, the craft accelerates downward at 0.39 m/s2. What is the weight of the landing craft in the vicinity of Callisto's surface
Answer:
The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.
Explanation:
The engine of the craft provides an upward thrust of [tex]3480 N[/tex] so that the space craft descends at a constant speed.
This implies that the net force on the space craft is zero.
The upward thrust will be equal to the downward gravitational pull by Callisto.
So the weight of the craft near the vicinity will be 3480 N.
A proton traveling along the x-axis enters a region at x = 0 where the x-component of the electric field is given by E = Ao/x1/2. How much work is done on the proton by the electric field as it moves along the x-axis from x = 0 to x = 7 m?
.Answer:
The value of the work done is [tex]\bf{ 5.29 qA_{0}}[/tex].
Explanation:
When a charged particle having charge [tex]q[/tex] is moving through an electric field [tex]E[/tex], the net force ([tex]F[/tex]) on the charge is
[tex]F = qE~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
and the work done ([tex]W[/tex]) by the particle is
[tex]W = \int\limits^x_0 {F} \, dx ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Given, [tex]E = \dfrac{A_{0}}{x^{1/2}}[/tex].
Substitute the value of electric field in equation (1) and then substitute the result in equation (2).
[tex]W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}[/tex]
Which statements describe the wave? Check all that
apply.
The wave is a mechanical wave.
The wave is an electromagnetic wave.
The wave moves energy through matter.
The wave moves eney through space without
matter,
The wave transfers energy parallel to the motion of
the wave.
The wave transfers energy perpendicular to the
motion of the wave.
Answer:the wave is an electromagnetic wave.
Explanation:
Answer:
A: The wave is a mechanical wave.
C: The wave moves energy through matter.
F: The wave transfers energy perpendicular to the motion of the wave.
Hope it works!
Explanation:
Charge can be transferred several
ways. In the clothes dryer example
given or in the experiment using the
paper and plastic spoon, charge is
transferred by_____
Answer: B
Explanation:
Friction
Answer:
Friction
Explanation:
Leading edge flaps can be used to decrease (or eliminate) the leading edge suction peak at a desired lift coefficient. When airfoils are designed for cruise performance, however, a better strategy is to design an airfoil that produces the correct lift with no suction peak using a cambered airfoil (i.e. without including leading edge flaps). To see that this is possible, we will consider the NACA 44XX airfoils. Also, p is the location of the maximum camber and is second digit/10. Apply thin airfoil theory to answer the following questions: (a) Determine the angle of zero lift for the 44XX airfoils (b) Determine the angle at which the suction peak is eliminated. We will call this the design angle of attack for the 44XX airfoils (c) What is the design lift coefficient for the 44XX airfoils (i.e. the lift coefficient at the design angle of attack)?
For the NACA 44XX series airfoils, using thin airfoil theory, the angle of zero lift is determined to be -1.6 degrees. The design angle of attack, which eliminates the suction peak, coincides with this angle. The design lift coefficient is 0.04, equivalent to the maximum camber of the airfoil.
The question provided revolves around the aerodynamic properties of the NACA 44XX series airfoils, specifically in terms of their performance at cruise without the need for leading edge flaps. Given the NACA numbering system, where the second digit represents the location of the maximum camber (p) as a fraction of the chord, we can determine several characteristics for the 44XX airfoils.
(a) The angle of zero lift for the 44XX airfoils can be found using thin airfoil theory, which tells us that for a symmetrical airfoil (e.g., NACA 0012), the angle of zero lift is 0 degrees. For cambered airfoils, the angle of zero lift becomes negative and is proportional to the camber. Specifically, the angle of zero lift is -4p degrees for cambered airfoils according to thin airfoil theory. Therefore, for the 44XX series with p value of 0.4 (from the second digit), the angle of zero lift is -4 * 0.4 = -1.6 degrees.(b) The design angle of attack, at which the suction peak is eliminated, would ideally coincide with the angle of zero lift, as this is where the airfoil is designed to operate without a strong suction peak. Thus, for the NACA 44XX airfoils, the design angle of attack would also be -1.6 degrees.(c) The design lift coefficient (CL design) for the NACA 44XX airfoils at the design angle of attack can be approximated using the lift curve slope for a thin airfoil, which is typically about 2pi per radian. Considering the zero angle of lift coefficient for a 4% cambered airfoil is equivalent to the maximum camber (which is 0.04 for this series), the lift coefficient at the design angle of attack would be the lift coefficient at zero angle of attack plus the product of the lift curve slope and the design angle of attack in radians, albeit the design angle of attack for these airfoils is the angle of zero lift, resulting in a design lift coefficient equivalent to the camber value, which is 0.04 or 0.4 in non-dimensional terms.
(a) Angle of zero lift: 0 degrees. (b) Design angle: ~1.376°. (c) Design lift coefficient: ~0.047. Utilized thin airfoil theory.
To address these questions, we can utilize thin airfoil theory, which provides approximate solutions for the aerodynamic characteristics of airfoils at small angles of attack. Let's go through each question:
(a) Determine the angle of zero lift for the NACA 44XX airfoils:
The angle of zero lift, also known as the zero-lift angle of attack [tex](\(\alpha_{L=0}\))[/tex], can be calculated using thin airfoil theory. For symmetrical airfoils like the NACA 44XX series, the zero-lift angle of attack is simply 0 degrees.
(b) Determine the angle at which the suction peak is eliminated. We will call this the design angle of attack:
The suction peak is eliminated when the airfoil is operating at its design lift coefficient [tex](\(C_{L_{\text{design}}}\))[/tex]. In thin airfoil theory, the lift coefficient is directly proportional to the angle of attack [tex](\(\alpha\))[/tex]. For cambered airfoils, this design lift coefficient is achieved at an angle of attack where the camber is aligned with the oncoming flow, resulting in no suction peak.
For NACA 44XX airfoils, the maximum camber is located at [tex]\(p = \frac{4}{10} = 0.4\)[/tex]. The angle of attack at which the camber aligns with the flow is called the camber line slope angle [tex](\(\alpha_{camber}\))[/tex]. For NACA 44XX series airfoils, the camber line slope angle can be approximated by:
[tex]\[\alpha_{camber} \approx 0.06 \times p = 0.06 \times 0.4 = 0.024 \text{ radians}\][/tex]
To convert this to degrees:
[tex]\[\alpha_{camber} \approx 0.024 \times \frac{180}{\pi} \approx 1.376^\circ\][/tex]
So, the design angle of attack where the suction peak is eliminated is approximately [tex]\(1.376^\circ\).[/tex]
(c) What is the design lift coefficient for the NACA 44XX airfoils?
At the design angle of attack, the lift coefficient [tex](\(C_L\))[/tex] can be calculated using thin airfoil theory. For symmetrical airfoils, [tex]\(C_L\)[/tex] is given by:
[tex]\[C_L = 2\pi \alpha\][/tex]
Substituting the design angle of attack [tex]\(\alpha_{camber}\)[/tex], we get:
[tex]\[C_{L_{\text{design}}} = 2\pi \times 1.376^\circ = 0.047\][/tex]
So, the design lift coefficient for the NACA 44XX airfoils is approximately [tex]\(0.047\).[/tex]