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.
A 0.74-m diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 m*N which accelerates it uniformly from rest through a total of 160 revolutions in 14.0 s.
1. What is the mass of the sphere?
Answer:
Mass of the sphere is 19.2 kg
Explanation:
We have given diameter of the sphere d = 0.74 m
So radius r = 0.37 m
Initial angular velocity [tex]\omega _i=0rad/sec[/tex]
Time t = 14 sec
Angular displacement [tex]\Theta =160revolution=160\times 2\pi =1004.8rad[/tex]
From second equation of motion
[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2[/tex]
So [tex]1004.8=0\times t+\frac{1}{2}\times \alpha \times 14^2[/tex]
[tex]\alpha =10.25rad/sec^2[/tex]
Torque is given [tex]\tau =10.8Nm[/tex]
Torque is equal to [tex]\tau =I\alpha[/tex], here I is moment of inertia and [tex]\alpha[/tex] angular acceleration
So [tex]10.8=10.25\times I[/tex]
[tex]I=1.053kgm^2[/tex]
Moment of inertia of sphere is equal to [tex]I=\frac{2}{5}Mr^2[/tex]
So [tex]1.053=\frac{2}{5}\times M\times 0.37^2[/tex]
M = 19.23 kg
So mass of the sphere is 19.23 kg
A barge floating in fresh water (p = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and height H = 2.0 m. When empty the
bottom of the barge is located H0 = 0.45 m below the surface of the water. When fully
loaded with coal the bottom of the barge is located H; = 1.05 m below the surface. Part (a) Find the mass of the coal in kilograms.
Numeric : A numeric value is expected and not an expression.
m1 =
Part (b) How far would the barge be submerged (in meters) if m; = 450000 kg of coal had been placed on the empty barge? Numeric : A numeric value is expected and not an expression.
Answer:
a) [tex]\Delta m = 330000\,kg[/tex], b) [tex]h = 1.268\,m[/tex]
Explanation:
a) According the Archimedes' Principle, the buoyancy force is equal to the displaced weight of surrounding liquid. The mass of the coal in the barge is:
[tex]\Delta m \cdot g = \rho_{w}\cdot g \cdot \Delta V[/tex]
[tex]\Delta m = \rho_{w}\cdot \Delta V[/tex]
[tex]\Delta m = \left(1000\,\frac{kg}{m^{3}} \right)\cdot (550\,m^{2})\cdot (1.05\,m-0.45\,m)[/tex]
[tex]\Delta m = 330000\,kg[/tex]
b) The submersion height is found by using the equation derived previously:
[tex]\Delta m = \rho_{w}\cdot \Delta V[/tex]
[tex]450000\,kg = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (550\,m^{2})\cdot (h-0.45\,m)[/tex]
The final submersion height is:
[tex]h = 1.268\,m[/tex]
Answer:
a) m = 330000 kg = 330 tons
b) H3 = 1.268 meters
Explanation:
Given:-
- The density of fresh-water, ρ = 1000 kg/m^3
- The base area of the rectangular prism boat, A = 550 m^2
- The height of the boat, H = 2.0 m ( empty )
- The bottom of boat barge is H1 = 0.45 m of the total height H under water. ( empty )
- The bottom of boat barge is H2 = 1.05 m of the total height H under water
Find:-
a) Find the mass of the coal in kilograms.
b) How far would the barge be submerged (in meters) if m; = 450000 kg of coal had been placed on the empty barge?
Solution:-
- We will consider the boat as our system with mass ( M ). The weight of the boat "Wb" acts downward while there is an upward force exerted by the body of water ( Volume ) displaced by the boat called buoyant force (Fb):
- We will apply the Newton's equilibrium condition on the boat:
Fnet = 0
Fb - Wb = 0
Fb = Wb
Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V1 ). The expression of buoyant force (Fb) is given as:
Fb = ρ*V1*g
Where,
V1 : Volume displaced when the boat is empty and the barge of the boat is H1 = 0.45 m under the water:
V1 = A*H1
Hence,
Fb = ρ*A*g*H1
Therefore, the equilibrium equation becomes:
ρ*A*g*H1 = M*g
M = ρ*A*H1
- Similarly, apply the Newton's equilibrium condition on the boat + coal:
Fnet = 0
Fb - Wb - Wc = 0
Fb = Wb + Wc
Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V2 ). The expression of buoyant force (Fb) is given as:
Fb = ρ*V2*g
Where,
V2 : Volume displaced when the boat is filled with coal and the barge of the boat is H2 = 1.05 m under the water:
V2 = A*H2
Hence,
Fb = ρ*A*g*H2
Therefore, the equilibrium equation becomes:
ρ*A*g*H2 = g*( M + m )
m+M = ρ*A*H2
m = ρ*A*H2 - ρ*A*H1
m = ρ*A*( H2 - H1 )
m = 1000*550*(1.05-0.45)
m = 330000 kg = 330 tons
- We will set the new depth of barge under water as H3, if we were to add a mass of coal m = 450,000 kg then what would be the new depth of coal H3.
- We will use the previously derived result:
m = ρ*A*( H3 - H1 )
H3 = m/ρ*A + H1
H3 = (450000 / 1000*550) + 0.45
H3 = 1.268 m
Three point charges lie in a straight line along the y-axis. A charge of q1 = − 9.0 μC is at y = 6.0 m, and a charge of q2 = −8.0 μC is at y = − 4.0m. The net electric force on the third point charge is zero. Where is this charge located?
The third point charge must be located at the point where the net electric force on it is zero.
How can the charge be located?The net electric force on a point charge is the vector sum of the electric forces exerted on it by all the other point charges. The electric force exerted by a point charge q1 on a point charge q2 is given by Coulomb's law:
F = k * |q1| * |q2| / r^2
where k is Coulomb's constant, |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.
The direction of force will be opposite for both charges, and since net force is zero, the force due to both charges will be equal and opposite.
So, |q1| * (1/r1^2) = |q2| * (1/r2^2)
Where r1 is the distance between point charge 1 and point charge 3, r2 is the distance between point charge 2 and point charge 3.
Solving for position of point charge 3, we can say that it is located at y = 0.
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To find the position where a third charge experiences zero net force due to two other charges along the y-axis, we must consider the magnitudes and signs of all charges and use Coulomb's Law. The third charge will be or negative and situated closer to the smaller magnitude charge to balance the forces. The exact position, however, cannot be determined without the magnitude of the third charge.
Explanation:The question pertains to the concept of electric forces and charge equilibrium in Physics. We are asked to find the position where a third charge would experience a net electric force of zero due to the presence of two other point charges arranged along the y-axis. The charges are q1 = − 9.0 μC at y = 6.0 m and q2 = − 8.0 μC at y = − 4.0 m.
Let's denote the position of the third charge as y3 where the net force is zero. To ensure a net force of zero on the third charge, the electric force due to q1 must be equal in magnitude and opposite in direction to the force due to q2. Since both q1 and q2 are negative, the third charge must also be negative (so it will be repelled by both). The third charge must be placed between q1 and q2.
Likewise, since q1 is larger in magnitude than q2, the third charge must be placed closer to q2 to balance the forces (since electric force is inversely related to the square of the distance). Mathematically, we can set up the equation where the magnitude of the forces due to each charge on the third charge is equal. This investigation involves Coulomb's Law (F = k · |q1 · q3| / r^2), where F is the force between charges, k is Coulomb's constant, q3 is the third charge, and r is the distance between the charges involved.
Through setting up the equations and solving for y3, we can determine the precise location. Unfortunately, the problem does not provide the magnitude of the third charge, so while we can describe the process, we can't compute a specific answer without this information. In a real scenario though, any negative third charge would suffice, and its magnitude would not affect the balance point.
Suppose we assume that Einstein and Zweistein should compute the same total force on the sphere. Would Zweistein compute a larger or smaller value of the electric field at the sphere than Einstein
Answer:
Forces observed by both is same as both
The frames are non-acceleration
Explanation:
Zweistein = Einstein
[tex](F_B + E_B)_Z[/tex] = [tex](F_B + E_B)_E[/tex]
as [tex](F_B + E_B)_E[/tex] tend to zero (0)
[tex](E_B)_Z[/tex] < [tex](F_B)_Z[/tex] as some quantity is subtracted from the force.
According to Einstein's Theory of Relativity, the laws of physics are the same for all observers, regardless of their motion. Applying this to the calculation of electric field strength around a sphere, both Einstein and Zweistein, assuming they are not accelerating, should compute the same value of the electric field.
Explanation:If we consider the principles of physics consistently at play regardless of the observer, both Einstein and Zweistein must compute the same total force on the sphere. This is following Einstein's postulate stating the laws of physics remain the same in all inertial frames. Therefore, the electric field computed at the sphere by both Einstein and Zweistein should also be the same.
The principle of relativity implies that the laws of physics—including those of electric field strength—are the same for all observers, regardless of their speed or direction as long as they are not accelerating. Einstein's Theory of Relativity extends this to include accelerated motion and gravity.
This principle, combined with the equivalence of inertial and gravitational mass, is enough to show that the force of gravity perceived by an observer in an accelerating spacecraft must be the same as if the observer were stationary on the Earth. So, if Einsteins's Theory of Relativity applies to electric fields in the same way it does for gravitational fields, then the electric field strength calculated by Einstein and Zweistein should be the same.
To clarify, in the context of this question, the 'electric field' refers to the region around a charged particle or object where another charge placed within it would experience a force either of attraction or repulsion.
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Difference Between Photometry and Spectrophotometry ?
Answer:
Spectrophotometry is the quantitative measurement of light spectra reflection and transmission properties of materials as function of the wavelength. Photometry measures the total brightness as seen by the human eye
Explanation:
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A beam of light starts at a point 8.0 cm beneath the surface of a liquid and strikes the surface 7.0 cm from the point directly above the light source. Total internal reflection occurs for the beam. What can be said about the index of refraction of the liquid?
Answer:
Explanation:
Given that,
The light starts at a point 8cm beneath the surface
It strikes the surface 7cm directly above the light
The angle of incidence from the given distances is
Using trigonometry
Tan I = opp / adj
tan I= (7.0 cm)/(8.0 cm) = 0.875
I = arctan(0.875)
I = 41.1859
According to Snell's law
n(water) sin I = n(air) sin R
Here R =90° for total reflection to occur
nair = 1
n(water) sin I = n(air) sinR
n(water) = (1)sin 90
n(water) = 1
n(water) = 1/sin I
n(water) =1/ sin(41.1859o)
= 1.52
The refractive index index of the liquid is 1.52
To find the index of refraction of the liquid, we calculate the angle of incidence using the given distances and apply Snell's law. While specific numeric calculations aren't provided, the assumption is the liquid's refractive index is greater than air, as total internal reflection occurs.
A beam of light undergoes total internal reflection when it moves from a medium of higher refractive index to one of lower refractive index, and the angle of incidence is greater than the critical angle. In this case, we are given that a beam of light in a liquid undergoes total internal reflection. To determine the index of refraction of the liquid, we can use the information given about the distances involved to find the angle of incidence and then apply Snell's law.
First, we use the distances given to find the angle of incidence. The light source is 8.0 cm beneath the surface and strikes the surface 7.0 cm from the point directly above it. This forms a right triangle, allowing us to calculate the angle of incidence using trigonometry. The tangent of the angle of incidence (θ) is opposite over adjacent, or θ = tan⁻¹(8.0/7.0). The angle of incidence calculated is therefore tan⁻¹(8/7).
For total internal reflection to occur, the angle of incidence must be greater than the critical angle, which is defined by Snell's law as sin⁻¹(n₂/n₁) where n₂<n₁. Since we know the reflection happens at the interface with air (where n₂ = 1.00 for air), we can find the refractive index of the liquid. However, without exact values for angles given in this scenario, the specific calculation steps cannot be completed, but we understand that the liquid's refractive index is greater than 1, which is typical for most transparent materials compared to air.
A rocket is attached to a toy car that is confined to move in the x-direction ONLY. At time to = 0 s, the car is not moving but the rocket is lit, so the toy car accelerates in the +x-direction at 5.35 m/s2. At t; = 3.60 s, the rocket's fuel is used up, and the toy car begins to slow down at a rate of 1.95 m/s2 because of friction. A very particular physics professor wants the average velocity for the entire trip of the toy car to be +6.50 m/s. In order to make this happen, the physics professor plans to push the car (immediately after it comes to rest by friction) with a constant velocity for 4.50 sec. What displacement must the physics professor give the car (immediately after it comes to rest by friction) in order for its average velocity to be +6.50 m/s for its entire trip (measured from the time the rocket is lit to the time the physics professor stops pushing the car)?
Answer:
What displacement must the physics professor give the car
= 12.91 METERS
Explanation:
Check the attached file for explanation
Final answer:
To find the displacement that the physics professor must give the car, we can break down the different stages of the car's motion and use the equations of motion. By adding the displacements from each stage, we can find the total displacement of the car.
Explanation:
To find the displacement that the physics professor must give the car in order for its average velocity to be +6.50 m/s for the entire trip, we can break down the different stages of the car's motion and use the equations of motion.
1. From time t=0 to t=3.60s: The car is accelerated in the +x direction at 5.35 m/s². We can use the equation v = u + at to find the final velocity v1 at t=3.60s, where u is the initial velocity, a is the acceleration, and t is the time.
2. From time t=3.60s to t=8.10s: The car slows down at a rate of 1.95 m/s² due to friction. We can use the equation v = u + at to find the initial velocity u2 at t=3.60s, where v is the final velocity, a is the acceleration, and t is the time.
3. From time t=8.10s to t=12.60s: The physics professor pushes the car with a constant velocity for 4.50s. The average velocity during this time is +6.50 m/s. We can calculate the displacement during this time using the equation d = vt, where v is the velocity and t is the time.
By adding the displacements from each stage, we can find the total displacement of the car, which is the answer to the question.
A square current-carrying loop is placed in a uniform magnetic field B with the plane of the loop parallel to the magnetic field (see the drawing). The dashed line is the axis of rotation. The magnetic field exerts _____________ answer
Answer:
The magnetic field exerts net force and net torque on the loop.
which is a tool used by scientists to determine the composition of objects?
Answer:
A spectrometer
Explanation:
A plane electromagnetic wave varies sinusoidally at 86.0 MHz as it travels through vacuum along the positive x-direction. The peak value of the electric field is 2.30 mV/m, and it is directed along the positive y-direction. Determine the average power per unit area (the intensity) this wave carries through space. (Be careful with your units here and make sure to submit your answer in μW/cm2)
Answer:
The intensity is [tex]I = 0.0003053 \mu W/cm^2[/tex]
Explanation:
From the question we are told
The frequency of the electromagnetic wave is [tex]f = 86.0 Hz[/tex]
The peak value of the electric field is [tex]E_o = 2.30 mV/m = \frac{2.30}{1000 } = 2.30 *10^{-3} V/m[/tex]
Generally the intensity of this wave is mathematically represented as
[tex]I = c * \frac{1}{2} * \epsilon_o E^2_o[/tex]
Where c is the speed of light with value [tex]c = 3 *10^8 m/s[/tex]
[tex]\epsilon_o[/tex] is the permittivity of free space with value [tex]\epsilon _o = 8.85 *10^{-12} C^2 /Nm^2[/tex]
Substituting values into equation for intensity
[tex]I = 3.0 *10^8 * 0.5 * 8.85 *10^{-12} * 2.30*10^{-3}[/tex]
[tex]I = 3.053 *10^{-6} W/m^2[/tex]
Converting to [tex]cm^2[/tex] we divide by 10,000
[tex]I = \frac{3.053 *10^{-6}}{10000} W/cm^2[/tex]
[tex]= 3.053 *10^{-10} W/cm^2[/tex]
[tex]= 0.0003053 *10^{-6} W/cm^2[/tex]
[tex]I = 0.0003053 \mu W/cm^2[/tex]
We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in solenoid 2 is three times that in solenoid 1. How does the field B2 at the center of solenoid 2 compare to B1 at the center of solenoid 1?
Answer:
the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.
Explanation:
Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:
[tex]B=\mu_0\, \frac{N}{L} I[/tex]
Then, if we assign the subindex "1" to the quantities that define the magnetic field ([tex]B_1[/tex]) inside solenoid 1, we have:
[tex]B_1=\mu_0\, \frac{N_1}{L_1} I_1[/tex]
notice that there is no dependence on the diameter of the solenoid for this formula.
Now, if we write a similar formula for solenoid 2, given that it has :
1) half the length of solenoid 1 . Then [tex]L_2=L_1/2[/tex]
2) twice as many turns as solenoid 1. Then [tex]N_2=2\,N_1[/tex]
3) three times the current of solenoid 1. Then [tex]I_2=3\,I_1[/tex]
we obtain:
[tex]B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1[/tex]
The magnetic field at the center of solenoid 2 (B2) will be twelve times larger than the magnetic field at the center of solenoid 1 (B1), due to having four times the number of turns per unit length and three times the current.
Explanation:The magnetic field inside a solenoid is given by the formula B = µnI, where B is the magnetic field, µ (mu) is the magnetic permeability of the medium, n is the number of turns per unit length, and I is the current through the solenoid. Given solenoid 2 has twice the diameter of solenoid 1, half the length, and twice as many turns, with the current being three times that of solenoid 1, several factors will influence the magnetic field in solenoid 2 (B2) compared to solenoid 1 (B1).
The number of turns per unit length for solenoid 2 is four times that of solenoid 1, since it has twice as many turns and half the length. Additionally, the current in solenoid 2 is three times that in solenoid 1. Therefore, B2 will be twelve times larger than B1, since the magnetic field inside a solenoid is directly proportional to both the number of turns per unit length of the solenoid and the current through it (B2 = 12 * B1).
After driving a portion of the route, the taptap is fully loaded with a total of 25 people including the driver, with an average mass of 62 kgkg per person. In addition, there are three 15-kgkg goats, five 3- kgkg chickens, and a total of 25 kgkg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed
Answer:
0.4455 m
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
Total mass is
[tex]m=62\times 25+15\times 3+5\times 3+25\\\Rightarrow m=1635\ kg[/tex]
Here the spring constant is not given so let us assume it as [tex]k=36000\ N/m[/tex]
Here, the forces are balanced
[tex]mg=kx\\\Rightarrow 1635\times 9.81=36000\times x\\\Rightarrow x=\dfrac{1635\times 9.81}{36000}\\\Rightarrow x=0.4455\ m[/tex]
The springs are compressed by 0.4455 m
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach the surface of a 15-fmfm-diameter 238U238U nucleus
Answer:
[tex]\Delta V = 1.8 \times 10^7 V[/tex]
Explanation:
GIVEN
diameter = 15 fm =[tex]15 \times 10^{-15}[/tex]m
we use here energy conservation
[tex]K_{i}+U_{i} =K_{f}+U_{f}[/tex]
there will be some initial kinetic energy but after collision kinetic energy will zero
[tex]K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}[/tex]
on solving these equations we get kinetic energy initial
[tex]KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}[/tex]
[tex]KE_{i} = 35.33[/tex] J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the [tex]{238}_U[/tex] nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V
[tex]\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V[/tex]
An electron is moving at a speed of 2.20 ✕ 104 m/s in a circular path of radius of 4.3 cm inside a solenoid. The magnetic field of the solenoid is perpendicular to the plane of the electron's path. The solenoid has 25 turns per centimeter.(a) Find the strength of the magnetic fieldinside the solenoid.(b) Find the current in the solenoid.
Answer:
a) 2.90*10^-6 T
b) 0.092A
Explanation:
a) The magnitude of the magnetic field is given by the formula for the calculation of B when it makes an electron moves in a circular motion:
[tex]B=\frac{m_ev}{qR}[/tex]
me: mass of the electron = 9.1*10^{-31}kg
q: charge of the electron = 1.6*10^{-19}C
R: radius of the circular path = 4.3cm=0.043m
v: speed of the electron = 2.20*10^4 m/s
By replacing all these values you obtain:
[tex]B=\frac{(9.1*10^{-31}kg)(2.20*10^4 m/s)}{(1.6*10^{-19}C)(0.043m)}=2.90*10^{-6}T=2.9\mu T[/tex]
b) The current in the solenoid is given by:
[tex]I=\frac{B}{\mu_0 N}=\frac{2.90*10^{-6}T}{(4\pi*10^{-7}T/A)(25)}=0.092A=92mA[/tex]
To find the strength of the magnetic field inside the solenoid, we can use the formula for the magnetic field produced by a solenoid. To find the current in the solenoid, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field.
Explanation:To find the strength of the magnetic field inside the solenoid, we can use the formula for the magnetic field produced by a solenoid: B = µ0nI, where B is the magnetic field, µ0 is the permeability of free space (4π × 10-7 T·m/A), n is the number of turns per unit length, and I is the current flowing through the solenoid.
Given that the solenoid has 25 turns per centimeter and the radius of the circular path is 4.3 cm, we can find the number of turns per unit length: n = 25 × 100 cm = 2500 turns/m.
The strength of the magnetic field inside the solenoid is then: B = (4π × 10-7 T·m/A) × (2500 turns/m) × (I).
To find the current in the solenoid, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field: F = qvB, where F is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field.
Given that the velocity of the electron is 2.20 × 104 m/s and the radius of the circular path is 4.3 cm, we can find the magnetic force experienced by the electron: F = (1.6 × 10-19 C) × (2.20 × 104 m/s) × (B).
Since the magnetic force is centripetal, it can also be expressed as F = mv2/r, where m is the mass of the electron, v is its velocity, and r is the radius of the circular path.
Combining these two equations, we can find the current in the solenoid: I = (mv2/r) / (vB).
A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a .. 29.47 particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) W
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis? (d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?
Given Information:
Radius of parallel-plate capacitor = 4.10 cm = 0.041 m
Conduction current = Ic = 0.276 A
Required Information:
a) Displacement current density = JD = ?
b) Rate of change of electric field = dE/dt = ?
c) Magnetic field between plates at r = 2 cm = ?
d) Magnetic field between plates at r = 1 cm = ?
Answer:
a) Displacement current density = JD = 52.27 A/m²
b) Rate of change of electric field = dE/dt = 5.904×10¹² V/m.s
c) Magnetic field between plates at r = 2 cm = 6.568×10⁻⁷ Tesla
d) Magnetic field between plates at r = 1 cm = 3.284×10⁻⁷ Tesla
Explanation:
a) What is the displacement current density JD in the air space between the plates?
Displacement current density is given by
JD = Id/A
Where Id is the conduction current and A is the area of capacitor given by
A = πr²
A = π(0.041)²
A = 0.00528 m²
As you can notice in the diagram, conduction current has equal displacement between the capacitor plates therefore, Id = Ic
JD = 0.276/0.00528
JD = 52.27 A/m²
b) What is the rate at which the electric field between the plates is changing?
The rate of change of electric field is given by
dE/dt = JD/ε₀
Where JD is the displacement current density and ε₀ is the permittivity of free space and its value is 8.854×10⁻¹² C²/N.m²
dE/dt = 52.27/8.854×10⁻¹²
dE/dt = 5.904×10¹² V/m.s
c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis?
The induced magnetic field between the plates can be found using Ampere's law
B = (μ₀/2)*JD*r
Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷ T.m/A, JD is the displacement current density and r is the distance from the axis.
B = (4π×10⁻⁷/2)*52.27*0.02
B = 6.568×10⁻⁷ Tesla
d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?
B = (μ₀/2)*JD*r
B = (4π×10⁻⁷/2)*52.27*0.01
B = 3.284×10⁻⁷ Tesla
This physics-based question involves understanding the properties and behaviors of a parallel-plate, air-filled capacitor. Specifically, it focuses on finding the displacement current density and the rate at which the electric field between the plates is changing as the capacitor is charged.
Explanation:The question revolves around a parallel-plate capacitor, a system of two identical conducting plates separated by a distance. This specific case involves an air-filled capacitor where the plates have a certain radius, and a conduction current is defined.
First, let's address part (a): What is the displacement current density jD in the air space between the plates? The displacement current density, noted as 'jD', can be found by using Maxwell's equation which correlates the time rate of change of electric flux through a surface to the displacement current passing through the same surface.
For (b): What is the rate at which the electric field between the plates is changing? As the current is introducing charge onto the plates, the electric field (which is proportional to the charge on the capacitor's plates) will be changing. This change can be calculated as the derivative of the electric field's magnitude with respect to time.
Without specific numerical values or more information, I'm unable to provide a more detailed analysis or a concrete answer.
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Who's Faster, Sonic or Flash!?
Answer:
flash
Explanation:
Answer:
Sonic obviously
Explanation:
Sonic's body was made to run, the flashes is just a normal human body.
A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole’s symmetry axis. The rope makes a constant angle of 70.0° with respect to the vertical. What is the tangential speed of the ball?
Answer:
The tangential speed of the ball is 11.213 m/s
Explanation:
The radius is equal:
[tex]r=2.3*sin70=2.161m[/tex] (ball rotates in a circle)
If the system is in equilibrium, the tension is:
[tex]Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}[/tex]
Replacing:
[tex]\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}[/tex]
Replacing:
[tex]v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s[/tex]
Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at 40.0∘∘ to that of the first. Determine the intensity of the beam after it has passed through the second polarizer. g
Answer:
[tex]0.293I_0[/tex]
Explanation:
When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.
Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:
[tex]I_1=\frac{I_0}{2}[/tex]
Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd 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 polarizer
Here we have
[tex]\theta=40^{\circ}[/tex]
So the intensity after the 2nd polarizer is
[tex]I_2=I_1 (cos 40^{\circ})^2=0.587I_1[/tex]
And substituting the expression for I1, we find:
[tex]I_2=0.587 (\frac{I_0}{2})=0.293I_0[/tex]
What force is responsible for the
creation of the galaxies and stars?
Answer:
Gravity
Explanation:
I know because im a verified expert
Answer:
gravity
Explanation:
1. gravity is the force that pulls us to the surface of the Earth, keeps the planets in orbit around the Sun and causes the formation of planets, stars and galaxies.
2. electromagnetism is the force responsible for the way matter generates and responds to electricity and magnetism.
A bullet with a mass of 4.5 g is moving with a speed of 300 m/s (with respect to the ground) when it collides with a rod with a mass of 3 kg and a length of L = 0.25 m. The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass which is located exactly halfway along the rod. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating together. What is the magnitude of the angular velocity (in rad/s) of the rotation (with respect to the ground) immediately after the collision? You may treat the bullet as a point particle.
The magnitude of the angular velocity immediately after the collision is
[tex]\( 14.4 \, \text{rad/s} \)[/tex].
To find the angular velocity immediately after the collision, we can apply the principle of conservation of angular momentum. The angular momentum before the collision equals the angular momentum after the collision.
The angular momentum before the collision is zero since the rod is at rest. After the collision, the bullet-rod system rotates together. We can calculate the moment of inertia [tex]\( I \)[/tex] of the system and then use the equation [tex]\( L = I\omega \)[/tex], where [tex]\( L \)[/tex] is the angular momentum and [tex]\( \omega \)[/tex] is the angular velocity.
First, calculate the moment of inertia of the system:
[tex]\[ I = \frac{1}{3}mL^2 + md^2 \][/tex]
[tex]\[ I = \frac{1}{3}(3)(0.25)^2 + (0.0045)(0.25/4)^2 \][/tex]
[tex]\[ I = 0.015625 \, \text{kg m}^2 \][/tex]
Now, apply conservation of angular momentum:
[tex]\[ I\omega = mvd \][/tex]
[tex]\[ \omega = \frac{mvd}{I} \][/tex]
[tex]\[ \omega = \frac{(0.0045)(300)(0.25/4)}{0.015625} \][/tex]
[tex]\[ \omega = 14.4 \, \text{rad/s} \][/tex]
So, the magnitude of the angular velocity immediately after the collision is [tex]\( 14.4 \, \text{rad/s} \)[/tex].
Show your work and reasoning for below.
After watching a news story about a fire in a high rise apartment building, you and your friend decide to design an emergency escape device from the top of a building.
To avoid engine failure, your friend suggests a gravitational powered elevator. The design has a large, heavy turntable (a horizontal disk that is free to rotate about its center) on the roof with a cable wound around its edge. The free end of the cable goes horizontally to the edge of the building roof, passes over a heavy vertical pulley, and then hangs straight down.
A strong wire cage which can hold 5 people is then attached to the hanging end of the cable. When people enter the cage and release it, the cable unrolls from the turntable lowering the people safely to the ground.
To see if this design is feasible you decide to calculate the acceleration of the fully loaded elevator to make sure it is much less than "g."
Your friend's design has the radius of the turntable disk as 1.5m and its mass is twice that of the fully loaded elevator. The disk which serves as the vertical pulley has 1/4 the radius of the turntable and 1/16 its mass. The moment of inertia of the disk is 1/2 that of a ring.
Answer:
Explanation:
Given that,
His friend design has a turnable disk of radius 1.5m
R = 1.5m
The mass is twice the fully loaded elevator.
Let the mass of the full loaded elevator be M
Then, mass of the turn able
Mt = ½M
Radius of the disk that serves as a vertical pulley is ¼ radius of turntable and 1/16 of the mass.
Mass of pulley is
Mp = 1 / 16 × Mt
Mp = 1 / 16 × M / 2
Mp = M / 32
Also, radius of pulley
Rp = ¼ × R = ¼ × 1.5
Rp = 0.375m
The moment of inertia of the disk of a ring is
I = ½MR²
To calculate the moment of the turntable, we can use the formula
I_t = ½Mt•R²
I_t = ½ × ½M × 1.5²
I_t = 0.5625 M
I_t = 9M / 16
Also, the moment of inertia of the vertical pulley
I_p = ½Mp•Rp
I_p = ½ × (M/16) × 0.375
I_p = 0.01171875 M
I_p = 3M / 256
Let assume that, the tension in the cable between the pulley and the elevator is T1 and Let T2 be the tension between the turntable and the pulley
So, applying newton second law of motion,
For the elevator
Fnet = ma
Mg - T1 = Ma
a = (Mg-T1) / M
For vertical pulley,
The torque is given as
τ_p = I_p × α_p = (T2—T1)•r
τ_p = 3M/256 × α_p = (T2-T1)•r
For turntable
The torque is given as
τ_t = I_t × α_t = T2•r
τ_t = 9M/16 × α_t = T2•r
So, the torque are equal
τ_t = τ_p
9M/16 × α_t = 3M/256 × α_p
M cancel out
9•α_t / 16 = 3•α_p / 256
Cross multiply
9•α_t × 256 = 3•α_p × 16
Divide both sides by 48
48•α_t = α_p
α_t = α_p / 48
Then, from,
τ_t = 9M/16 × α_t = T2•r
T2•r = 9M / 16 × α_p / 48
T2•r = 3Mα_p / 16
Also, from
τ_p = 3M/256 × α_p = (T2-T1)•r
3M/256 × α_p = T2•r - T1•r
T1•r = T2•r - 3M/256 × α_p
T1•r = 3Mα_p / 16 - 3M/256 × α_p
T1•r = 3Mα_p / 16 - 3Mα_p/256
T1•r = 45Mα_p / 256
T1 = 45Mα_p / 256R
Then, from
a = (Mg-T1) / M
a = Mg - (45Mα_p / 256R) / M
a = g - 45α_p / 256
From the final answer, it is show that the acceleration is always less than acceleration due to gravity due to the subtraction of 45α_p / 256 from g
Find A and B
I need help like asap so pleasee help
Answer:
I don't understand?
Explanation:
Could you please elaborate?
A loop of wire is at the edge of a region of space containing a uniform magnetic field B⃗ . The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. That is, dAdt=−c, with c>0.Part A
The induced emf in the loop is measuredto be V. What is the magnitude B of the magnetic field that the loop was in?
Part B
For the case of a square loop of sidelength L being pulled out of the magneticfield with constant speed v, what is the rate of change ofarea c = -\frac{da}{dt}?
The question is not clear enough. So i have attached a copy of the correct question.
Answer:
A) B = V/c
B) c = Lv
Explanation:
A) we know that formula for magnetic flux is;
Φ = BA
Where B is magnetic field and A is area
Now,
Let's differentiate with B being a constant;
dΦ/dt = B•dA/dt
From faradays law, the EMF induced is given as;
E = -dΦ/dt
However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.
Thus, V = -dΦ/dt
Thus, we can now say that;
-V = B•dA/dt
Now from the question, we are told that dA/dt = - c
Thus;
-V = B•-c
So, V = Bc
Thus, B = V/c
B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:
Thus;
E =LvB or can be written as; V = LvB
Where;
V is EMF
L is length of bar
v is velocity
From the first solution, we saw that;
V = Bc
Thus, equating both of the equations, we have;
Bc = LvB
B will cancel out to give;
c = Lv
The magnitude of the magnetic field B is calculated using Faraday's Law and is equal to V/-c. For a square loop being pulled out from the field, the rate of change of area is proportional to the speed of withdrawal and the length of the side still in the field. Hence, c = Lv.
Explanation:Part A: The magnitude of the magnetic field B can be calculated using Faraday's Law of electromagnetic induction, which states that the induced voltage in a circuit is equal to the negative rate of change of magnetic flux. Mathematically, it is expressed as V = -dB/dt, where V is the induced emf, and dB/dt is the rate of change of magnetic field. Because the area is decreasing at a constant rate, the magnetic field B is V/-c.
Part B: The rate of change of the area, denoted as c = -da/dt, for a square loop of side length L being pulled out of the magnetic field with constant speed v, could be calculated using the formula c = Lv. This is because as the square loop is pulled out with a constant velocity, the rate at which its area decreases is proportional to its speed and the length of the side that is still in the field.
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A distance-time graph indicates that an object moves 20 m in 10 s and then remains at rest for 20s. What is the average speed of the object?
Answer:
v = 2 m/s
Explanation:
The object moves 20 m in 10 s initially. Then it remains at rest for 20 s. It is required to find the average speed of the object.
Fro 10 s it covers 20 m. For that time, speed is :
[tex]v=\dfrac{d}{t}\\\\v=\dfrac{20\ m}{10\ s}\\\\v=2\ m/s[/tex]
For 20 s, it was at rest means distance covered is 0. So, speed is 0.
The average speed of the object is equal to the sum of speed for 10 s and for 20 s i.e.
v = 2 m/s + 0 = 2 m/s
So, the average speed of the object is 2 m/s.
A ray of laser light travels through air and enters an unknown material. The laser enters the material at an angle of 36 degrees to the normal. The refracted angle is 27.5 degrees. If the index of refraction of air is n = 1.00, what is the index of refraction of the unknown material? g
Answer:
n = 1.27
Explanation:
just took test :)
Answer:
N= 1.27
Explanation:
An object moving with a speed of 21 m/s and has a kinetic energy of 140 J, what is the mass of the object.
Answer:2940
Explanation:
Answer:
0.635 kg
Explanation:
Recall that Kinetic energy is defined as:
K.E. = (1/2) mv²,
where
K.E = Kinetic energy = given as 140J
m = mass (we are asked to find this)
v = velocity = given as 21 m/s
Substituting the above values into the equation:
K.E. = (1/2) mv²
140 = (1/2) m (21)² (rearranging)
m = (140)(2) / (21)²
m = 0.635 kg
Unpolarized light with an average intensity of 845 W/m2 moves along the x-axis when it enters a Polarizer A with a vertical transmission axis (along the y-axis). The transmitted light then enters a second polarizer, B at an angle in the y-z plane . The light that exits the second polarizer is found to have an average intensity of 275 W/m2. What is the orientation angle of the second polarizer (B) relative to the first one (A)
Answer:
θ = 36.2º
Explanation:
When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law
I = I₀ cos² tea
The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization
I₁ = I₀ / 2
I₁ = 845/2
I₁ = 422.5 W / m²
In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of
I = 275 W / m ²
Cos² θ = I / I₁
Cos θ = √ I / I₁
Cos θ = √ (275 / 422.5)
Cos θ = 0.80678
θ = cos⁻¹ 0.80678
θ = 36.2º
This is the angle between the two polarizers
A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an angular displacement of 13 rad. What is its displacement θ at a time of t = 5.0 s?
Answer:
52 rad
Explanation:
Using
Ф = ω't +1/2αt²................... Equation 1
Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.
Since the object states from rest, ω' = 0 rad/s.
Therefore,
Ф = 1/2αt²................ Equation 2
make α the subject of the equation
α = 2Ф/t².................. Equation 3
Given: Ф = 13 rad, t = 2.5 s
Substitute into equation 3
α = 2(13)/2.5²
α = 26/2.5
α = 4.16 rad/s².
using equation 2,
Ф = 1/2αt²
Given: t = 5 s, α = 4.16 rad/s²
Substitute into equation 2
Ф = 1/2(4.16)(5²)
Ф = 52 rad.
Answer:
[tex]\theta = 52 rads[/tex]
Explanation:
The rotational kinetic equation is given by the formula:
[tex]\theta = w_{0} t + 0.5 \alpha t^{2}[/tex]..............(1)
The object is starting from rest, angular speed, w = 0 rad/s
At t = 2.5 sec, angular displacement, θ = 13 rad
Inserting these parameters into equation (1)
[tex]13 = (0 * 2.5) + 0.5 \alpha 2.5^{2}\\\alpha = \frac{13}{3.125} \\\alpha = 4.16 rad/s^{2}[/tex]
At time, t = 5.0 sec, we substitute the value of [tex]\alpha[/tex] into the kinematic equation in (1)
[tex]\theta = (0*5) + 0.5 *4.16* 5^{2}\\\theta = 52 rad[/tex]
The sound level produced by one singer is 70.8 dB. What would be the sound level produced by a chorus of 37 such singers (all singing at the same intensity at approximately the same distance as the original singer)
Answer:
86.48 dB . Ans
Explanation:
Sound level by one singer = 70.8 dB
= 7.08 B
If I be its intensity in terms of watt/m²
log ( I / 10⁻¹² ) = 7.08 ( given )
I / 10⁻¹² = 10⁷°⁰⁸
I = 10⁻¹² x 10⁷°⁰⁸
= 10⁻⁴°⁹²
= .000012022 W / m²
intensity by 37 singers
= 37 x .000012022 W /m²
= .0004448378
= 4448.378 x 10⁻⁷
intensity level
= log I / 10⁻¹² B
= log 4448.378 x 10⁻⁷ / 10⁻¹²
= log 444837800 B
log 4448378 + 2 B
= 6.6482 +2 B
8.6482 B
intensity level = 8.6482 B
= 86.48 dB . Ans
Answer:
86.48 dB
Explanation:
86.48 dB
Explanation:
sound level intensity, β = 70.8 dB
Io = 10^-12 W/m²
Let the intensity if I at the level of 70.8 dB
Use the formula
7.08 = log I + 12
log I = - 4.92
I = 1.2 x 10^-5 W/m²
There are 37 singers, so the total intensity is I'.
I' = 37 x I
I' = 37 x 1.2 x 10^-5 = 4.45 x 10^-4 W/m²
Let the sound intensity is β' in decibels.
Use the formula
β' = 86.48 dB
An object with mass 3M is launched straight up. When it reaches its maximum height, a small explosion breaks the object into two pieces with masses M and 2M. The explosion provides an impulse to each object in the horizontal direction only. The heavier of the two pieces is observed to land a distance 53 m from the point of the original launch. How far from the original launch position does the lighter piece land
Using conservation of momentum, the lighter piece in the explosion scenario lands 53 meters away from the launch point in the opposite direction in which the heavier piece landed.
The question revolves around a scenario where conservation of momentum applies. An object of mass
3M explodes into two pieces of mass M and 2M. The conservation of momentum dictates that the center of mass of the two pieces and the original object all follow the same horizontal trajectory. If the heavier piece lands 53 meters away from the point of launch, then, due to the center of mass being conserved during the explosion, the lighter piece would have landed at the same point of launch.
Utilizing the conservation law, which states that the total momentum before the explosion must be equal to the total momentum after the explosion, the pieces must move symmetrically about the center of mass of the original object. Given that the larger piece is observed to land 53 meters away from the launch point and has twice the mass of the smaller piece, the smaller piece would have landed 53 meters on the other side of the launch point, making a total separation of 106 meters between the two pieces. Therefore, the lighter piece would have landed 53 meters away in the opposite direction from where the heavier piece landed.