Answer:
0.48 W
Explanation:
Given that:
the mass of the steel ball = 0.0367 kg
C = 0.270 N
g (acceleration due to gravity) = 9.8
Now;
At Terminal Velocity Weight is balance by drag force
mg =Cv
Making v the subject of the formula:we have:
[tex]v = \frac {mg}{C}[/tex]
[tex]v = \frac {0.0367*9.8}{0.270}[/tex]
v = 1.332 m/s
Thus, the Rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity is:
P=Fv
P = mgv
[tex]P = (0.0367*9.8)*1.332[/tex]
P=0.48 W
Final answer:
To find the rate at which gravitational energy is converted to thermal energy at terminal velocity, equate the drag force to the gravitational force on the ball and solve for power as the product of force and velocity.
Explanation:
The question asks for the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity in a vat of corn syrup. To find this rate, we must recognize that at terminal velocity, all the gravitational force acting on the ball (minus any buoyant force, which is ignored here) is converted into thermal energy due to viscous drag. The viscous drag force [tex]F_{drag}[/tex] is given by [tex]F_{drag[/tex] = -Cv, where C is the drag coefficient, and v is the velocity of the ball. The gravitational force acting on the ball can be calculated using [tex]F_{gravity[/tex] = mg, where m is the mass of the ball, and g is the acceleration due to gravity. At terminal velocity, [tex]F_{drag[/tex] = [tex]F_{gravity[/tex], hence the power (rate of energy conversion) P can be found as P = Fv = mgv. Assuming standard gravity (9.81 m/s2), the mass of the ball (0.0367 kg), and solving for v using [tex]F_{drag[/tex], one can determine the rate of gravitational energy conversion to thermal energy.
Arc welding uses electric current to make an extremely hot electric arc that can melt metal. The arc emits ultraviolet light that can cause sunburn and eye damage if a welder is not wearing protective gear. Why does the arc give off ultraviolet light
Answer: The arc gives off ultraviolet light as a result of the high temperature of the arc which can be as high as above 3000°c.
Explanation: arc welding employs or uses electric current to generate heat for the purpose of joining metals. Metals being joined usually have high melting points above 3000°c, This high melting points of metals means the welding arc needs to attain a higher temperature to be able to join the metals. In the process of attaining that temperature needed to join the metals the arc gives of ultraviolet light (UV).
5.19 Identify the force present and explain whether work is being performed in the following cases: (a) You lift a pencil off the top of a desk. (b) A spring is compressed to half its normal length.
Work in physics is done when a force causes displacement in the direction of the force. Lifting a pencil off a desk and compressing a spring are both examples of work being done because in both cases there is a force and a displacement.
Explanation:In physics, work is defined as the product of force, displacement, and the cosine of the angle between them. It's represented by the formula W = Fd cos(θ). Let's apply this definition to two scenarios:
Lifting a pencil: When you lift a pencil off the top of a desk, the force you apply is against gravity, causing an upward displacement. Work is being performed since there is a displacement in the direction of the force.Compressing a spring: When a spring is compressed to half its normal length, work is done on the spring. This work is negative because the force applied is in the opposite direction of the spring's natural expansion.Considering the examples from the given reference, we understand that work requires both a force and a displacement. No work is done if either the force or the displacement is zero, or if the force is perpendicular to the direction of the displacement.
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The reaction converting glycerol to glycerol-3-phosphate is energetically unfavorable by 9.2 kJkJ. The reaction of ATP with water to yield ADP, hydrogen phosphate ion, and hydrogen ion is energetically favorable by 30.5 kJkJ. These reactions are coupled so that the overall process is energetically favorable. What is ΔGΔGDelta G for the overall process?
Answer:
- 21.3 kJ
Explanation:
For the favorable reaction, ΔG₁ = + 9.2 kJ (ΔG > 0 for unfavorable reaction)
For the favorable reaction, ΔG₂ = - 30.5 kJ (ΔG < 0 for favorable reaction)
For the overall reaction, ΔG = ΔG₁ + ΔG₂ = + 9.2 kJ + (- 30.5 kJ)
ΔG = + 9.2 kJ - 30.5 kJ = - 21.3 kJ
Lindsay is standing in the middle of the aisle of a bus that’s at rest at a stop light. The light turns green and the bus speeds up. Without grabbing or hanging on to anything, Lindsay manages to remain stationary with respect to the floor. While the bus is speeding up, the net force on Lindsay points in which direction?
a. Toward the front of the bus b. There is no net force on Lindsay c. Toward the back of the bus
Answer:
b. There is no net force on Lindsay.
Explanation:
Since Lindsay manages to stay stationary then this means that either there is no force acting on Lindsay or all the forces acting on Lindsay are balanced and produces no net force on Lindsay with respect to the bus.
Answer:
No net force acting on Lindsay.
Explanation:
Solution:-
- Lets assume the mass of the bus to be = M
- Assume mass of Lindsay = m
- The initial speed of bus and lindsay was vi = 0 m/s
- We will consider the system ( Lindsay + Bus ) to be isolated with no fictitious unbalanced forces acting on the system.
- Since, the system can be isolated the application of principle of conservation of linear momentum is completely valid.
- We will assume the direction of bus moving to be positive.
- The principle of conservation of momentum states:
Pi = Pf
vi*(m+M) = m*vL + M*vb
Where,
vL : The velocity of Lindsay after bus starts moving
vb : The velocity with which bus moves
Since, vi = 0 m/s.
0 = m*vL + M*vb
vL = - ( M / m )*vb
- The lindsay moves in opposite direction of motion of bus. We will apply the Newton's second law of motion on Lindsay:
Impulse = F*t = m*( vL - vi )
Impulse = F*t = m*( - ( M / m )*vb - 0 )
Impulse = F*t = -M*vb
F = -M*vb/t
- We see that a force ( F ) acts on Lindsay as soon as the bus starts moving. The negative sign shows the direction of force which is opposite to the motion of bus. So the force acts on Lindsay towards the back of the bus.
- However, if we consider the system of ( bus + Lindsay ), the net force exerted on Lindsay remains zero as the impulse force ( F ) becomes an internal force of the system and is combated by the friction force ( Ff ) between Lindsay's shoes and the aisle floor.
- The magnitude of Friction force ( Ff ) is equivalent to the impulse created by force ( F ) by Newton's third Law of motion. Every action has its equal but opposite reaction. Hence,
Fnet = F + Ff
Fnet = F - F
Fnet = 0
- However, The force ( F ) created by the rate of change of momentum of bus must be less than Ff which is equivalent to weight of Lindsay:
Ff = m*g*us
Where,
us: The coefficient of static friction.
- So for forces to remain balanced, with no resultant force then:
Ff ≥ F
m*g*us ≥ M*ab
us ≥ (M/m)*(ab/g)
- The coefficient of static friction should be enough to combat the acceleration of bus ( ab ).
A toroidal coil has a mean radius of 16 cm and a cross-sectional area of 0.25 cm2; it is wound uniformly with 1000 turns. A second toroidal coil of 750 turns is wound uniformly over the first coil. Ignoring the variation of the magnetic field within a toroid, determine the mutual inductance of the two coils.
Answer:
Explanation:
Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .
magnetic field in a torroid B = μ₀ n I , n is number of turns per unit length and I is current .
B = 4π x 10⁻⁷ x (1000 / 2π x .16 )x 1 ( current = 1 A)
flux in the secondary coil
= B x area of face of coil x no of turns of secondary
= 4π x 10⁻⁷ x (1000 /2π x .16 ) .25 x 10⁻⁴ x 750
= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹
2343.75 x 10⁻⁸
= 23.43 x 0⁻⁶ H.
.
Answer:
2.5 x 10^-5 henry
Explanation:
The mutual inductance between the toroids is same.
mean radius of the toroid, r = 16 cm = 0.16 m
Area of crossection, A = 0.25 cm²
Number of turns in the first toroid, N1 = 1000
Number of turns in the second toroid, N2 = 750
The formula for the mutual inductance is given by
[tex]M =\frac{\mu_{0}N_{1}N_{2}A}{l}[/tex]
Where, l is the length
l = 2 x 3.14 x r = 2 x 3.14 x 0.16 = 1.0048 m
[tex]M =\frac{4\pi\times 10^{-7}\times 1000\times 750\times 0.25\times 10^{-4}}{1.0048}[/tex]
M = 2.5 x 10^-5 henry
Thus, the mutual inductance between the two toroid is 2.5 x 10^-5 henry.
The point is at the edge of the disk and the component bodies are:
a. A uniform disk of radius and mass .
b. A uniform rod of length and mass .
c. A uniform rectangle with side lengths and , and mass .
d. A point mass at with mass .
What is the moment of inertia about the axis through the point?
Answer:
a. A uniform disk of radius and mass .
Explanation:
The moment of inertia I of an object depends on a chosen axis and the mass of the object. Given the axis through the point, the inertia will be drawn from the uniform disc having a radius and the mass.
.
Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.
Recall that k = 8.99 × 109 N•meters squared per Coulomb squared.
What is the force applied between q1 and q2?
Answer:
F = 10.788 N
Explanation:
Given that,
Charge 1, [tex]q_1=6\ \mu C=6\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=2\ \mu C=2\times 10^{-6}\ C[/tex]
Distance between charges, d = 0.1 m
We know that there is a force between charges. It is called electrostatic force. It is given by :
[tex]F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N[/tex]
So, the force applied between charges is 10.788 N.
Answer:
10.8 N
Explanation:
The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a) Through how many radians does the wheel turn during the first 2.00 s of its motion? (b) What is the angular acceleration (in rad/s2) of the wheel at the end of the first 2.00 s of its motion?
Answer:
a
The number of radians turned by the wheel in 2s is [tex]\theta= 8\ radians[/tex]
b
The angular acceleration is [tex]\alpha =14 rad/s^2[/tex]
Explanation:
The angular velocity is given as
[tex]w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3[/tex]
Now generally the integral of angular velocity gives angular displacement
So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec
This is mathematically evaluated as
[tex]\theta(t ) = \int\limits^2_0 {2t + t^3} \, dt[/tex]
[tex]= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.[/tex]
[tex]= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0[/tex]
[tex]= 4 +4[/tex]
[tex]\theta= 8\ radians[/tex]
Now generally the derivative of angular velocity gives angular acceleration
So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration
This is mathematically evaluated as
[tex]\frac{dw}{dt} = \alpha (t) = 2 + 3t^2[/tex]
so at t=2
[tex]\alpha (2) = 2 +3(2)^2[/tex]
[tex]\alpha =14 rad/s^2[/tex]
The wheel turns through 8 radians in the first 2 seconds of its motion. The angular acceleration of the wheel at the end of the first 2 seconds is 14.00 rad/s².
Explanation:The questions are related to the concepts of angular velocity and acceleration in the realm of Physics. To solve parts (a) and (b) we would use the principles of rotational motion.
(a) The number of radians the wheel turns in first 2 seconds is given by the integral of the angular velocity function from 0 to 2. The equation becomes ∫ₒ² ω(t) dt = ∫ₒ² ((2.00 rad/s²)t + (1.00 rad/s⁴)t ³ dt) = [t² + 0.25t⁴]ₒ² = 4 rad + 4 rad = 8 rad.
(b) The angular acceleration is given by the derivative of the angular velocity with respect to time which is ω'(t) = 2.00 rad/s² + 3(1.00 rad/s⁴)t². Evaluating at t = 2s gives an angular acceleration of 2.00 rad/s² + 3(1.00 rad/s⁴)(2s)² = 2.00 rad/s² + 12.00 rad/s² = 14.00 rad/s².
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A new metal alloy is found to have a specific heat capacity of 0.321 J/(g⋅∘C). First, 44.5 g of the new alloy is heated to 100. ∘C. Then, it is placed in an ideal constant-pressure calorimeter containing 175 g of water (Vs,water=4.184 J/(g⋅∘C)) at an initial temperature of 20.0 ∘C. What will the final temperature of the mixture be after it attains thermal equilibrium
Answer:
The final temperature is 21.531°c
Explanation:
Heat gained by the water = heat lost by the metal
Heat gained = (m)(c)(∆tita)
Where m = mass
C = specific heat capacity
∆tita = temperature change
X = equilibrium temperature
So ....
Heat gained by water
= 175*4.185*(x-20)
= 732.2x - 14644
Heat lost by metal
= 44.5*0.321*(100-x)
=1428.45 -14.2845x
So....
1428.45 - 14.2845x = 732.2x - 14644
1428.45+14644= 732.2x + 14.2845x
16072.45= 746.4845x
16072.45/746.4845= x
21.531 = x
The final temperature is 21.531°c
Two hollow cylinders have the same inner and outer diameters and the same mass, but they have different lengths because one is made of low density wood and the other of high density brass. Which cylinder has the greater moment of inertia about its cylindrical center axis?
Answer:
Both cylinders will have the same moment of inertia.
Explanation:
The moment of inertia of a rigid body depends on the distribution of mass around the axis of rotation, i.e at what radius from the axis how much mass is located. What the moment of of inertia DOES NOT depend in is the distribution of mass parallel to the axis of rotation. This means that two hollow cylinders of the same mass but with different lengths will have the same moment of inertia!
For completeness, the momentum of inertia of a hollow cylinder is
[tex]I \approx mR^2[/tex]
from which we clearly see that [tex]I[/tex] only depends on the mass and the radius of the hollow cylinder and not on its height; Hence, both the wooden and the brass cylinders will have the same moment of inertia.
The two (2) hollow cylinders will have the same moment of inertia about their cylindrical center axis.
What is moment of inertia?Moment of inertia is also referred to as the mass moment of inertia and it can be defined as a measure of the rotational inertia of an object or its resistance to angular acceleration about a reference axis.
Mathematically, the moment of inertia of a hollow cylinder is given by the formula:
[tex]I=mr^2[/tex]
Where:
I is the moment of inertia.m is the mass.r is the radius.From the above formula, we can deduce that the moment of inertia of a hollow cylinder is highly dependent on the distribution of mass around its axis of rotation (radius) rather than its height.
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On a day when the speed of sound is 340 m/s, a ship sounds its whistle. The echo of the sound from the shore is heard at the ship 10.0 s later. How far is the ship from the shore
The distance of the ship from the shore is 1020m
Explanation:
Given:
Speed, s = 340 m/s
Time, t = 10s
Distance, x = ?
The sound is going to have to go to shore, then come back.
The total round-trip distance is
D = speed X time
D = (340 m/s) * (6.0 s)
D = 2,040 m
But as previously stated, the sound had to get there, then come back. So the actual ship-to-shore distance is only half that.
x = D/2
[tex]x = \frac{2040}{2} \\\\x = 1020m[/tex]
Therefore, the distance of the ship from the shore is 1020m
A long solenoid has n turns per unit length, has a radius R1, and carries a current R2. A circular coil with radius R2 and with N total turns is coaxial with the solenoid and equidistant from its ends. (a) Find the magnetic flux through the coil ifR2 > R1 (b) Find the magnetic flux through the coil if R2 < R1.
Answer:
a) Фm = μ₀*I*n*π*N*R₁²
b) Фm = μ₀*I*n*π*N*R₂²
Explanation:
Given
n = number of turns per unit length of the solenoid
R₁ = radius of the solenoid
I = current passing through the solenoid
R₂ = radius of the circular coil
N = number of total turns on the coil
a) Фm = ? If R₂ > R₁
b) Фm = ? If R₂ < R₁.
We can use the formula
Фm = N*B*S*Cos θ
where
N is the number of turns
B is the magnetic field
S is the area perpendicular to magnetic field B.
The magnetic field outside the solenoid can be approximated to zero. Therefore, the flux through the coil is the flux in the core of the solenoid.
We know that the magnetic field inside the solenoid is uniform. Thus, the flux through the circular coil is given by the same expression with R2 replacing R1 (from the area).
a) Then, the flux through the large circular loop outside the solenoid (R₂ > R₁) is obtained as follows:
Фm = N*B*S*Cos θ
where
B = μ₀*I*n
S = π*R₁²
θ = 0°
⇒ Фm = (N)*(μ₀*I*n)*(π*R₁²)*Cos 0°
⇒ Фm = μ₀*I*n*π*N*R₁²
b) The flux through the coil when R₂ < R₁ is
Фm = (N)*(μ₀*I*n)*(π*R₂²)*Cos 0°
⇒ Фm = μ₀*I*n*π*N*R₂²
Answer:
a) ∅ [tex]= u_0*n*I*N*pi*(R_1)^2[/tex]
b) ∅ [tex]= u_0 * n*I*N * pi * (R_2)^2 [/tex]
Explanation:
The magnetic field inside the solenoid is uniform and the flux in the coil is the same as the flux in the solenoid.
The magnetic field outside can be said to be zero which also means that the flux through the coil is the same as flux in the solenoid.
Thus, we can say
R2 = R1
Where:
N = total turns on coil
n = number of turns per unit length of solenoid
I = current
R1 = radius of solenoid
R2 = circular coil radius
a) The magnetic flux through the coil if R2 > R1 is giben as:
∅ [tex]= N(u_0 *n*I)(pi*(R_1 )^2) Cos 0[/tex]
Solving further, we have:
∅ [tex]= u_0*n*I*N*pi*(R_1)^2[/tex]
Here,
[tex] u_0*n*I [/tex] is the magnetic field
(Pi*R1²) is the area pependicular to magnetic field
N is the number of turns
b) The magnetic flux through the coil if R2 < R1
∅ [tex] = N(u_0*n*I)(pi*(R_2)^2)Cos 0 [/tex]
∅ [tex]= u_0 * n*I*N * pi * (R_2)^2 [/tex]
Here,
[tex] u_0*n*I [/tex] is the magnetic field
(Pi*R2²) is the area pependicular to magnetic field
N is the number of turns
Describe a situation in which an electron will be affected by an external electric field but will not be affected by an external magnetic field. Is it possible that an electron is affected by an external magnetic field but not by an external electric field?
Answer:
Explanation:
Situations in which an electron will be affected by an external electric field but will not be affected by an external magnetic field
a ) When an electron is stationary in the electric field and magnetic field , he will be affected by electric field but not by magnetic field. Magnetic field can exert force only on mobile charges.
b ) When the electron is moving parallel to electric field and magnetic field . In this case also electric field will exert force on electron but magnetic field field will not exert force on electrons . Magnetic field can exert force only on the perpendicular component of the velocity of charged particles.
Situations when electron is affected by an external magnetic field but not by an external electric field
There is no such situation in which electric field will not affect an electron . It will always affect an electron .
Final answer:
In certain conditions, an electron is affected by an external electric field but not by an external magnetic field. Conversely, an electron can be influenced by an external magnetic field but not by an external electric field.
Explanation:
An electron will be affected by an external electric field but will not be affected by an external magnetic field when it is at rest in a magnetic field as it experiences no force.
It is possible for an electron to be affected by an external magnetic field but not by an external electric field, as seen in the case of a moving charged particle forming a magnetic field around wires carrying electrical currents.
Two sheets of polarizing material are placed such that their polarizing axes are 90° to each other and no light passes through the combination. Can you place a third sheet of polarizing material between the two and cause light to pass through this three-sheet combination? Explain why if this is possible.
Answer:
Explanation:
Polarization In this case angle of incidence is not equal to angle of polarization, hence reflected light is partially polarized and transmitted light is also partially polarized. by reflection is explained by Brewster's law,
According to this when unpolarized light incident on glass plate at an angle is called as angle of polarizing the reflected light is plane polarized, and transmitted light is partially polarized. The plane of vibration of polarized light is having plane of vibrations perpendicular to plane of incidence.
When two sheets of polarizing material with perpendicular axes are placed together, no light passes through. However, if a third sheet of polarizing material is placed between the first two sheets at an angle, some light can pass through the combination.
Explanation:When two sheets of polarizing material are placed with their polarizing axes at right angles to each other, no light passes through the combination because the second sheet blocks the light that is polarized by the first sheet. However, if a third sheet of polarizing material is placed between the first two sheets, at an angle between 0° and 90° to the axis of the first sheet, some light can pass through the three-sheet combination.
This is because the third sheet allows a fraction of the previously blocked light to pass through. As the angle between the first and third sheets approaches 90°, more light is transmitted by the combination.
For example, in the case of Figure 27.41, when the axes of the first and second filters are aligned (parallel), all of the polarized light passed by the first filter is also passed by the second filter. When the second filter is rotated to make the axes perpendicular, no light is passed by the second filter.
A 3.0 kg object is moving along the x-axis in a region where its potential energy as a function of x is given as U(x) = 4.0x2 , where U is in joules and x is in meters. When the object passes the point x = -0.50 m, its velocity is +2.0 m/s. All forces acting on the object are conservative. Calculate the total mechanical energy of the object Calculate the x-coordinate of any points at which the object has zero kinetic energy. Calculate the magnitude of the momentum of the object at x = 0.60 m. Calculate the magnitude of the acceleration of the object as it passes x = 0.60 m. On the axes below, sketch graphs of the object’s position x versus time t and kinetic energy K versus time t. Assume that x = 0 at time t = 0 . The two graphs should cover the same time interval and use the same scale on the horizontal axes.
Answer:
a) [tex]E_{tot} = 7\,J[/tex], b) [tex]x = \pm\sqrt{\frac{7}{4} }[/tex]c) [tex]p = 5.775\,\frac{kg\cdot m}{s}[/tex]
Explanation:
a) The total energy of the object is:
[tex]E_{tot} = U + K[/tex]
[tex]E_{tot} = 4\cdot x^{2} + \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]E_{tot} = 4\cdot (-0.50\,m)^{2} + \frac{1}{2}\cdot (3\,kg)\cdot (2\,\frac{m}{s} )^{2}[/tex]
[tex]E_{tot} = 7\,J[/tex]
b) The total energy of the object is:
[tex]E_{tot} = U[/tex]
[tex]7\,J = 4\cdot x^{2}[/tex]
[tex]x = \pm\sqrt{\frac{7}{4} }[/tex]
c) The speed of the object is clear in the total energy expression:
[tex]E_{total} = U + K[/tex]
[tex]K = E_{total}-U[/tex]
[tex]\frac{1}{2}\cdot m \cdot v^{2} = E_{total} - 4\cdot x^{2}[/tex]
[tex]v^{2} = \frac{2\cdot (E_{total}-4\cdot x^{2})}{m}[/tex]
[tex]v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }[/tex]
[tex]v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0.6\,m)^{2}]}{3\,kg} }[/tex]
[tex]v \approx 1.925\,\frac{m}{s}[/tex]
The magnitude of the momentum is:
[tex]p = (3\,kg)\cdot (1.925\,\frac{m}{s} )[/tex]
[tex]p = 5.775\,\frac{kg\cdot m}{s}[/tex]
d) Before calculating the acceleration experimented by the object, it is required to determine the net force exerted on it. There is a relationship between potential energy and net force:
[tex]F = -\frac{dU}{dx}[/tex]
[tex]F = -8\cdot x[/tex]
Acceleration experimented by the object is:
[tex]a = -\frac{8\cdot x}{m}[/tex]
[tex]a = -\frac{8\cdot (0.6\,m)}{3\,kg}[/tex]
[tex]a = -1.6\,\frac{m}{s^{2}}[/tex]
e) The position of the object versus time is found by solving the following differential equation:
[tex]\frac{d^{2}x}{dt} +\frac{8\cdot x}{m} = 0[/tex]
[tex]s^{2}\cdot X(s)- s\cdot v(0) - x(0) + \frac{8}{m}\cdot X(s) = 0[/tex]
[tex](s^{2} + \frac{8}{m})\cdot X(s) = s\cdot v(0)+x(0)[/tex]
[tex]X(s) = \frac{s\cdot v(0)+x(0)}{(s^{2}+\frac{8}{m} )}[/tex]
[tex]X(s) = v(0)\cdot \frac{s}{s^{2}+\frac{8}{m} } +\frac{m\cdot x(0)}{8} \cdot \frac{\frac{8}{m}}{s^{2}+\frac{8}{m}}[/tex]
[tex]x(t) = v(0) \cdot \cos \left(\frac{8}{m}\cdot t \right)+\frac{m\cdot x(0)}{8}\cdot \sin \left(\frac{8}{m}\cdot t \right)[/tex]
The velocity is obtained by deriving the previous expression:
[tex]v(t) = -\frac{8\cdot v(0)}{m}\cdot \sin \left(\frac{8}{m}\cdot t \right)+x(0)\cdot \cos \left(\frac{8}{m}\cdot t \right)[/tex]
Speed of the object at [tex]x = 0[/tex] is:
[tex]v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }[/tex]
[tex]v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0\,m)^{2}]}{3\,kg} }[/tex]
[tex]v \approx 2.160\,\frac{m}{s}[/tex]
The equation of the motion are:
[tex]x(t) = 2.160\cdot \cos \left(2.667\cdot t \right)[/tex]
[tex]v(t) = -5.76\cdot \sin (2.667\cdot t)[/tex]
The expression for the kinetic energy of the object is:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]K = \frac{1}{2}\cdot (3\,kg)\cdot [33.178\cdot \sin^{2}(2.667\cdot t)][/tex]
[tex]K = 49.767\cdot \sin^{2}(2.667\cdot t)[/tex]
Graphics are included below as attachments. (Position versus time, kinetic energy vs time).
Following are the responses to the given points:
Given:
object mass [tex](m)= 3.0\ kg \\\\[/tex]
Potential energy [tex]U(x)=4.0\ x^2\\\\[/tex]
[tex]\to x=-0.5 \ m\\\\ \to velocity \ (v)=2.0 \ \frac{m}{s}\\\\[/tex]
Solution:
For point (a)
Total Energy [tex](TE) = PE + KE \\\\[/tex]
[tex]\to PE = 4.0 (0.5 m)^2 = 1\ J\\\\ \to KE = \frac{1}{2} mv^2 = 0.5 \times 3.0\ kg (2.0 \frac{m}{s})^2 = 6\ J \\\\\to TE = 7\ J\\\\[/tex]
For point (b)
If an object's potential energy is 7 J, it has 0 kinetic energy.
[tex]\to U(x) = 4x^2 = 7\ J \\\\\to x=+1.2 \ m, -1.2 \ m[/tex]
For point (c)
[tex]\to x=0.60\ m \\\\ \to TE=4x^2 + \frac{1}{2}mv^2 = 7\ J\\\\ \to 4(0.60)^2 +0.5 \times 3.0 \ kg \times v^2 = 7\ J \\\\ \to v=1.92 \ \frac{m}{s}\\\\ \to Momentum (p) = mv \\\\\to p= 3.0\ kg \times 1.92 \ \frac{m}{s}\\\\ \to p= 5.76 kg \ \frac{m}{s}\\\\[/tex]
For point (d)
[tex]\to x_1= 0.6 \ m\\\\ \to v_1 = 1.92 \frac{m}{s}\\\\ \to x_2 = 1.2\ m[/tex] the velocity is found by
[tex]\to 4x^2 + \frac{1}{2} mv^2= 7\ J \\\\ \to 4(1.2)^2 +0.5 \times 3.0\ kg \times v^2 = 7\ J \\\\ \to v_2 = 0.9 \ \frac{m}{s}\\\\[/tex]
Calculating the acceleration:
[tex]\to V^2_{2}-v^2_{1}= 2a(x_2-x_1) \\\\\to (0.9 \ \frac{m}{s})^2 - (1.92\ \frac{m}{s})^2 = 2 a(1.2\ m - 0.6\ m) \\\\\to a= -2.4 \frac{m}{s^2}\\\\[/tex]
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The frequency of a sound wave is 300 Hz and the room temperature is 30 Celsius. What is the wavelength of this sound wave. Two children stretch a jump rope between them and send wave pulses back and forth. The rope is 4 meters long and its mass is .4 kg and the force exerted on it by the children is 50 N. What is the speed of the waves on the rope?
Answer:
Explanation:
Velocity of sound wave at 30 degree = 350 m /s
frequency of sound = 300 Hz .
wavelength of sound in air. = velocity / frequency
= 350 / 300
= 1.167 m
for wave formed in the rope :
velocity of wave in the rope
= [tex]\sqrt{\frac{T}{m} }[/tex]
T is tension in the rope and m is mass per unit length .
m = .4 / 4
= .1
Putting the given values in the equation above
v = [tex]\sqrt{\frac{50}{.1} }[/tex]
v = 22.36 m /s .
velocity of wave in the rope.
= 22.36 m /s .
Two charged particles, Q 1 and Q 2, are a distance r apart with Q 2 = 5 Q 1. Compare the forces they exert on one another when is the force Q 2 exerts on Q 1 and is the force Q 1 exerts on Q 2.
Describe how the horizontal position of The baseball varies overtime is the change in position constant variable? Why might this be so?
Answer:
When you hit a baseball with a bat, you are applying a force only in the time while the bat is in contact with the ball, after that, the horizontal speed of the ball is constant. (if you ignore the friction of the air)
This happens because of the second Newton's law:
Force is equal to mass times the acceleration or:
F = m*a
If we do not have a force, then we do not have acceleration.
This means that the change in position for a fixed amount of time is constant, so horizontal speed is constant.
A child's toy consists of a small wedge that has an acute angle θ. The sloping side of the wedge is frictionless, and an object of mass m on it remains at constant height if the wedge is spun at a certain constant speed. The wedge is spun by rotating, as an axis, a vertical rod that is firmly attached to the wedge at the bottom end. Show that, when the object sits at rest at a point at distance L up along the wedge, the speed of the object must be v = (gL sin(θ))1/2.
Answer:
see explanation
Explanation:
Net force along x axis is
[tex]\sum F_x = F \sin \theta= \frac{v^2}{R} ----(1)[/tex]
Net force along y axis is
[tex]\sum F_y = F \cos \theta= mg ----(2)[/tex]
The object can along accelerate down the stamp.
Thus F(net) is at the angle down the stamp
[tex]F_{net}=F_c\\\\F_{net}= \sin \theta mg\\\\F_c = \frac{mv^2}{r} \\[/tex]
where r = L in the direction of acceleration
[tex]\sin \theta mg = \frac{mv^2}{L}[/tex]
[tex]v^2 = gL \sin \theta[/tex]
[tex]v = \sqrt{gL \sin \theta}[/tex]
[tex]v = (gL \sin \theta )^{1/2}[/tex]
The relationship between the distance of the object and speed of the object is [tex]v = \sqrt{gL sin(\theta)}[/tex].
The given parameters:
inclination of the wedge, = θlet the constant speed = vThe relationship between the distance of the object, speed of the object can be determined by the net force on the toy is calculated follows;
[tex]Fsin(\theta) = \frac{mv^2}{L} \\\\mgsin(\theta) = \frac{mv^2}{L} \\\\gsin(\theta) = \frac{v^2}{L} \\\\v^2 = gL sin(\theta)\\\\v = \sqrt{gL sin(\theta)}[/tex]
Thus, the relationship between the distance of the object and speed of the object is [tex]v = \sqrt{gL sin(\theta)}[/tex].
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A square metal plate 2.5m on each side is pivoted about an axis though point O at its center and perpendicular to the plate. Calculate the net torque due to the three forces if the magnitudes of the forces are F1=18N, F2=20N, and F3=11N..
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The net torque [tex]\tau = 21.95N \cdot m[/tex]
Explanation:
From the question we are told that
The length of each side is [tex]L = 2.5 m[/tex]
The first force is [tex]F_1 = 18N[/tex]
The second force is [tex]F_2 = 20 N[/tex]
The third force is [tex]F_3 = 11N[/tex]
The free body diagram for the question is shown on the second uploaded image
Generally torque is mathematically represented as
[tex]\tau = r * F[/tex]
Where [tex]\tau[/tex] is the torque
r is the length from the rotating point to the point the force is applied, this is also the radius of the circular path made
F is the force causing the rotation.
looking at the free body diagram we can deduce that L is the diameter of the circular path made as a result of toque
Now for the torque due to force [tex]F_1[/tex]
[tex]\tau_1 = - F_1 * r_1[/tex]
The negative sign is because the direction of [tex]F_1[/tex] is clockwise
=> [tex]\tau_1 = - F_1 * \frac{L}{2}[/tex]
Substituting value
[tex]\tau_1 = - 18 * \frac{2.5}{2 }[/tex]
[tex]\tau_1 = - 22.5 N \cdot m[/tex]
The torque as a result of the second force is mathematically evaluated as
[tex]\tau_2 = F_2 * r_2[/tex]
[tex]\tau_2 = F_2 * \frac{L}{2}[/tex]
[tex]= 20 * \frac{2.5}{2}[/tex]
[tex]\tau_ 2 = 25 \ N \cdot m[/tex]
The torque as a result of the third force is mathematically evaluated as
[tex]\tau_3 =r_3 (F_3 sin \theta + F_3 cos \theta )[/tex]
[tex]\tau_3 = \frac{L}{2} (F_3 sin \theta + F_3 cos \theta )[/tex]
Where the free body diagram [tex]\theta = 45^o[/tex]
[tex]\tau_3 = \frac{2.5}{2} (11 * sin (45) +11 cos (45) )[/tex]
[tex]\tau_ 3 = 19.45 \ N \cdot m[/tex]
The net torque the mathematically
[tex]\tau = \tau_1 + \tau_2 + \tau_3[/tex]
substituting value
[tex]\tau = -22.5 + 25 + 19.45[/tex]
[tex]\tau = 21.95N \cdot m[/tex]
A change in velocity means a change in?
A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.260 T/s .
(a) What is the magnitude of the electric field induced in the ring?
(b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?
The magnitude of the induced electric field can be found using Faraday's law of electromagnetic induction. The current flows clockwise as viewed by someone on the south pole of the magnet.
Explanation:To find the magnitude of the electric field induced in the ring, we can use Faraday's law of electromagnetic induction. According to the law, the magnitude of the induced electric field is equal to the rate of change of magnetic flux through the ring. The magnetic flux can be calculated by multiplying the magnetic field strength by the area of the ring. In this case, the magnetic field is decreasing at a rate of 0.260 T/s, and the area of the ring is π(2.50 cm)2. Therefore, the magnitude of the induced electric field is 2π(2.50 cm)2(0.260 T/s).
As for the direction of the current flow, it can be determined using the right-hand rule for induced currents. If you point your right thumb in the direction of the magnetic field (from north to south), the induced current will flow in the direction of your curled fingers, which in this case will be clockwise as viewed by someone on the south pole of the magnet.
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A uniform meter stick is suspended from the ceiling of an elevator on one end and is swinging back and forth. The elevator is accelerating upwards with acceleration a.
The period of this meter stick, in terms of its length L, g and a is ______.
Answer:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g + a} }[/tex]
Explanation:
It is known that stick is experimenting a Simple Harmonic Movement and, to be exactly, can be modelled as a simple pendulum. The period of oscilation of the stick is:
[tex]T = \frac{2\pi}{\omega}[/tex]
The pendulum is modelled by the Newton's Laws. The Free Body Diagram is presented below:
[tex]\Sigma F_{r} = T - m\cdot g \cdot \cos \theta = m\cdot (\omega^{2}\cdot l + a\cdot \cos \theta)[/tex]
[tex]\Sigma F_{t} = m\cdot g \cdot \sin \theta = m\cdot (l\cdot \alpha - a \cdot \sin \theta)[/tex]
Let assume that pendulum is just experimenting small oscillations, so that:
[tex]\theta \approx \sin \theta[/tex]
Then:
[tex]m\cdot g \cdot \theta = m\cdot (l\cdot \alpha - a\cdot \theta)[/tex]
[tex]g\cdot \theta = l\cdot \alpha - a\cdot \theta[/tex]
[tex](g + a)\cdot \theta = l\cdot \alpha[/tex]
[tex]\alpha = \frac{g+a}{l}\cdot \theta[/tex]
Where [tex]\omega =\sqrt{\frac{g + a}{l} }[/tex].
Finally, the period is:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g + a} }[/tex]
Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T. (a) What is the energy difference between the magnetic moment orientations of the silver atoms in the two subbeams? (b) What is the frequency of the radiation that would induce a transition between these two states? (c) What is the wavelength of this radiation, and (d) to what part of the electromagnetic spectrum does it belong?
Answer:
Explanation
Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T
A. To calculate the energy difference in the magnetic moment orientation
∆E = 2μB
For example, any electron's magnetic moment is measured to be 9.284764×10^−24 J/T
Then
μ = 9.284764 × 10^-24 J/T
∆E = 2μB
∆E = 2 × 9.284764 × 10^-24 × 0.21
∆E = 3.8996 × 10^-24 J
Then, to eV
1eV = 1.602 × 10^-19J
∆E = 3.8996 × 10^-24 J × 1eV / 1.602 × 10^-19J
∆E = 2.43 × 10^-5 eV
B. Frequency?
To determine the frequency of radiation hitch would induce the transition between the two states is,
∆E = hf
Where h is plank constant
h = 6.626 × 10-34 Js
Then, f = ∆E / h
f = 3.8996 × 10^-24 / 6.626 × 10^-34
f = 5.885 × 10^9 Hz
f ≈ 5.89 GHz
C. The wavelength of the radiation
From wave equation
v = fλ
In electromagnetic, we deal with speed of light, v = c
And the speed of light in vacuum is
c = 3 × 10^8 m/s
c = fλ
λ = c / f
λ = 3 × 10^8 / 5.885 × 10^9
λ = 0.051 m
λ = 5.1 cm
λ = 51 mm
D. It belongs to the microwave
From table
Micro waves ranges from
•Wavelength 10 to 0.01cm
Then we got λ = 5.1 cm, which is in the range.
•Frequency 3GHz to 3 Thz
Then, we got f ≈ 5.89 GHz, which is in the range
•Energy 10^-5 to 0.01 eV
We got ∆E = 2.43 × 10^-5 eV, which is in the range of the microwave
The value above is in microwave range
The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string
Answer:
given
y=6.0sin(0.020px + 4.0pt)
the general wave equation moving in the positive directionis
y(x,t) = ymsin(kx -?t)
a) the amplitude is
ym = 6.0cm
b)
we have the angular wave number as
k = 2p /?
or
? = 2p / 0.020p
=1.0*102cm
c)
the frequency is
f = ?/2p
= 4p/2p
= 2.0 Hz
d)
the wave speed is
v = f?
= (100cm)(2.0Hz)
= 2.0*102cm/s
e)
since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction
f)
the maximum transverse speed is
umax =2pfym
= 2p(2.0Hz)(6.0cm)
= 75cm/s
g)
we have
y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]
= -2.0cm
The amplitude of the wave is 6.0 cm, the wavelength is 100π cm, the frequency is approximately 0.637 Hz, the speed is approximately 200π cm/s, the direction of propagation is the positive x-direction, and the maximum transverse speed of a particle in the string is 120π cm/s.
Explanation:The given equation representing a transverse wave along a very long string is y = 6.0 sin(0.020px - 4.0pt). We can extract various information from this equation:
The amplitude of the wave is 6.0 cm. The amplitude represents the maximum displacement of the particles in the string from their equilibrium position.The wavelength of the wave can be determined by comparing the equation to its standard form y = A sin(kx - ωt), where k represents the wave number. From the equation given, we can find k = 0.020p. The wave number is related to the wavelength by the equation λ = (2π)/k. Substituting the value of k, we get λ = (2π)/(0.020p), which simplifies to 100π cm.The frequency of the wave can be determined from the equation ω = 2πf, where ω is the angular frequency. In the given equation, comparing it to the standard form, we find ω = 4.0 pt. Solving for f, we get f = (ω/2π) = 4.0 pt/(2π) ≈ 0.637 Hz.The speed of the wave can be determined by the equation v = fλ, where v is the speed. Substituting the frequency and wavelength values, we get v = (0.637 Hz)(100π cm) ≈ 200π cm/s.The direction of propagation of the wave can be determined from the sign of the coefficient of x in the equation. In this case, the coefficient is positive, indicating that the wave is propagating in the positive x-direction.The maximum transverse speed of a particle in the string can be determined by taking the derivative of the displacement equation with respect to time, which gives the velocity equation. Taking the derivative of the given equation, we get v = -120πe^(-0.020px) cos(0.020px - 4.0pt). The maximum transverse speed will occur when the cos(0.020px - 4.0pt) term is equal to 1, resulting in a maximum speed of 120π cm/s.Learn more about Properties of Transverse Waves here:https://brainly.com/question/32780857
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You push your physics book 1.50 m along a horizontal tabletop with a horizontal push of 2.40 N while the opposing force of friction is 0.600 N. How much work does each of the following forces do on the book: (a) your 2.40-N push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?
Answer:
a) 3.6 J
b) 0.9 J
c) 0 J
d) 0 J
e) 2.7 J
Explanation:
Work = force F x diaplament in direction of force d
Given that,
Displacement in direction of push force d = 1.5 N
Push force = 2.4 N
Friction = 0.6 N
a) work done by push force = 2.4 x 1.5 = 3.6 J
b) word done by friction = -0.6 x 15 = -0.9 J (minus sign shows the work is opposite that done by the push force)
c) work done by normal force from table top ( force acting upwards to oppose gravity) = 0 J since there is no vertical displacement
d) work done by gravity (doward force) = 0 J since there is no vertical displacement
e) net work done on book = 3.6 J + (-0.9J) + 0 J + 0 J = 2.7 J
The work done by each force is calculated using the formula Work = Force × Distance. The net work done on the book is the sum of the work done by each force. (a) 3.6 J (b) 0.9 J (c) 0 J (d) 0 J (e) 4.5 J
Explanation:To calculate the work done by each force, we use the formula:
Work = Force × Distance
(a) The work done by your push is:
Work = 2.40 N × 1.50 m = 3.6 J
(b) The work done by the friction force is:
Work = 0.600 N × 1.50 m = 0.9 J
(c) The normal force from the tabletop does no work because it is perpendicular to the direction of motion.
(d) The work done by gravity is zero since the book doesn't move vertically and gravity acts vertically.
(e) The net work done on the book is the sum of the work done by all forces:
Net Work = Work by your push + Work by friction + Work by normal force + Work by gravity = 3.6 J + 0.9 J + 0 J + 0 J = 4.5 J
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A long, straight solenoid with a cross-sectional area of 8.00cm28.00cm2 is wound with 9090turns of wire per centimeter, and the windings carry a current of 0.350A0.350A. A second winding of 1212 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400s0.0400s. What is the average induced emf in the second winding?
Answer: The induced emf is 9.50 × 10^-4 V
Explanation: Please see the attachments below
The average induced emf in the second winding of a solenoid given the changing magnetic field, number of turns, and cross-sectional area, can be calculated using Faraday's Law.
Explanation:The induced emf in the second winding of a solenoid can be determined using Faraday's Law, which states that the induced emf is equal to the change in the magnetic flux over time. First, we need to calculate the magnetic field in the solenoid using the formula B = µonI, where µo is the permeability of free space, n is the number of turns per unit length, and I is the current. Then, we calculate the magnetic flux by multiplying the magnetic field by the cross-sectional area. Since the current is changing, we will experience a changing magnetic field, so we find the change in flux and divide by the change in time to get our induced emf.
From Faraday's Law, we know that the magnetic flux through each loop is equal to the magnetic field times the cross-sectional area, or Φ = BA. Finally, we multiply our induced emf by the number of turns in the second winding to account for each loop having an induced emf.
A critical factor to note here is that the induced emf is directly proportional to the rate of change of the magnetic field and the number of turns in the second winding, while being inversely proportional to the cross-sectional area of the solenoid.
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The hair breaks under a 1.2 N tension. What is the cross-sectional area of a hair?
I need an explanation or the calculation too, so I will not give points for only the answer.
Answer:
bow
Explanation:
Which appliances uses the larger current, a toaster or a desk lamp?
Answer:
DeskLamp
Explanation:
You're average toaster uses so much more electricity than the average desk lamp. 2 slice toasters I believe uses about 1000 watts? And 4 slice uses 1600. The brightest lightbulbs tend to run on 100 watts so just imagine a lightbulb with 1600 watts. (There's a video of a man cranking the watts to a 1000 watt lightbulb up to millions until it pops) Lightbulbs have a very far range in the amounts of energy they can take in. Here is a simplified answer-
On average- the toaster would use the most watts in terms of use
In other circumstances- The lightbulb would be the case.
So yeah,
It's Desklamps. They're really powerful.
A common physics demonstration is to drop a small magnet down a long, vertical aluminum pipe. Describe the motion of the magnet through the pipe and explain the physical processes that cause this motion.
Answer and Explanation:
This experiment is known as Lenz's tube.
The Lenz tube is an experiment that shows how you can brake a magnetic dipole that goes down a tube that conducts electric current. The magnet, when falling, along with its magnetic field, will generate variations in the magnetic field flux within the tube. These variations create an emf induced according to Faraday's Law:
[tex]\varepsilon =-\frac{d\phi_B}{dt}[/tex]
This emf induced on the surface of the tube generates a current within it according to Ohm's Law:
[tex]V=IR[/tex]
This emf and current oppose the flux change, therefore a field will be produced in such a direction that the magnet is repelled from below and is attracted from above. The magnitude of the flux at the bottom of the magnet increases from the point of view of the tube, and at the top it decreases. Therefore, two "magnets" are generated under and above the dipole, which repel it below and attract above. Finally, the dipole feels a force in the opposite direction to the direction of fall, therefore it falls with less speed.
Answer:
Check below for the answer and explanation
Explanation:
According to Faraday's law of electromagnetic induction, if a conductor is exposed to changing magnetic flux, an emf and hence a current is induced in the conductor. The strength of the induced emf is directly proportional to the rate of change of the magnetic flux.
Induced emf, [tex]e =-N\frac{d \phi}{dt}[/tex]
Induced current, I = e/R
In this example, as magnet is dropped down the aluminium pipe, the magnetic flux changes, and current is induced in the pipe.
According to Lenz's law, the direction of the induced current in the conductor opposes the direction of the magnetic flux that produces it.
Based on these stated laws, current is induced in this aluminium pipe and the direction of this induced current opposes the magnetic flux change. The magnetic field is repelled and falls slowly.