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?

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:

The 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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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).

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.

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:

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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.

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 .

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

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 ).

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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Answer:

Explanation:

The mass difference is

Δm = m2 - m1

56.936296 u - 56.935399 u

= 0.000897‬ u

The energy released by the electron-capture decay

E = Δmc²

  ( 0.000897‬ u) c² ( 931. 5 MeV /c²÷ 1 u)

= 0.8355555 MeV

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.

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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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.

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.

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.

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:

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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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]

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]

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.

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:

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.

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:

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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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

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

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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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.

16.7 x 10²V

The work-energy theorem states that the change in kinetic energy of a particle, Δ[tex]K_{E}[/tex], results in work done by the particle, W. i.e;

Δ[tex]K_{E}[/tex] = W                     ------------------(i)

But:

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex]mv² -

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex]m [v² - u²]    ------------------(ii)

Where;

m = mass of particle (proton in this case) = 1.67 x 10⁻²⁷kg

v = final velocity of the particle = 6 x 10⁵m/s

u = initial velocity of the particle = 2 x 10⁵m/s

Substitute these values into equation (ii) as follows;

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(6 x 10⁵)² - (2 x 10⁵)²]

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(36 x 10¹⁰) - (4 x 10¹⁰)]

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(36 - 4) x 10¹⁰]

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [32 x 10¹⁰]

Δ[tex]K_{E}[/tex] = (1.67 x 10⁻²⁷) [16 x 10¹⁰]

Δ[tex]K_{E}[/tex] = 26.72 x 10⁻¹⁷J

Also:

W = qΔV         ----------------(iii)

Where;

q = charge of the particle (proton) = 1.6 x 10⁻¹⁹C

ΔV = change in potential difference of the particle = V (from the question)

Substitute these values into equation (iii) as follows;

W = 1.6 x 10⁻¹⁹ x V        

W = 1.6 x 10⁻¹⁹V         -------------------(iv)

Now:

Substitute the values of Δ[tex]K_{E}[/tex] = 26.72 x 10⁻¹⁷ and W in equation(iv) into equation (i)

26.72 x 10⁻¹⁷ = 1.6 x 10⁻¹⁹V

Solve for V;

V = (26.72 x 10⁻¹⁷) / (1.6 x 10⁻¹⁹)

V = 16.7 x 10²V

Therefore, the potential difference through which the proton fell was 16.7 x 10²V

The potential contrast between different locations represents the work or energy dissipated in the transmission of the unit amount of voltage from one point to another.

Following are the calculation of the potential difference:

Change in energy [tex], e^{v}=\frac{1}{2} \ m(v_2^2-v_1^2)[/tex]

[tex]1.602 \times ^{-19}=\frac{1}{2} \times 1.67 \times 10^{-27} (6^2-2^2)\times 10^{10}\\\\[/tex]

[tex]\to v \times 1.602 \times 10^{-19}=0.835 \times 10^{-17}\times 32\\\\[/tex]

[tex]\to v = \frac{0.835 \times 10^{-17}\times 32}{1.602 \times 10^{-19}}\\\\[/tex]

        [tex]= \frac{0.835 \times 10^{2}\times 32}{1.602 }\\\\ = \frac{26.72 \times 10^{2} }{1.602 }\\\\=16.67\times 10^{2}[/tex]

Therefore the final answer is "[tex]\bold{16.67 \times 10^2}[/tex]".

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Answer:

bow

Explanation:

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

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.

.

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.

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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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.

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

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