The two spheres are rigidly connected to the rod of negligible mass and are initially at rest on the smooth horizontal surface. A force F is suddenly applied to one sphere in the y-direction and imparts an impulse of 9.8 N·s during a negligibly short period of time. As the spheres pass the dashed position, calculate the magnitude of the velocity of each one.

Answer:

100.49 mm

Explanation:

Given that :

The vertical shear force = 1200 N

Allowable shearing force in each nail = 740 N

From the diagram attached below, We first determine the horizontal force per unit length on the lower surface of the upper flange

[tex]q = \frac{VQ_{1-1}}{I_{NA}} \ \ \ N/mm[/tex]

where;

[tex]I_{NA} = \frac{50*100^3}{12}+2[\frac{150*50^3}{12}+ (150*50)*( \bar {y}^2][/tex]

[tex]I_{NA} = \frac{50*100^3}{12}+2[\frac{150*50^3}{12}+ (150*50)*( 75)^2][/tex]

[tex]I_{NA} = 91.667*10^6 \ mm^4[/tex]

Also;

[tex]Q_{1-1} = A \bar{y}[/tex]

where A = area above (1-1)

[tex]Q_{1-1} = 50*150*75[/tex]

[tex]Q_{1-1} =56.25*10^4 \ \ \ mm^3[/tex]

Therefore ;

[tex]q = \frac{1220*56.25*10^4}{91.667*10^6}[/tex]

[tex]q = 7.3636 \ N/mm[/tex]

Now; the largest permissible spacing s between the nails.  [tex]S_{max} = \frac{740}{7.3636} = 100.49 \ \ \ mm[/tex]

To determine the largest permissible spacing between the nails in a beam, calculate the total shearing force the nails can withstand and divide it by the area of contact. The largest permissible spacing is 14.8 cm.

To determine the largest permissible spacing between the nails, we need to calculate the total shearing force that the beam can withstand. Since each nail can withstand a maximum shearing force of 740 N, the total shearing force that the nails can withstand is 3 multiplied by 740 N, which equals 2220 N. We can set up an equation to find the largest permissible spacing, s, by dividing the total shearing force by the area of contact between the nails and the beam. The area of contact can be calculated by multiplying the thickness of the beam by the length of the nails, so it is 3 multiplied by 50 mm (or 0.05 m) multiplied by s. So, we have the equation:

2220 N = 0.05 m x 3 x s

Simplifying the equation, we get

2220 N = 0.15 m x s

Dividing both sides of the equation by 0.15 m, we find:

s = 2220 N / 0.15 m

s = 14800 mm (or 14.8 cm)

Therefore, the largest permissible spacing between the nails is 14.8 cm.

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Answer:

a) The change in the tire’s angular velocity is 5 rads/sec

b) The tire’s average angular acceleration is 4 rad/s²

Explanation:

initial angular speed , ω (initial) = - 2.5 rad/s

final speed , ω (final) = 2.5 rad/s

time = 1.25 s

a) The change in tire's angular speed

ω = ω (final) - ω (initial)

ω = 2.5 - ( -2.5) rad/s

ω = 2.5 + 2.5 (rad/s)

ω = 5 rad/s

b) The average angular acceleration

a = ω ÷ t

a = 5 rad/s ÷ 1.25 s

a = 4 rad/s²

a) The change in the tire’s angular velocity is 5 rads/sec

b) The tire’s average angular acceleration is 4 rad/s

initial angular speed , ω (initial) = - 2.5 rad/s

final speed , ω (final) = 2.5 rad/s

time = 1.25 s

a) The change in tire's angular speed

ω = ω (final) - ω (initial)

ω = 2.5 - ( -2.5) rad/s

ω = 2.5 + 2.5 (rad/s)

ω = 5 rad/s

b) The average angular acceleration

a = ω ÷ t

a = 5 rad/s ÷ 1.25 s

a = 4 rad/s²

Answer:

(d) Zero

Explanation:

Given:

Side of the square, = L

Magnetic field, = B

According to Faraday, the net force acting on a conductor is equal to product of magnetic field, current, length of conductor and sine of the angle between the field and current.

F = BILsinΦ

Where:

B is magnetic field

I is current in the loop

L is length of the loop

Φ is the angle between the magnetic field and current

Φ = 0, since the current and magnetic field are directed in opposite directions.

F = BILsin(0)

F = 0

Therefore, the net magnetic force acting on the loop is zero.

The correct answer is (d) Zero

The correct option is Option d (zero). The net magnetic force acting on the loop is zero because the forces on the sides of the loop cancel each other out. Thus, the correct answer is d. Zero.

In a uniform magnetic field, the net force on a current-carrying loop in a magnetic field can be determined using the equation F = I \times B. We need to apply this equation to each side of the loop.

Step-by-Step Explanation:

Therefore, the net magnetic force acting on the loop is zero, making the correct answer d. Zero.

Answer:

1117.51N/C

Explanation:

The magnitude of the electric force is given by:

[tex]|\vec{F}|=|k\frac{q_1q_2}{r^2}|[/tex]

k: Coulomb's constant = 8.98*10^9Nm^/2C^2

r: distance between the charges = 0.15m

By replacing the values of q1, q2, k and r you obtain:

[tex]|\vec{F}|=|(8.98*10^9Nm^2/C^2)\frac{(-4.0*10^{-5}C)(7.0*10^{-5}C)}{(0.15m)^2}|=1117.51\frac{N}{C}[/tex]

hence, the force between the charges is 1117.51N/C

Answer:

The force of attraction between the charges is 1120 N

Explanation:

Given:

positive charge of the particle, q1 = 7 x 10^-5 C

negative charge of the particle, q2 = -4 x 10^-5 C

distance between the charges, r = 0.15 m

The force of attraction between the charges will be calculated using Coulomb's law:

F = (k|q1q2|) / r^2

Where:

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 Nm^2/c^2

|q1| is magnitude of charge 1

|q2| is magnitude of charge 2

F = (9 x 10^9 x 7 x 10^-5 x 4 x 10^-5) / (0.15 x 0.15)

F = 1120 N

Thus, the force between the charges is 1120 N

Answer:

0.0367

Explanation:

The loss in kinetic energy results into work done by friction.

Since kinetic energy is given by

KE=0.5mv^{2}

Work done by friction is given as

W= umgd

Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.

Making u the subject of the formula then we deduce that

[tex]u=\frac {v^{2}}{2gd}[/tex]

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

[tex]u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367[/tex]

Therefore, the coefficient of kinetic friction is approximately 0.0367

The coefficient of kinetic friction on the floor is approximately 0.037. This is found by equating the work done by friction, which is equal to the change in kinetic energy of the suitcase, to the frictional force times the distance over which it acts.

To solve this problem we need to understand that the work done by friction is equal to the change in kinetic energy of the suitcase, as per the work-energy theory. The initial kinetic energy of the suitcase is given by (1/2)*m*v^2, where m is the mass (15 kg) and v is the velocity (1.2 m/s). The final kinetic energy is 0 because the suitcase comes to a stop. Thus, the work done by friction is equal to the initial kinetic energy.

The work done by friction can also be expressed as the force of friction times the distance over which it acts, which in this case, is the distance the suitcase traveled (2.0 m). We know that the frictional force is equal to the product of the coefficient of kinetic friction and the normal force. Here, the normal force is equal to the weight of the suitcase, which is weight = m*g, where g is the gravitational acceleration (9.8 m/s^2).

Setting the two expressions for the work done by friction equal to each other gives us the equation (1/2)*m*v^2 = u*m*g*d. Solving this equation for u (the coefficient of kinetic friction) gives us: u = [(1/2)*v^2] / (g*d).

Substituting the given values into this equation, we find that the coefficient of kinetic friction on the floor is approximately 0.037.

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Answer:

a)  the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b) The maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c)  The maximum magnitude of the acceleration of the sphere is = 30.93 [tex]m/s^2[/tex]

The maximum acceleration occurs at maximum excursion from equilibrium.

Explanation:

Given that :

Frequency (f) = 4.80 Hz

Amplitude (A) = 3.40 cm

a)

The total distance traveled by the sphere during one cycle of simple harmonic motion is:

d = 4A   (where A is the Amplitude)

d = 4(3.40 cm)

d = 13.60 cm

Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b)

As we all know that:

[tex]x = Asin \omega t[/tex]

Differentiating the above expression with respect to x ; we have :

[tex]\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)[/tex]

[tex]v = A \omega cos \omega t[/tex]

Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;

Then:

[tex]v_{max} = A \omega[/tex]

We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0  i.e at maximum excursion from equilibrium

substituting [tex]2 \pi f[/tex] for [tex]\omega[/tex] in the above expression;

[tex]v_{max} = A(2 \pi f)[/tex]

[tex]v_{max} = 3.40 cm (2 \pi *4.80)[/tex]

[tex]v_{max} = 102.54 \ cm/s[/tex]

Therefore, the maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c) Again;

[tex]v = A \omega cos \omega t[/tex]

By differentiation with respect to  t;

[tex]\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)[/tex]

[tex]a =- A \omega^2 sin \omega t[/tex]

The maximum acceleration of the sphere is;

[tex]a_{max} =A \omega^2[/tex]

where;

[tex]w = 2 \pi f[/tex]

[tex]a_{max} = A(2 \pi f)^2[/tex]

where A= 3.40 cm = 0.034 m

[tex]a_{max} = 0.034*(2 \pi *4.80)^2[/tex]

[tex]a_{max} = 30.93 \ m/s^2[/tex]

The maximum magnitude of the acceleration of the sphere is = 30.93 [tex]m/s^2[/tex]

The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of [tex]x = \pm A[/tex]

The sphere moves a total distance of 6.80 cm in one cycle. Its maximum speed is 102.5 cm/s as it passes through equilibrium and the maximum acceleration is at maximum excursion from equilibrium, with a magnitude of 18.5 m/s².

The sphere's total distance moved in one cycle is twice the amplitude, as it moves to the maximum amplitude and back again. Therefore, in the case of this sphere, it moves through a total distance of 6.80 cm during one cycle. The maximum speed of an object in simple harmonic motion occurs as it passes through equilibrium. You can calculate this speed using the formula v = ωA, where ω is the angular frequency and A is the amplitude.

The angular frequency is ω = 2πf, where f is the frequency. Thus, the maximum speed of the sphere is approximately 102.5 cm/s. Lastly, the maximum magnitude of acceleration occurs at maximum excursion from equilibrium, and can be calculated with the formula derived from Newton's second law a = ω²A, thus the acceleration is 18.5 m/s².

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Answer:

C. Gaming and simulation

Explanation:

In a presentation, you are just going to read from something and resume that to present. So there is not much higher order thinking or problem solving.

Word processing you will just be writing a test, so not much to learn in these aspects.

However

In gaming, there are different strategies your team may assume(which means that the students have to use high order thinking to choose the best strategy), you have to work well as a group, depending on the game(which is a needed skill on the real-world), and other things.

So the correct answer is:

C. Gaming and simulation

Answer:

C. Gaming and Simulation

Explanation:

Problem solving and high order thinking refers to the ability to use  knowledge, facts, and data to effectively solve problems. This  means the ability to assess problems and  find suitable solutions. It is the ability to develop a well thought out solution  within a reasonable time frame.

Integrating games and simulation into the learning process has a positive effect in developing the cognitive and behavioral abilities of students. Simulations promote the use of critical and evaluative thinking. Because they are ambiguous or open-ended, they encourage students to contemplate the implications of a scenario. It allows the teacher to properly engage the students while assuming a real world situation.

Due to constant practice, it allows the student to grasp the concept of a subject matter. The students are placed in a world defined by the teacher and they have enough motivation to learn because they now catch fun with activities that involve critical thinking.

Other options stated such as presentation and word processing only helps the students to give reports of activities and researches carried out. Gaming and simulation are the only ones that develop the critical thinking and problem solving skills of the students.

Answer:

Frequency of vibration will be equal to 720 Hz

Explanation:

It is given that car moves with a speed of 14.4 m/sec

Its mean that car cover 14.4 m in 1 sec

We have to find the frequency of vibration corresponding to this velocity

It is given that car has a tread pattern with a crevice every 2 cm

So number of crevices moved in 1 sec will be equal to [tex]=\frac{14.4}{0.02}=720[/tex]

So frequency of these vibration will be equal to 720 Hz

Answer:

The spacing between the slits is    [tex]d = 0.00145m[/tex]                

Explanation:

From the question we are told that

  The wavelength of the light is [tex]\lambda = 406.192nm = 406.192*10^{-9} m[/tex]

   The distance of the slit from the screen is [tex]D = 5.937 \ m[/tex]

    The number of bright fringe is [tex]n = 24[/tex]

     The  length the fringes span is   [tex]L = 39.835 mm = \frac{39.835 }{1000} = 0.0398 m[/tex]

The fringe width (i.e the distance of between two successive bright or dark fringe) is mathematically represented as

             [tex]\beta = \frac{\lambda D}{d}[/tex]

Where d is  the distance between the  slits

            [tex]\beta[/tex] is the fringe width which can also be evaluated as

                         [tex]\beta = \frac{L}{n}[/tex]

Substituting values

                        [tex]\beta = \frac{0.0398}{24}[/tex]

                          [tex]\beta = 1.660 *10^{-3}[/tex]

Making d the subject of formula in the above equation

                [tex]d = \frac{\lambda D}{\beta }[/tex]

Substituting values

                [tex]d = \frac{406.192 *10^{-9} * 5.937 }{1.660 *10^{-3}}[/tex]

                    [tex]d = 0.00145m[/tex]                

Answer:

Explanation:

Given:

Number of turns [tex]N = 200[/tex]

Cross sectional area [tex]A = 10^{-3}[/tex] [tex]m^{2}[/tex]

Rate of increasing magnetic field [tex]\frac{dB}{dt} = 6[/tex] [tex]\frac{T}{s}[/tex]

From the faraday's law,

Induced emf is given by,

   [tex]\epsilon = -N\frac{d \phi}{dt }[/tex]

Where [tex]\phi =[/tex] magnetic flux

  [tex]\phi = AdB \cos(0)[/tex]          ( because angle between normal coil and field is zero)

Where [tex]A =[/tex] area of coil

Put the value of [tex]\phi[/tex] in above equation,

Here we neglect minus sign

   [tex]\epsilon = NA\frac{dB}{dt}[/tex]

   [tex]\epsilon = 200 \times 10^{-3} \times 6[/tex]

   [tex]\epsilon = 1.2[/tex] V

Therefore, 1.2 volt induced in coil

Answer:

[tex]80.6\mu m[/tex]

Explanation:

When light passes through a narrow slit, it produces a diffraction pattern on a distant screen, consisting of several bright fringes (constructive interference) alternated with dark fringes (destructive interference).

The formula to calculate the position of the m-th maximum in the diffraction pattern produced in the screen is:

[tex]y=\frac{m\lambda D}{d}[/tex]

where

y is the distance of the m-th maximum from the central maximum (m = 0)

[tex]\lambda[/tex] is the wavelength of light used

D is the distance of the screen from the slit

d is the width of the slit

In this problem, we have:

[tex]\lambda=578.0 nm = 578\cdot 10^{-9} m[/tex] is the wavelength

D = 62.5 cm = 0.625 m is the distance of the screen

We know that the distance between the third order maximum (m=3) and the central maximum is 1.35 cm (0.0135 m), which means that

[tex]y_3 = 0.0135 m[/tex]

For

m = 3

Therefore, rearranging the equation for d, we find the width of the slit:

[tex]d=\frac{m\lambda D}{y_3}=\frac{(3)(578\cdot 10^{-9})(0.625)}{0.0135}=80.3\cdot 10^{-6} m=80.6\mu m[/tex]

The width a of the slit will be "80.6 μm". To understand the calculation, check below.

According to the question,

Light's wavelength, λ = 578.0 nm

Screen's distance, D = 62.5 or,

                                   = 0.625 m

Third order maximum, m = 3

Central maximum = 1.35 cm or,

                               = 0.0135 m

We know the relation,

→ y = [tex]\frac{m \lambda D}{d}[/tex]

or,

→ Width, d = [tex]\frac{m \lambda D}{y_3}[/tex]

By substituting the values,

                 = [tex]\frac{3\times 578.10^{-9}\times 0.625}{0.0135}[/tex]

                 = 80.3 × 10⁻⁶ m or,

                 = 80.6 μm    

Thus the above answer is correct.

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Answer: 0.137 m

Explanation:

Given

Mass of brick, m = 3 kg

Angle of inclination, Φ = 34°

Force constant of the spring, k = 120 N/m

The force of the brick, F can be gotten using the relation

F = mg

F = 3 * 9.8

F = 29.4 N

Now, the force parallel to the incline, F(p) can be gotten using the formula,

F(p) = F sinΦ, so that

F(p) = 29.4 * sin 34

F(p) = 29.4 * 0.559

F(p) = 16.4 N

The stretch distance then is,

d = F(p) / k * 1 m

d = 16.4 / 120

d = 0.137 m

Thus, the spring stretched by a distance of 0.137 m

Answer:

The transverse displacement is   [tex]y(1.51 , 0.150) = 0.055 m[/tex]    

Explanation:

 From the question we are told that

     The generally equation for the mechanical wave is

                    [tex]y(x,t) = Acos (kx -wt)[/tex]

     The speed of the transverse wave is [tex]v = 8.25 \ m/s[/tex]

     The amplitude of the transverse wave is [tex]A = 5.50 *10^{-2} m[/tex]

     The wavelength of the transverse wave is [tex]\lambda = 0540 m[/tex]

      At t= 0.150s , x = 1.51 m

 The angular frequency of the wave is mathematically represented as

          [tex]w = vk[/tex]

Substituting values  

         [tex]w = 8.25 * 11.64[/tex]

        [tex]w = 96.03 \ rad/s[/tex]

The propagation constant k is mathematically represented as

                  [tex]k = \frac{2 \pi}{\lambda}[/tex]

Substituting values

                  [tex]k = \frac{2 * 3.142}{0. 540}[/tex]

                   [tex]k =11.64 m^{-1}[/tex]

Substituting values into the equation for mechanical waves

      [tex]y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03 * 0.150))[/tex]

       [tex]y(1.51 , 0.150) = 0.055 m[/tex]    

To calculate the absolute pressure at a depth in a fluid, sum the atmospheric pressure and the pressure due to the fluid, while considering the fluid's density, gravitational acceleration, and depth. To find the force exerted by the fluid on a circular window at that depth, multiply the pressure by the area of the window.

To begin with, you need to realize that the absolute pressure at a certain depth in a fluid is the sum of the atmospheric pressure and the pressure due to the fluid itself. This is because of Pascal's principle. In this case, the absolute pressure (Pabs) can be calculated as follows:

Pabs = Patm + density * g * h (height or depth in the fluid)

Here, Patm (atmospheric pressure) = 101.3 kPa = 101300 Pa (since 1 kPa = 1000 Pa), the density of water = 1.00 ✕ 103 kg/m3, acceleration due to gravity (g) = 9.81 m/s2 (approx), and h = 29.0 m

Using these values in the formula gives the absolute pressure in the fluid at a depth of 29.0 m:

Next, to determine the force exerted by the fluid on the circular window of the instrument probe at this depth, we have to understand that this force is actually the pressure at that depth multiplied by the area over which the pressure is exerted (Force = Pressure * Area) and the area of a circle = π * (d/2)²

Plug the determined pressure and given diameter into this equation to get the force in Newtons (N). Here, d = 3.90 cm = 0.039 m (since 1 cm = 0.01 m).

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The absolute pressure in the fluid at a depth of 29.0 m is 2.83 × 10^4 Pa. The force exerted by the fluid on the window of the instrument probe at this depth is 10.9 N.

To determine the absolute pressure in the fluid at a depth of 29.0 m, we need to consider the hydrostatic pressure. The hydrostatic pressure is given by the equation p = hρg, where p is the pressure, h is the depth, ρ is the density of the fluid, and g is the acceleration due to gravity. In this case, the density of the water is 1.00 × 103 kg/m3. We can substitute these values into the equation and solve for the pressure:

p = (29.0 m)(1.00 × 103 kg/m3)(9.81 m/s2) = 2.83 × 104 Pa

Therefore, the absolute pressure in the fluid at a depth of 29.0 m is 2.83 × 104 Pa.

To determine the force exerted by the fluid on the window of an instrument probe at this depth, we can use the formula F = PA, where F is the force, P is the pressure, and A is the area. The area of the circular window can be calculated using the formula A = πr2, where r is the radius of the window. Given that the diameter of the window is 3.90 cm, the radius is half of the diameter, so r = 1.95 cm = 0.0195 m. Substituting the values into the formulas, we can find the force:

F = (2.83 × 104 Pa)(π(0.0195 m)2) = 10.9 N

Therefore, the force exerted by the fluid on the window of the instrument probe at a depth of 29.0 m is 10.9 N.

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a)

i) 120 s

ii) 1.57 m/s

b)

i) See attachment

ii) Up

c) [tex]N=mg+m\frac{v_b^2}{R}[/tex]

d) Greater than

Explanation:

The problem is incomplete: find the complete text in attachments.

a)

i) The period of revolution of the book is equal to the total time taken by the book to complete one revolution.

Looking at the graph, the period of revolution can be estimated by evaluating the difference in time between two consecutive points of the motion of the book that have the same shape.

For instance, we can evaluate the period by calculating the difference in time between two consecutive crests. We see that:

- The first crest occur at t = 90 s

- The second crest occurs at t = 210 s

Therefore, the period of revolution is

T = 210 - 90 = 120 s

ii)

The tangential speed of the book is given by the ratio between the distance covered during one revolution (so, the perimeter of the wheel) and the period of revolution.

Mathematically:

[tex]v_b=\frac{2\pi R}{T}[/tex]

where

R is the radius of the wheel

T = 120 s is the period

From the graph, we see that the maximum position of the book is x = +30 m, while the minimum position is x = -30 m, so the diameter of the wheel is

d = +30 - (-30) = 60 m

So the radius is

R = d/2 = 30 m

So, the speed is

[tex]v_b=\frac{2\pi (30)}{120}=1.57 m/s[/tex]

b)

i) See in attachment the free-body diagram of the book at its lowest position.

There are 2 forces acting on the book at the lowest position:

- The weight of the book, of magnitude

[tex]W=mg[/tex]

where m is the mass of the book and g the acceleration due to gravity. This force acts downward

- The normal force exerted by the bench on the book, of magnitude N. This force acts upward

ii)

When the book is at its lowest position, it is moving horizontally at constant speed.

However, the book is accelerating. In fact, acceleration is the rate of change of velocity, and velocity is a vector, so it has both a speed and a direction; here the speed is not changing, however, the direction is changing (upward), so the book has an upward net acceleration.

According to Newton's second law of motion, the net vertical force on the book is proportional to its net vertical acceleration:

[tex]F=ma[/tex]

where F is the net force, m is the mass, a is the acceleration. Therefore, since a is different  from zero, the book has a net vertical force, in the same direction of the acceleration (so, upward).

c)

As we said in part b), there are two forces acting on the book at its lowest position:

- The weight, [tex]W=mg[/tex], downward

- The normal force of the bench, N, upward

Since the book is in uniform circular motion, the net force on it must be equal to the centripetal force [tex]m\frac{v_b^2}{R}[/tex], so we can write:

[tex]N-mg=m\frac{v_b^2}{R}[/tex]

where

[tex]v_b[/tex] is the speed of the book

R is the radius of the path

Therefore, we find an expression for the normal force:

[tex]N=mg+m\frac{v_b^2}{R}[/tex]

d)

As we said in part c) and d):

- The normal force acting on the book at its lowest position is

[tex]N=mg+m\frac{v_b^2}{R}[/tex]

- The weight (force of gravity) of the book is

[tex]W=mg[/tex]

By comparing the two equations above, we observe that

[tex]N>W[/tex]

Therefore, we can conclude that the normal force exerted by the bench on the book is greater than the weight of the book.

To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations

[tex]\text{Mass of the child} = m = 25kg[/tex]

[tex]\text{Acceleration due to gravity} = g = 9.81m/s^2[/tex]

[tex]\text{Height lifted} = h = 0.80m (Upward)[/tex]

Work done to upward the object

[tex]W = mgh[/tex]

[tex]W = (25)(9.81)(0.8)[/tex]

[tex]W = 196.2J[/tex]

Horizontal Force applied while carrying 10m,

[tex]F = 0N[/tex]

[tex]W = 0J[/tex]

Height descended in setting the child down

[tex]h' = -0.8m (Downwoard)[/tex]

[tex]W = mgh'[/tex]

[tex]W = (25)(9.81)(-0.80)[/tex]

[tex]W = -196.2J[/tex]

For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.

To calculate the work done on the child during different parts of the trip and for the whole trip, we need to consider the weight of the child and the vertical distances involved. The work done in lifting the child up and setting the child back down is equal to the weight of the child multiplied by the vertical distance. The work done in carrying the child horizontally is zero.

In this problem, we need to calculate the work done on the child during different parts of the trip and for the whole trip. Work, W, is defined as the product of force and displacement, W = Fd. Since the child is being lifted vertically, the work done in lifting the child up is equal to the weight of the child multiplied by the vertical distance lifted, Wup = mgh. The work done in carrying the child horizontally is zero since the displacement is perpendicular to the force. Finally, the work done in setting the child back down is also equal to the weight of the child multiplied by the vertical distance lowered, Wdown = mgh. Therefore, the total work done on the child for the whole trip is the sum of the work done in lifting the child up and setting the child back down, Wtotal = Wup + Wdown.

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Answer:

Induced emf in first coil is 0.986 [tex]\mu T[/tex] and in second case 0.396 [tex]\mu T[/tex]

Explanation:

Given:

Number of turns per centimeter [tex]n = 40[/tex]

Current [tex]I = 0.240[/tex] A

Current rate [tex]\frac{dI}{dt} = \frac{0.240}{0.70} = 0.343[/tex] [tex]\frac{A}{s}[/tex]

The magnetic field in solenoid is given by,

  [tex]B = \mu _{o} nI[/tex]

Where [tex]\mu _{o} = 4\pi \times 10^{-7}[/tex]

We write,

  [tex]\frac{dB}{dt} = \mu_{o} n \frac{dI}{dt}[/tex]

  [tex]\frac{dB}{dt} = 4\pi \times 10^{-7} \times 40 \times 0.343[/tex]

  [tex]\frac{dB}{dt} = 172.3 \times 10^{-7}[/tex]

(A)

Number of turns [tex]N = 48[/tex]

Radius of coil [tex]r = \frac{d}{2} = 1.95 \times 10^{-2}[/tex] m

From faraday's law

   [tex]\epsilon = NA \frac{dB}{dt}[/tex]

Where [tex]A = \pi r^{2} = 3.14 (1.95 \times 10^{-2} ) ^{2} = 11.93 \times 10^{-4}[/tex] [tex]m^{2}[/tex]

   [tex]\epsilon = 48 \times 11.93 \times 10^{-4} \times 172.3 \times 10^{-7}[/tex]  

   [tex]\epsilon = 98665.87 \times 10^{-11}[/tex]

   [tex]\epsilon = 0.986 \mu T[/tex]

(B)

Number of turns [tex]N = 96[/tex]

Radius of coil [tex]r = \frac{d}{2} = 3.9 \times 10^{-2}[/tex] m

From faraday's law

   [tex]\epsilon = NA \frac{dB}{dt}[/tex]

Where [tex]A = \pi r^{2} = 3.14 (3.9 \times 10^{-2} ) ^{2} = 47.76 \times 10^{-4}[/tex] [tex]m^{2}[/tex]

   [tex]\epsilon = 48 \times 47.96 \times 10^{-4} \times 172.3 \times 10^{-7}[/tex]  

   [tex]\epsilon = 396648.38 \times 10^{-11}[/tex]

   [tex]\epsilon = 0.396 \mu T[/tex]

Therefore, induced emf in first coil is 0.986 [tex]\mu T[/tex] and in second case 0.396 [tex]\mu T[/tex]

The substance that can be separated into several different elements is air. The correct option is A.

Air, which is a mixture mainly composed of nitrogen and oxygen, among other gases, can thus be separated into its component elements. Substances like Iron (Fe), Hydrogen (H), and Nickel (Ni) are elements and cannot be broken down into simpler substances by ordinary chemical means.

Answer:

To oppose applied magnetic field current will flow in anticlockwise direction

Explanation:

Given:

Uniform magnetic field directed vertically upward.

Current will flow in clockwise direction

Here loop of wire is suddenly contracts to half so diameter of loop is reduced.

Hence less number of magnetic field line pass through the loop. This change in magnetic field lines lead to flow of current.

Now from lenz law flow of induced current will oppose the cause of its production

Therefore, to oppose applied magnetic field current will flow in anticlockwise direction.

Answer:

The speed of the box at the top of the hill will be 5.693m/s.

Explanation:

The kinetic energy of the box at the bottom of the hill is

[tex]K.E = \dfrac{1}{2}mv^2[/tex]

putting in [tex]m =24kg[/tex] and  [tex]v = 12.1m/s[/tex] we get

[tex]K.E = \dfrac{1}{2}(24kg)(12.1)^2\\\\K.E = 1756.92J[/tex]

Now, the potential energy this box gains as it rises [tex]h =5.7m[/tex] up the hill is

[tex]P.E = mgh[/tex]

[tex]P.E = (24kg)(10ms/s^2)(5.7m)\\\\P.E = 1368[/tex]

Therefore, the energy left [tex]E_{left}[/tex] in the box at the top if the hill will be

[tex]E_{left} =K.E - P.E = 1756.92J-1368J\\[/tex]

[tex]\boxed{E_{left} = 388.92J}[/tex]

This left-over energy must appear as the kinetic energy of the box at the top of the hill (where else could it go? ); therefore,

[tex]\dfrac{1}{2}mv_t^2= 388.92J[/tex]

putting in numbers and solving for [tex]v_t[/tex] we get:

[tex]\boxed{v_t = 5.693m/s.}[/tex]

Thus, the speed of the box at the top of the hill is 5.693m/s.

Answer:

[tex]\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875[/tex]

Explanation:

We name M the larger mas, and m the smaller mass, so base on this we write the following conservation of momentum equation for the collision:

[tex]P_i=P_f\\M\,v_i -m\,v_i=(M+m)\,v_f\\(M-m)\,v_i = (M+m)\,v_f\\\\\frac{v_i}{v_f} = \frac{(M+m)}{(M-m)}[/tex]

We can write this in terms of what we are looking for (the quotient of masses [tex]\frac{m}{M}[/tex]:

[tex]\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}[/tex]

We use now the information about  Kinetic Energy of the system being reduced in half after the collision:

[tex]K_i=2\,K_f\\\frac{1}{2} (M+m)\,v_i^2= 2*\frac{1}{2} (M+m)v_f^2\\v_i^2=2\,v_f^2\\\frac{v_i^2}{v_f^2} =2[/tex]

We can combine this last equation with the previous one to obtain:

[tex]\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}\\(\frac{v_i}{v_f})^2 = \frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}\\2=\frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}[/tex]

where solving for the quotient m/M gives:

[tex]\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875[/tex]

Answer:

Explanation:

check attachment  for the solution.

(a) The network done is equal to zero.

(b) The change in the potential energy is equal to the 5.54 ×10⁶ J

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

The network done is given as;

[tex]\rm W= Fd cos \alpha \\\\ \rm W= F \times D cos 90^0 \\\\\rm W=0\ J[/tex]

Hence the net work done is equal to zero.

(b) The change in the potential energy is equal to the 5.54 ×10⁶ J

[tex]\rm \triangle U_g= mg (h+x sin\theta)\\\\ \rm \triangle U=1200 \times 9.81(45+3.5sin 37.8^0)\\\\ \rm \triangle U_g=5.5443 \times 10^6 \ J[/tex]

Hence the changes in the potential energy are equal to the 5.54 ×10⁶ J.

(c) The change in the thermal energy will be 3.3425×10⁶ J.

The formula for the thermal energy change is found as;

[tex]\rm \triangle E_{thermal} =\mu_k mgcos \theta (\frac{h-L}{sin \theta} +x )\\\\ \rm \triangle E_{thermal} =0.6 \times 1200 \times 9.81 \times 37.8^0 ( (\frac{45-18.4}{sin 37.8^0} +3.5 )\\\\ \rm \triangle E_{thermal} =3.342 \times 10^6 \ J[/tex]

Hence the changes in the thermal energy will be 3.3425×10⁶ J.

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Answer:

a)  x = 0.33 A , b)   K = ¾ Em , c) the kinetic energy increases 9 times

Explanation:

a) In a simple harmonic motion the mechanical energy is conserved and is expressed by the relation

       Em = ½ k A²

At all points of movement the mechanical energy and

       Em = K + U

remember that potential energy is

       U = ½ k x²

they ask us for the point where

        U = 1/9 Em

  we substitute

       ½ k x² = 1/9 (½ k A²)

       x = √1/9     A

       x = 0.33 A

b) At all points the equation for mechanical energy is

      Em = K + U

      K = Em - U

      K = ½ k A² - ½ k x²

      K = ½ k (A² -x²)

at point x = ½ A

     K = ½ k (A² - ¼ A²)

     K = (½ k A²) ¾

     K = ¾ Em

     U = ½ k (½A) 2

     U = ¼ Em

the fraction of energy e

   U / K = 1/3

c) Kinetic energy is

       K = ½ k v²

the system is described by the expression

        x = A cos (wt + Ф)

speed is defined

         v = dx / dt

         v = - A w sin (wt + Ф)

we substitute

         K = ½ k (- A w sin ( wt + Ф))²

We write this equation for the initial amplitude A

        K₀ = ½ k (w sin (wt + Ф))² A²

we write it for the new amplitude A´ = 3 A

    K = ½ k (w sin (wt +Ф))² (3A)²

the relationship between this energies is

    K / K₀ = 9

whereby the kinetic energy increases 9 times

The displacement for 1/9 of mechanical energy is A/3. When the displacement is half amplitude, 3/4 of total mechanical energy becomes kinetic energy. If the amplitude increases by a factor of 3, the maximum kinetic energy increases by a factor of 9.

In this context, we know that the total mechanical energy (E) of the system is constant and is the sum of kinetic and potential energy. The potential energy (U) of a mass-spring system is given by the formula U = 1/2 kx².

Part (a)

We want to find the displacement x such that the potential energy is 1/9 total mechanical energy, thus we set U = E/9. From substituting U = 1/2kx² and E = 1/2kA² in the equation,

we find that x²/A² = 1/9, which gives x = A/3.

Part (b)

We know that when the particle is at x = A/2 the potential energy is U = 1/2 k(A/2)² = 1/8 kA².

The total energy E = 1/2 kA², so the kinetic energy (K) is given by E - U = 1/2kA² - 1/8kA² = 3/8 kA².

The fraction of the total energy which is kinetic is (K/E) = (3/8)/(1/2) = 3/4.

Part (c)

The maximum kinetic energy is the total mechanical energy (since potential energy is zero at maximum kinetic energy), which is E = 1/2*kA².

If we triple the amplitude, the new energy E' = 1/2*k(3A)² = 9/2*kA².

The factor by which the maximum kinetic energy changes is (E'/E) = (9/2)/(1/2) = 9.

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Answer:

The least and second least thicknesses of the film are 0.115 um and 0.346 um respectively.

Explanation:

Optical path length ===> 2n * t = (m + 0.5) * λ

λ = 614 nm , n = 1.33

Substitute in the parameters in the equation.

∴ 2(1.33) * t = (m + 0.5) * 614

  2.66 * t = 614m + 307

  t = (614m + 307) / 2.66 .............(1)

(a) for m = 0

    t = (614m + 307) / 2.66

    t = (614(0) + 307) / 2.66

    t = 307 / 2.66

    t = 115 nm == 0.115 um

(b) for m = 1

    t = (614(1) + 307) / 2.66

    t = (614 + 307) / 2.66

    t = 921 / 2.66

    t = 346.24 nm = 0.346 um

Answer:

The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]

The magnitude of the force at A is 832.56N

Explanation:

Here, m is the mass of the beam and l is the length of the beam.

[tex]I =\frac{1}{3} ml[/tex]

[tex]I=\frac{1}{3} \times110\times4^2\\I=586.67kgm^2[/tex]

Take the moment about point A by applying moment equilibrium equation.

[tex]\sum M_A =I \alpha[/tex]

[tex]P \sin45^0 \times 3 =I \alpha[/tex]

Here, P is the force applied to the attached cable and [tex]\alpha[/tex] is the angular acceleration.

Substitute 350 for P and 586.67kg.m² for I

[tex]350 \sin 45^0 \times3=568.67 \alpha[/tex]

[tex]\alpha =1.3056rad/s^2[/tex]

The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]

Find the acceleration along x direction

[tex]a_x = r \alpha[/tex]

Here, r is the distance from center of mass of the beam to the pin joint A.

Substitute 2 m for r and 1.3056rad/s² for [tex]\alpha[/tex]

[tex]a_x = 2\times 1.3056 = 2.6112m/s^2[/tex]

Find the acceleration along the y direction.

[tex]a_y = r \omega ^2[/tex]

Here, ω is angular velocity.

Since beam is initially at rest,ω=0

Substitute 0 for ω

[tex]a_y = 0[/tex]

Apply force equilibrium equation along the horizontal direction.

[tex]\sum F_x =ma_x\\A_H+P \sin45^0=ma_x[/tex]

[tex]A_H + 350 \sin45^0=110\times2.6112\\\\A_H=39.75N[/tex]

Apply force equilibrium equation along the vertical direction.

[tex]\sum F_y =ma_y\\A_V-P \cos45^0-mg=ma_y[/tex]

[tex]A_v +350 \cos45^0-110\times9.81=0\\A_V = 831.61\\[/tex]

Calculate the resultant force,

[tex]F_A=\sqrt{A_H^2+A_V^2} \\\\F_A=\sqrt{39.75^2+691.61^2} \\\\= 832.56N[/tex]

The magnitude of the force at A is 832.56N

Answer:

a) Initial angular acceleration of the beam = 1.27 rad/s²

b) [tex]F_{A} = 851.11 N[/tex]

Explanation:

[tex]tan \theta = \frac{opposite}{Hypothenuse} \\tan \theta = \frac{3}{3} = 1\\\theta = tan^{-1} 1 = 45^{0}[/tex]

Force applied to the attached cable, P = 350 N

Mass of the beam, m = 110-kg

Mass moment of the inertia of the beam about point A = [tex]I_{A}[/tex]

Using the parallel axis theorem

[tex]I_{A} = I_{G} + m(\frac{l}{2} )^{2} \\I_{G} = \frac{ml^{2} }{12} \\I_{A} = \frac{ml^{2} }{12} + \frac{ml^{2} }{4} \\I_{A} = \frac{ml^{2} }{3}[/tex]

Moment = Force * Perpendicular distance

[tex]\sum m_{A} = I_{A} \alpha\\[/tex]

From the free body diagram drawn

[tex]\sum m_{A} = 3 Psin \theta\\ 3 Psin \theta = \frac{ml^{2} \alpha }{3}[/tex]

P = 350 N, l = 3+ 1 = 4 m, θ = 45°

Substitute these values into the equation above

[tex]3 * 350 * sin 45 = \frac{110 * 4^{2}* \alpha }{3} \\\alpha = 1.27 rad/s^{2}[/tex]

Linear acceleration along the x direction is given by the formula

[tex]a_{x} = r \alpha[/tex]

r = 2 m

[tex]a_{x} = 2 * 1.27\\a_{x} = 2.54 m/s^{2}[/tex]

the linear acceleration along the y-direction is given by the formula

[tex]a_{y} = r w^{2}[/tex]

Since the beam is initially at rest, w = 0

[tex]a_{y} = 0 m/s^{2}[/tex]

General equation of motion along x - direction

[tex]F_{x} + Psin \theta = ma_{x}[/tex]

[tex]F_{x} + 350sin45 = 110 * 2.54\\F_{x} = 31.913 N[/tex]

General equation of motion along y - direction

[tex]F_{x} + Pcos \theta - mg = ma_{y}[/tex]

[tex]F_{y} + 350cos45 - 110*9.8 = m* 0\\F_{y} = 830.513 N[/tex]

Magnitude [tex]F_{A}[/tex] of the force supported by the pin at A

[tex]F_{A} = \sqrt{F_{x} ^{2} + F_{y} ^{2} } \\F_{A} = \sqrt{31.913 ^{2} + 850.513 ^{2} } \\F_{A} = 851.11 N[/tex]

Answer:

       F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v²

          Em₀ = Emf

          e ΔV = ½ m v²

          v =√ 2 e ΔV / m

          v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

           v = √(1.8075 10¹⁶)

           v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

        F = 1.6 10⁻¹⁹  1.344 10⁸ 0.4

       For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Emo = U = q DV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v2

          Emo = Emf

          .e DV = ½ m v2

          .v = RA 2 e DV / m

          .v = RA (2 1.6 10-19 51400 / 9.1 10-31)

           .v = RA (1.8075 10 16)

           .v = 1,344 108 m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

       F = 1.6 10-19 1,344 108 0.4

       F = 8.6 10-12 N

Answer: The meteorologist can report the intensity of a tornado only after it has occurred because (Tornado intensity is based on damage done.)

Explanation:

HURRICANE is a type of violent storm called topical cyclone which is characterised by a large rotating storm with high speed winds that forms over warm waters in tropical areas. They are form over the warm ocean water of the tropics. When warm moist air over the water rises, it is replaced by cooler air. The cooler air will then warm and start to rise. This cycle causes huge storm clouds to form. These storm clouds will begin to rotate with the spin of the Earth forming an organized system. If there is enough warm water, the cycle will continue and the storm clouds and wind speeds will grow causing a hurricane to form.

Hurricane can cause alot of damages such as flooding and storm surge. And they are also capable of developing to TORNADO.

The intensity of any tornado that occurred can only be measured through the damage it causes. This is because the correct intensity can only be determined while on site or remote sensing of the tornado which is quite impractical for wire scale use, therefore damages as a result of the tornado is used to quantify it.

A television meteorologist can inform about the approaching hurricane but not a tornado because of its formation and movements.

A television meteorologist can inform about the approaching hurricane because it has t its pattern, which continues from the oceans and when it enters the land.

Meteorologist focuses their attention on these patterns for hurricanes in the Atlantic Ocean, and most of the time, their movement is predictable.

Tornados happen when the air pressure is strong in the ground for creating it.

The formation of a tornado can be sudden within minutes.

The direction of a tornado movement is unpredictable and based on its air pressure.

We have also seen meteorologists chasing a tornado and getting caught in it.

Therefore we can conclude that both have different patterns based on climate and temperature.

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Answer:

[tex]P_{c[/tex]= 5.74W

[tex]P_{w}[/tex]=0.27 W

Explanation:

d= 2.6cm =>0.026m (for both the ceramic and the carpet )

Thermal conductivity of wool '[tex]k_{w}[/tex]'= [tex]k_{carpet}[/tex] = [tex]k_{wool}[/tex]=  0.04J/(sm °C)

Thermal conducticity of carpet '[tex]k_c}[/tex]' = 0.84 J/(sm °C)

Area 'A'= [tex]A_{carpet}[/tex]= [tex]A_{ceramic}[/tex]= 77.2cm²=> 77.2 x [tex]10^{-4}[/tex]m²

[tex]T_h}[/tex]=33.0°C

[tex]T_c}[/tex]=10.0°C

Average Power [tex]P_{avg}[/tex] is determined by dividing amount of energy'Q' by time taken for the transfer't':

[tex]P_{avg}[/tex] = Q/Δt

Due to conductivity, heat of flow rate will be P= dQ/dt

P=[tex]\frac{dQ}{dt}[/tex] = [tex]\frac{kA[T_{h}-T_{c} ]}{d}[/tex]

For CERAMIC:

[tex]P_{c[/tex]=[tex]\frac{k_{c} A[T_{h}-T_{c} ]}{d}[/tex] => [0.84 x 77.2 x [tex]10^{-4}[/tex](33-10) ]/0.026

[tex]P_{c[/tex]= 5.74W

For WOOL CARPET:

[tex]P_{w}[/tex]= [tex]\frac{k_{w} A[T_{h}-T_{c} ]}{d}[/tex]=> [0.04 x 77.2 x [tex]10^{-4}[/tex](33-10) ]/0.026

[tex]P_{w}[/tex]=0.27 W

Answer:

Distance will be 2.81 m

Explanation:

Detailed explanation and calculation is shown in the image below.

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

Answer:

The relative velocity of the queen is -vy

Explanation:

If the collision is elastic, thus e = 1. The expression is equal:

[tex]e=\frac{relative-velocity-of-approach}{relative-velocity-of-separation} \\relative-velocity-of-approach=relative-velocity-of-separation[/tex]

The relative velocity of separation is:

relative velocity of separation = 0 - vy = -vy

This expression means that:

velocity of queen - velocity of strikes = -vy

Thus the relative velocity of the queen with respect to the striker is equal to -vy