A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10 cm/s and no damping is applied.
(a) Determine the position u of the mass at any time t. Use 9.8 m/s as the acceleration due to gravity. Pay close attention to the units.
(b) When does the mass first return to its equilibrium position?

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

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:

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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Final answer:

Using the conservation of momentum for an inelastic collision, Ender's mass is calculated to be approximately 51.43 kg when the given velocities and Shen's mass are considered.

Explanation:

The subject of this question is physics, specifically dealing with the concept of conservation of momentum during collisions. We are asked to find Ender's mass given the velocities and mass involved in a collision in space. Since the collision here is inelastic (Ender and Shen hold on to each other), the total momentum before the collision must be equal to the total momentum after the collision.

According to the conservation of momentum, m1 × v1 + m2 × v2 = (m1 + m2) × v3, where m1 and m2 are the masses of Ender and Shen respectively, and v1, v2, and v3 are their velocities.

We know Shen's mass (m2) is 45 kg, Ender's velocity (v1) is 12 m/s, Shen's velocity (v2) is 9 m/s, and the combined velocity after the collision (v3) is 6.4 m/s. Using the momentum conservation formula, we can solve for Ender's mass (m1) as follows:

m1 × 12 m/s + 45 kg × 9.0 m/s = (m1 + 45 kg) × 6.4 m/s

Expanding this and rearranging the terms, we get:

12m1 = 6.4m1 + 6.4 × 45

12m1 - 6.4m1 = 6.4 × 45

m1(12 - 6.4) = 6.4 × 45

m1 = (6.4 × 45) / (12 - 6.4)

m1 = 288 / 5.6

m1 = 51.43 kg

Therefore, Ender's mass is approximately 51.43 kg.

a)

i) Negative

ii) 160 J

b) 1280 N/m

c) [tex]32 m/s^2[/tex]

d) 1.27 Hz

e)

i) See attached plot

ii) See attached plot

Explanation:

a)

i) The work done by a force is given by

[tex]W=Fx cos \theta[/tex]

where

F is the force

x is the displacement of the object

[tex]\theta[/tex] is the angle between the direction of the force and the direction

Here we have:

- The force that the spring exerts on the box is to the left (because the box is moving to the right, trying to compress the spring)

- The displacement of the box is to the right

So, F and x have opposite direction, and so [tex]\theta=180^{\circ}[/tex] and [tex]cos \theta=-1[/tex], which means that the work done is negative.

ii)

According to the work-energy theorem, the work done by the spring is equal to the change in kinetic energy of the box:

[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]

where

[tex]K_i[/tex] is the initial kinetic energy of the box

[tex]K_f[/tex] is the final kinetic energy

m = 20 kg is the mass of the box

u = 4.0 m/s is its initial speed

v = 0 m/s is the final speed (the box comes to rest)

Therefore,

[tex]W=\frac{1}{2}(20)(0)^2-\frac{1}{2}(20)(4.0)^2=-160 J[/tex]

So, the magnitude is 160 J.

b)

The elastic energy stored in a spring when it is compressed is given by

[tex]U=\frac{1}{2}kx^2[/tex]

where

k is the spring constant

x is the stretching/compression of the spring

Due to the law of conservation of energy, the kinetic energy lost by the box is equal to the elastic energy gained by the spring, so:

[tex]|W|=U=\frac{1}{2}kx^2[/tex]

We have

[tex]|W|=160 J[/tex]

x = 0.50 m is the maximum compression of the spring

Solving for k:

[tex]k=\frac{2U}{x^2}=\frac{2(160)}{(0.50)^2}=1280 N/m[/tex]

c)

The magnitude of the force exerted on the box is given by

[tex]F=kx[/tex]

where

k = 1280 N/m is the spring constant

x = 0.50 m is the compression of the spring

Substituting,

[tex]F=(1280)(0.50)=640 N[/tex]

Now we can find the maxmum acceleration using Newton's second law:

[tex]a=\frac{F}{m}[/tex]

where

F = 640 N is the maximum force

m = 20 kg is the mass of the box

So,

[tex]a=\frac{640}{20}=32 m/s^2[/tex]

d)

The frequency of oscillation of a spring-mass system is given by

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

where

k is the spring constant

m is the mass

Here we have:

k = 1280 N/m is the spring constant of this spring

m = 20 kg is the mass of the box

So, the frequency of this system is:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{1280}{20}}=1.27 Hz[/tex]

e)

i)

Here we want to sketch the kinetic energy of the box as a function of the position, x: find this graph in attachment.

In a spring-mass oscillating system, the kinetic energy is zero when the system is at the extreme position, i.e. when the spring is maximum compressed/stretched. In this problem, this happens when x = - 0.50 m and x = +0.50 m (we called x = 0 the position of equilibrium of the spring). In these positions in fact, the mass has zero speed, so its kinetic energy is zero.

On the other hand, the box has maximum speed when x = 0 (because it's the moment where all the elastic energy is converted into kinetic energy, which is therefore maximum, and so the speed is also maximum).

ii)

Here we want to plot the acceleration of the box as a function of the position x: find the graph in attachment.

In a spring-mass system, the acceleration is proportional to the negative of the displacement, since the restoring force

[tex]F=-kx[/tex]

By rewriting the force using Newton's second Law, we have

[tex]ma=-kx \\a=-\frac{k}{m}x[/tex]

Which means that acceleration is proportional to the displacement, but with opposite sign: so, this graph is a straight line with negative slope.

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.

Answer:

R = 1.81 10² km

Explanation:

Let's start by looking for the power in the visible range emitted this is 10W, the energy of that power is one second is

         P = E₁ / t

         E₁ = P t

         E₁ = 10 J

 Let's find the energy of a photon with Planck's equation

          E = h f

          c = λ f

we substitute

         E = h c /λ

         E = 6.63 10⁻³⁴ 3 10⁸/580 10⁻⁹

         E = 3.42 10⁻¹⁹ J

we can use a direct proportions rule to find the number of photons in the energy E₁

          #_photon = E₁ / E

          #_photon = 10 / 3.42 10⁻¹⁹

          #_photon = 2.92 10¹⁹ photons

This number of photons is distributed on the surface of a sphere. Let's find what the distance is so that there are 500 photons in 3 mm = 0.003 m.

the area of ​​the sphere is

            A = 4π R²

area of ​​the circle is

           A´ = π r²

as the intensity is constant over the entire sphere

         P = #_photon / A = 500 / A´

       # _photon / 4π R² = 500 / π r²

       R² = #_photon r² / 4 500

       r = d / 2 = 0.003 / 2 = 0.0015 m

       R² = 2.92 10¹⁹ 0.0015 2/2000

       R = √ (3,285 10¹⁰)  

        R = 1.81 10⁵ m

       R = 1.81 10² km

Answer:

They are equal

Explanation:

Newton 3rd Law of motion states that for every force applied or action there is usually an equal and opposite force.

Impulse = Force * time

It is measured in Newton seconds.

The force and time of collision is the same which translates to an equal impulse by both scenarios.

The impulse imparted to the smaller mass by the larger mass is equal in magnitude to the impulse imparted to the larger mass by the smaller one, due to the conservation of momentum and Newton's third law. Both experience equal and opposite momentum transfers, ensuring the total momentum of the system remains constant.

In the context of collisions, impulse is defined as the change in momentum of an object when it is subjected to a force over a period of time. According to Newton's third law, 'For every action, there is an equal and opposite reaction,' meaning that the impulse imparted to the smaller mass by the larger mass is exactly equal in magnitude to the impulse imparted to the larger mass by the smaller one, although the direction of the impulses will be opposite. When considering the conservation of momentum, the total momentum before the collision must equal the total momentum after the collision if no external forces are acting on the system (assuming a closed system). Therefore, if two cars collide, such as described in the provided text, regardless of their masses, the momentum transfer will be the same for both, thus the total momentum of the system remains constant.

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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Answer:

Coefficient of static friction will be equal to 0.642  

Explanation:

We have given acceleration [tex]a=6.3m/sec^2[/tex]

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to [tex]a=\mu g[/tex], here [tex]\mu[/tex] is coefficient of static friction and g is acceleration due to gravity

So [tex]\mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642[/tex]

So coefficient of static friction will be equal to 0.642

Answer:

a) Seems true at first glance, but on further inspection, the statement is false.

b) True

c) False

d) True

e) True

f) False

g) False

Explanation:

Taking the statements one by one

a.) The Clausius statement denies the possibility of heat transfer from a cooler to a hotter body.

The Clausius Statement denies the possibility of heat transfer from a cooler to a hotter body without extra work. It does not outrightly state that there is no possibility of heat transfer from a cooler to a hotter body.

For example, an Air conditioner or refrigerator rejects heat from a cold reservoir to a hot reservoir.

So, this statement is false.

b) The COP of a reversible refrigeration cycles is equal or greater than the COP of an irreversible refrigeration cycle if both cycles operate between the same thermal reservoirs.

The Coefficient of Performance of a reversible cycle is the maximum efficiency possible. It is the efficiency of a Carnot Engine.

Hence, it is greater than or equal to the Coefficient of Performance of an irreversible cycle.

COP(reversible) ≥ COP(irreversible)

This statement is true.

c. Mass, energy, and temperature are the examples of intensive properties.

Intensive properties are properties of thermodynamic systems that do not depend on the extent of the system. They are the same for a particular size of substance and stay the same if the size of the substance is doubled or halved. Examples include temperature, specific capacity, specific volume, every specific property basically, etc.

Extensive properties depend on the extent of the system. They double or half when the size of the extent doubles or halves respectively.

Mass and Energy are Extensive properties.

Temperature is the only intensive property among these options.

This statement is false.

d. For reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs, the relation between their coefficients of performance is

COPHP = COPR + 1.

The coefficients of performance for reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs are indeed related through

COP(HP) = COP(R) + 1

COP of a heat pump = COP(HP) = (Qh/W)

COP of a refrigerator = COP(R) = (Qc/W)

But, Qh = Qc + W

Divide through by W

(Qh/W) = (Qc/W) + (W/W)

COP(HP) = COP(R) + 1 (Proved!)

This Statement is true.

e. For a reversible heat pump that operates between cold and hot thermal reservoirs at 350°C and 550°C, respectively, the COP is equal to 4.11.

The COP is given as (1/efficiency).

Efficiency = 1 - (Tc/Th)

Tc = temperature of cold thermal reservoir in Kelvin = 350°C = 623.15 K

Th = temperature of hot thermal reservoir in Kelvin = 550°C = 823.15 K

Efficiency = 1 - (623.15/823.15)

= 1 - 0.757 = 0.243

COP = (1/Efficiency) = (1/0.243) = 4.11

This statement is true.

f. In the absence of any friction and other irreversibilities, a heat engine can achieve an efficiency of 100%.

In the absence of friction and other irreversibilities and for a heat engine to have 100% efficiency, the temperature of its cold reservoir has to be 0 K or the tempersture of its hot reservoir has to be infinity.

Efficiency = 1 - (Tc/Th)

For efficiency to be 1,

(Tc/Th) = 0; that is, Tc = 0 or Th = infinity

These two aren't physically possible and for 100% efficiency to happen, the heat engine will have to violate the Kelvin-Planck's statement of the second law of thermodynamics.

According to Kelvin-Planck's statement of the second law of thermodynamics, net amount of work cannot be produced by exchanging heat with single reservoir i.e. there will be another reservoir to reject heat.

Hence, a heat engine cannot have an efficiency of 100%.

This statement is false.

g. A refrigerator, with a COP of 1.2, rejects 60 kJ/min from a refrigerated space when the electric power consumed by the refrigerator is 50 kJ/min. This refrigerator violates the first law of thermodynamics.

The COP of a refrigerator is given as

COP = (Qcold)/W

Qcold = Heat rejected from the cold reservoir = Heat rejected from refrigerated space = 60 KJ/min

W = work done on the system = electrical power consumed = 50 KJ/min

COP = (60/50) = 1.2

This system does not violate the first law of thermodynamics.

This statement is false.

Hope this Helps!!!

Answer:

[tex]2.08\cdot 10^9 A[/tex]

Explanation:

The magnetic dipole moment of a circular coil with a current is given by

[tex]\mu = IA[/tex]

where

I is the current in the coil

[tex]A=\pi r^2[/tex] is the area enclosed by the coil, where

[tex]r[/tex] is the radius of the coil

So the magnetic dipole moment can be rewritten as

[tex]\mu = I\pi r^2[/tex] (1)

Here we can assume that the magnetic dipole moment of Earth is produced by charges flowing in Earth’s molten outer core, so by a current flowing in a circular path of radius

[tex]r=3500 km = 3.5\cdot 10^6 m[/tex]

Here we also know that the Earth's magnetic dipole moment is

[tex]\mu = 8.0\cdot 10^{22} J/T[/tex]

Therefore, we can re-arrange eq (1) to find the current that the charges produced:

[tex]I=\frac{\mu}{\pi r^2}=\frac{8.00\cdot 10^{22}}{\pi (3.5\cdot 10^6)^2}=2.08\cdot 10^9 A[/tex]

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:

[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:

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

The momentum of the box at t = 1.90 s  is  - 2.586 i + 2.85j

And, the x and y component is - 2.586 and 2.85

Since we know that

impulse = force * time

Here impulse means the change in momentum

Now we can write as  in different way like

change in momentum = force * time

Also, Force = F = .285 t -.46t²

Due to variable force

change in momentum = ∫ F dt

So,

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

Sine t = 1.9

So,

change in momentum = .285 * 1.9² /2 i  -  .46 * 1.9³ / 3 j

= .514i - 1.05 j

And,

final momentum

= - 3.1 i + 3.9j +.514i - 1.05j

= - 2.586 i + 2.85j

So, finally

x component = - 2.586

y component = 2.85

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the momentum of the box at t = 1.90 s is approximately (-2.8435 kg·m/s)i^ + (1.4007 kg·m/s)j^.

To find the momentum of the box at t = 1.90 s, we can integrate the net force over time to get the change in momentum.

The net force is given as:

F⃗ = (0.285 N/s)ti^ + (-0.460 N/s^2)t^2j^

To find the change in momentum, we integrate this force with respect to time from t = 0 to t = 1.90 s:

Δp⃗ = ∫(0 to 1.90) F⃗ dt

Δp⃗ = ∫(0 to 1.90) (0.285ti^ - 0.460t^2j^) dt

Now, we integrate each component separately:

Δpx = ∫(0 to 1.90) (0.285t) dt

Δpx = 0.285 * (t^2/2)|[0, 1.90]

Δpx = 0.285 * (1.90^2/2 - 0)

Δpx = 0.2565 N·s

Δpy = ∫(0 to 1.90) (-0.460t^2) dt

Δpy = -0.460 * (t^3/3)|[0, 1.90]

Δpy = -0.460 * (1.90^3/3 - 0)

Δpy = -2.4993 N·s

Now, we need to add the change in momentum to the initial momentum to find the momentum at t = 1.90 s:

p⃗(t = 1.90 s) = p⃗(t = 0 s) + Δp⃗

p⃗(t = 1.90 s) = (-3.10 kg·m/s)i^ + (3.90 kg·m/s)j^ + (0.2565 N·s)i^ + (-2.4993 N·s)j^

Now, add the x and y components separately:

p⃗(t = 1.90 s) = (-3.10 + 0.2565)i^ + (3.90 - 2.4993)j^

p⃗(t = 1.90 s) = (-2.8435 kg·m/s)i^ + (1.4007 kg·m/s)j^

So, the momentum of the box at t = 1.90 s is approximately (-2.8435 kg·m/s)i^ + (1.4007 kg·m/s)j^.

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Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance [tex]R_e[/tex] defined by the formula:

[tex]\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}[/tex]

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

[tex]\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\[/tex]

and when this is combined with the third resistor in series, the equivalent resistance ([tex]R''_e[/tex]) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

[tex]R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}[/tex]

The problem states that the difference between the equivalent resistances in both circuits is given by:

[tex]R''_e=R_e+630 \,\Omega[/tex]

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

[tex]\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega[/tex]

Answer:

0.83 amps is what I believe is the answer.

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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The mass of the ion is 5.96 X 10⁻²⁵ kg

Explanation:

The electrical energy given to the ion Vq will be changed into kinetic energy [tex]\frac{1}{2}mv^2[/tex]

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force [tex]\frac{mv^2}{r}[/tex].

So,

[tex]Vq = \frac{1}{2}mv^2[/tex]

and

[tex]Bqv = \frac{mv^2}{r}[/tex]

Right from these eliminating v, we can derive

[tex]m = \frac{B^2r^2q}{2V}[/tex]

On substituting the value, we get:

[tex]m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\[/tex]

m = 5.96 X 10⁻²⁵ kg.

The mass of the ion is found using the principles of kinetic and magnetic forces, is approximately 9.65 x 10⁻²⁷ kg.

To find the mass of the ion, we use the principles of energy and magnetic force. When a singly charged positive ion is accelerated through a voltage (V), it gains kinetic energy equal to the electrical potential energy provided by the voltage:

Kinetic Energy (KE) = qV

Since the ion starts from rest, this kinetic energy is also given by:

Kinetic Energy (KE) = (1/2)mv²

Equating these two expressions for kinetic energy gives:

qV = (1/2)mv²

Solving for the velocity (v), we get:

v =√(2qV/m)

When the ion enters the magnetic field (B) and bends into a circular path, the centripetal force required is provided by the magnetic force:

Magnetic Force = qvB

The centripetal force is also given by:

Centripetal Force = mv²/r

Equating these two expressions for the forces, we get:

qvB = mv²/r

Solving for the mass (m) of the ion:

m = qBr/v

We already have v = √(2qV/m), substituting this into the equation above to eliminate v, we get:

m = (qBr) / √(2qV/m)

Square both sides to simplify and solve for m:

m² = (q²B²r²) / (2qV)

m = (qB²r²) / (2V)

Using the given values (q = 1.602 x 10⁻¹⁹ C as charge of a singly charged positive ion, V = 2000 V, B = 0.400 T, and r = 0.305 m), we get:

m = (1.602 x 10⁻¹⁹ C x (0.400 T)² x (0.305 m)²) / (2 x 2000 V)

Calculating this gives the mass of the ion approximately equal to:

m ≈ 9.65 x 10⁻²⁷ kg

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

The linear charge density on the wire is approximately 0.002 N/Cm.

The linear charge density on the wire can be calculated using the formula:

λ = E × r / 2πk

where λ is the linear charge density, E is the electric field, r is the distance from the wire, and k is the Coulomb's constant. Plugging in the known values of E = 20.0 N/C and r = 2.00 cm = 0.02 m, we can solve for λ:

λ = (20.0 N/C) × (0.02 m) / (2π × 8.99 × 10^9 Nm^2/C^2) ≈ 0.002 N/Cm

Therefore, the linear charge density on the wire is approximately 0.002 N/Cm.

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The work done by the force of gravity on the projectile is given by W = mgh. The change in kinetic energy of the projectile is given by ΔKE = (1/2)mv0^2. The final kinetic energy of the projectile as it hits the water is given by KEf = (1/2)mv0^2.

(a) Work done by the force of gravity:

The work done by the force of gravity on the projectile during its flight can be calculated using the formula:

W = mgh

where W is the work done, m is the mass of the sphere, g is the acceleration due to gravity, and h is the height above the water.

(b) Change in kinetic energy:

The change in kinetic energy of the projectile from the time it was fired until it hits the water can be calculated using the formula:

ΔKE = (1/2)mv0^2

where ΔKE is the change in kinetic energy, m is the mass of the sphere, and v0 is the launch speed.

(c) Final kinetic energy:

The final kinetic energy of the projectile as it hits the water can be calculated using the same formula as in part (b):

KEf = (1/2)mv0^2

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The answers of the all the questions are given as 1-a,2-b,3-b,4-c,5-c,6-a,7-b,8-c,9-b,10-b or c,11-b,

Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.

A whole branch of physics, thermodynamics, deals with how heat is transferred between different systems and how work is done in the process.

Almost every transfer of energy that takes place in real-world physical systems does so with an efficiency of less than 100% and results in some thermal energy.

This energy is usually in the form of low-level thermal energy. Here, low-level means that the temperature associated with the thermal energy is close to that of the environment.

It is only possible to extract work when there is a temperature difference, so low-level thermal energy represents 'the end of the road' of energy transfer. No further useful work is possible; the energy is now 'lost to the environment.

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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:

The magnitude of their velocities will be the same but their direction will be reversed.

Final answer:

In a completely elastic collision between two particles of identical mass and equal but opposite velocities, both particles simply swap their velocities as a result of the collision, conserving both momentum and kinetic energy.

Explanation:

In a completely elastic head-on collision between two particles with identical mass and equal but opposite speeds, the outcome is quite straightforward. Since the collision is elastic, both conservation of momentum and conservation of kinetic energy apply.

Before the collision, let's assume particle 1 with mass m has velocity v, and particle 2 also with mass m has velocity -v. The total momentum before collision would be m*v + m*(-v) = 0, and the total kinetic energy would be (1/2)*m*v² + (1/2)*m*(-v)².

For identical masses and equal and opposite velocities, the particles simply swap velocities after collision. Therefore, the first particle will have a velocity of -v and the second a velocity of v post-collision, as momentum and kinetic energy are conserved. They essentially 'bounce' off each other and move in the opposite directions with the same speed they had before the collision.

Answer:

Explanation:

Attached is the evaluation

The given integral represents the volume of a solid bounded by specific surfaces and coordinates. To sketch this solid, imagine a cylindrical shape centered around the z-axis with varying radius and height. To evaluate the integral, integrate with respect to z, then θ, and finally r.

First, let's consider the given integral: ∫_0^6 ∫_0^(2π) ∫_4^24 r dz dθ dr. This triple integral represents the volume of a solid. The limits of integration indicate that the solid is bounded by the surfaces r = 4 and r = 24, the angle θ ranges from 0 to 2π, and the z-coordinate extends from 0 to 6.

To sketch this solid, imagine a cylindrical shape centered around the z-axis, with a varying radius (r) extending from 4 to 24, and a height (z) ranging from 0 to 6. The angle θ represents rotation around the z-axis.

To evaluate the integral, we can first integrate with respect to z, then θ, and finally r. This involves applying the limits of integration and performing the calculations step by step. The final result will give us the volume of the solid.

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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:

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 speed of waves on the string is 32 m/s.

The speed of waves on a string can be calculated using the formula:

velocity = frequency × wavelength

In this case, the frequency is given as 80.0 Hz and the distance between adjacent antinodes is 20.0 cm. Since the wavelength is twice the distance between adjacent antinodes, we have:

wavelength = 2 × 20.0 cm = 40.0 cm = 0.4 m

Plugging these values into the formula, we get:

velocity = 80.0 Hz × 0.4 m = 32 m/s

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