v = Circular motion constant speed

a = Centripetal acceleration

f = Centripetal force

The kinetic frictional force on a wooden block is directly proportional to the normal force on the block. This means that as the normal force increases, so does the kinetic frictional force, and vice versa.

The kinetic frictional force is the force that opposes the motion of a sliding object. When an object, such as a wooden block, is in contact with a surface, there is a normal force acting on the object that is perpendicular to the surface. The normal force is the force that prevents the object from passing through the surface. The frictional force between the object and the surface is directly proportional to the normal force. This is known as Coulomb's law of friction. The coefficient of friction is a constant that determines the amount of frictional force between the object and the surface. The coefficient of kinetic friction is used to calculate the frictional force when the object is in motion. It is a ratio of the kinetic frictional force and the normal force. Therefore, the kinetic frictional force on a wooden block depends on the normal force on the block and the coefficient of kinetic friction between the block and the surface. As the normal force increases, the frictional force also increases proportionally.

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(a) the work done by tension before the block gets to the incline: Wt = 199.2 J. (b) the work done by friction as the block slides on the flat horizontal surface: Wk = - 45.5 J.

a)

consider the motion along the horizontal direction

T = tension force in the rope pulling the mass = 33 N

d = displacement of the mass before it gets to the inclined surface = 6.9 m

ϴ = angle between the tension force and displacement = 29

work done by tension force is given as

[tex]w_{t}[/tex] = work done by tension force = T d Cosϴ

[tex]w_{t}[/tex] = (33 x 6.9) Cos29

[tex]w_{t}[/tex] = 199.2 J

b)

for the mass, the force equation in the vertical direction is given as

T Sinϴ + Fn = mg

inserting the values

33 Sin29 + Fn = 15 x 9.8

Fn = 131 N

uk = Coefficient of kinetic friction = 0.05

the kinetic frictional force is given as

[tex]f_{k} = u_{k} f_{n}[/tex]

inserting the values

[tex]f_{k}[/tex] = (0.05) (131)

[tex]f_{k}[/tex]  = 6.6 N

Ф = angle between the displacement "d"  and frictional force "[tex]f_{k}[/tex]" = 180

Wk = [tex]f_{k}[/tex] d Cos\Ф

work done by the frictional force is given as inserting the values

Wk = (6.6) (6.9) Cos180

Wk = - 45.5 J

c)

v = speed gained by the mass before it gets to the incline

using work-change in kinetic energy theorem

work done by the external force = change in kinetic energy

Wt + Wk = (0.5)m v2

199.2 + (- 45.5) = (0.5) (15)  v2

v = 4.5 m/s

d)

a = acceleration of mass parallel to the incline

consider the motion parallel to the inclined surface

parallel to the incline, the force equation for the motion is given as

T - mg Sin29 = ma

33 - (15 x 9.8) Sin29 = 15 a

a = - 2.6 m/s2

D = distance traveled parallel to the incline before coming to rest

vi = initial velocity at the start of incline = v = 4.5 m/s

v = final velocity as it comes to stop = 0 m/s

using the kinematics equation

vf2 = vi2 + 2 a D

02 = 4.52 + 2(- 2.6) D

D = 3.89 m

e)

h = height gained by the mass on an incline

Sin29 = h/D

hence h = D Sin29

h = 3.89 Sin29

h = 1.89 m

Fg = force of gravity in downward direction = mg = 15 x 9.8 = 147 N

= angle between the force of gravity "Fg" in the down direction and displacement "h" in the upward direction = 180

Work done by gravity is given as

Wg = Fg d Cosα

Wg = (147) (1.89) (Cos180)

Wg = - 277.83 J

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the complete question is:

A mass m = 15 kg is pulled along a horizontal floor, with a coefficient of kinetic friction. k = 0.05, for a distance d = 6.9 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle. = 29?½ with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of? = 29?½ (thus on the incline it is parallel to the surface) and has a tension T = 33 1)What is the work done by tension before the block gets to the incline? 2)What is the work done by friction as the block slides on the flat horizontal surface? 3)What is the speed of the block right before it begins to travel up the incline? 4)How far up the incline does the block travel before coming to rest? 5)What is the work done by gravity as it comes to rest?

The core, radiative zone, convective zone, photosphere, chromosphere, and corona are the sun's regions, listed in order of increasing radius from the centre out.

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A child sleds down a steep, snow-covered hill with an acceleration of 2.82m/s squared. if her initial speed is 0.0 m/s and her final speed is 15.5 m/s, It takes the child 5.50 seconds to travel from the top of the hill to the bottom.

We can use the following kinematic equation to solve this problem:

v^2 = u^2 + 2as

Where:

v is the final speed

u is the initial speed

a is the acceleration

s is the distance traveled

We are given u = 0.0 m/s, a = 2.82 m/s^2, and v = 15.5 m/s. We need to find s.

Rearranging the equation gives:

s = (v^2 - u^2) / (2a)

Substituting the values gives:

s = (15.5^2 - 0.0^2) / (2 x 2.82) = 74.47 m

Now, we can use another kinematic equation to find the time taken:

v = u + at

Where t is the time taken.

Substituting the values gives:

15.5 = 0.0 + 2.82t

Solving for t gives:

t = 15.5 / 2.82 = 5.50 seconds

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The number of cards Aaron had are 19 when Aaron, Bonny and Connor are playing a game with cards which are a total of 101.

Given that Aaron, Bonny and Connor are playing a game with cards.

Let the number of cards Aaron had = x

Bonny has twice the cards of Aaron = 2x

Number of cards Connor had are 6 more than Bonny = 2x + 6

The total number of cards = 101

Then by adding the number of cards that Aaron, Bonny and Connor had we get the total number of cards.

Then Aaron cards + Bonny cards + Connor cards = 101

x + 2x + 2x + 6 = 101

5x = 101 - 6 = 95

x = 95/5 = 19

Hence the total number of cards Aaron had = 19

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Answer:

fusion of water = 80 cal/gm

11.8 g * 80 cal / g = 940 cal released by each orange

When falling, raindrops often develop electric charges, which is accurate. However, these charges are normally quite small, thus they are not likely to create significant forces between the droplets.

What are raindrops? Raindrops are drops of water that fall from the atmosphere, as they grow heavier due to gravity. As they fall, they may accumulate electric charges.

When drops are charged, they can attract one another or repel one another, depending on their relative charges.

What are electric charges? The electric charge is the fundamental property of matter that distinguishes it from other properties such as mass or energy. Objects that have opposite charges attract each other, whereas objects that have the same charge repel each other.

Suppose two 1.8 mg drops each have a charge of 29 pC, and the centers of the droplets are at the same height and 0.44 cm apart. We want to determine whether this creates noticeable forces between the droplets. We can use Coulomb's law to calculate the electric force between the two drops:

[tex]F_{electric} = \frac{kq_1q_2}{r^2}[/tex]

where [tex]k = 9\times10^9 N m^2/C^2[/tex]

k is Coulomb's constant,

q1 and q2 are the charges of the droplets, and

r is the distance between their centers. T

he masses of the droplets aren't essential for this calculation, as they do not affect the electric force much.

From Coulomb's law, F =  1.35 x [tex]10^{-16}[/tex] N.

This force is incredibly small, as expected.

Therefore, it is unlikely that there would be noticeable forces between the droplets due to their electric charges when falling through the air.

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Answer:

Explanation:substitute the values for equation PE=m/vg×

The electron will be forced to move in the direction of the electric field, which in this case is the positive x-axis direction.

The electric field exerts a force on the electron that is equal to the product of the charge of the electron and the magnitude of the electric field. The force will act to accelerate the electron in the direction of the electric field.
An electron that is stationary will be forced to move in the direction opposite to that of the electric field if it is placed in a uniform electric field that points along the x-axis. The direction of the electric field determines the direction of the force felt by the charged particle.A uniform electric field is one in which the electric field is the same at all points in space. The strength of the electric field, in this case, is independent of the position of the point in space. The electric field is a vector quantity that is directed along the direction of the force experienced by the charged particle that is placed in the field.

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To find the moments of inertia Ix, Iy, I0 for the given lamina, we first need to calculate its mass and centroid. The density at any point is proportional to the square of its distance from the origin,

so the mass element dm can be expressed as kr^2dA, where k is the coefficient of proportionality, r is the distance from the origin, and dA is the differential area element. Using polar coordinates, we can express the given region as 0 ≤ r ≤ 6 and 0 ≤ θ ≤ π/2. Integrating dm over this region, we get the total mass of the lamina as: M = ∫∫ kr^2dA = k ∫∫ r^2dA = k ∫θ=0..π/2 ∫r=0..6 r^2r dr d = k ∫θ=0..π/2 [r^4/4]_r=0..6 dθ = (3/5)πk(6^5) To find the centroid of the lamina, we can use the formulae: x_c = (1/M) ∫∫ xdm, y_c = (1/M) ∫∫ ydm Simplifying, we get: x_c = (1/M) k ∫∫ xr^2dA, y_c = (1/M) k ∫∫ yr^2dA.

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One can use the kinetic theory of gases to explain why the pressure of gas increases as the temperature rises in the following ways:At the molecular level, gases are made up of a large number of small particles that are in constant motion.

The movement of the gas particles produces kinetic energy. The total kinetic energy of the particles in a gas is proportional to the temperature of the gas.When the temperature of a gas is increased, the kinetic energy of its particles increases as well.

As the particles move faster, they collide more frequently and with greater force. The force exerted by the particles on the walls of the container is the pressure of the gas.In conclusion, as the temperature of the gas increases, the average kinetic energy of the gas particles increases, which in turn leads to more collisions between particles and with the walls of the container. As a result, the pressure of the gas increases.

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To store 2.70x10^-3 J of electrical energy, capacitance c3 in the network must be 2.67μF. This can be calculated using the formula for energy stored in a capacitor network.

A network of capacitors is a collection of capacitors wired into a circuit. The capacitors store electrical energy as an electric field between their plates when a voltage is applied across the network. The formula E = 1/2 * C * V2 may be used to determine the total energy held in a capacitor network, where E is the energy held, C is the network's total capacitance, and V is the applied voltage. The capacitances of c1, c2, and c4 in this instance are specified as 4.00 F, 4.00 F, and 8.00 F, respectively. It is possible to determine that the capacitance c3 has to be 2.67 F to store 2.70 x 10-3 J of electrical energy by rearranging the formula and inserting the numbers.

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A baseball and a glove make contact. What formula is used to determine how much force the glove applies to the ball when it collides is W= f x d.

The ball striking the glove would be the action force, and the glove applying force back to the ball would be the reaction force.

W = f x d

W = work done is the formula for work completed.

Force = F, and Distance = D

The forces are balanced because the baseball is at rest in the pitcher's glove. When the ball is moving during the pitcher's windup and release, the forces are out of balance. Then, as the ball continues to move in the air at a consistent speed, they balance themselves once more. When the ball slows down after hitting the catcher's gloves, the forces become out of balance. Once the ball is absolutely still in the catcher's glove, they balance out again.

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When an object with mass m is attached to the end of a spring with spring constant k and displaced a distance d from equilibrium and released, it undergoes simple harmonic motion. The speed v of the mass when it returns to the equilibrium position is given by: v = sqrt((kd²)/m)

The maximum potential energy of the system is given by:

U = (1/2)kx²

where x is the displacement of the object from its equilibrium position. At the maximum displacement d, the potential energy of the system is:

U = (1/2)kd²

As the object oscillates back and forth, the potential energy is converted into kinetic energy, given by:

K = (1/2)mv²

where v is the speed of the mass at any point during its motion.

At the equilibrium position, all of the potential energy has been converted into kinetic energy, so we can equate the two expressions:

(1/2)kd² = (1/2)mv²

Solving for v, we get:

v = sqrt((kd²)/m)

Therefore, the speed v of the mass when it returns to the equilibrium position is given by:

v = sqrt((kd²)/m)

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When a student runs at a speed of 5 km/h and then increases his or her speed to 10 km/h, the kinetic energy of the student increases by a factor of four. This is because kinetic energy is proportional to the square of the speed.

The formula for kinetic energy is:

K = 1/2mv²

where K is the kinetic energy, m is the mass of the object and v is the speed.

Since the mass of the student does not change, we can calculate the ratio of kinetic energy at the two speeds as follows:

K₁/K₂ = (1/2)m(v₁²/v₂²)

where v₁ is the initial speed of 5 km/h and v₂ is the final speed of 10 km/h.

K₁/K₂ = (1/2)m(5²/10²) = (1/2)m(1/4) = (1/8)K₂

Therefore, the kinetic energy at the final speed of 10 km/h is four times greater than the kinetic energy at the initial speed of 5 km/h. This means that the student increased their kinetic energy by a factor of four.

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The oscillatory behavior is a function of Φ (phi) instead of the period, T. Option A is correct answer.

Equation 9.4 describes the motion of a simple harmonic oscillator subjected to a periodic driving force. The term ? in the equation represents the frequency of the driving force, while in a simple harmonic oscillator, the oscillatory behavior is described by the period T. Therefore, the oscillatory behavior in Equation 9.4 is a function of frequency rather than the period. The amplitude Amod in Equation 9.4 is not twice the amplitude of a simple harmonic oscillator, nor is it time-dependent. Therefore, options B and C are incorrect. Option D is also incorrect since Equation 9.4 is different from the equation of a simple harmonic oscillator due to the presence of the driving force.

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The force that is most directly responsible for making current flow around the coil is an electric force.

An electric force is the force that arises between electrically charged objects, which includes protons, neutrons, and electrons. In order to drive an electric current through a wire, an electric field must be present. When a conductor, such as copper wire, is moved through a magnetic field, a current is generated inside the conductor in a phenomenon known as electromagnetic induction.

The current produced in the conductor is determined by the size of the magnetic field, the speed of the conductor, and the orientation of the magnetic field. An electric force is responsible for driving the current through the conductor, and the magnetic field is responsible for the creation of the current in the conductor. Therefore, it can be said that an electric force is most directly responsible for making current flow around the coil.

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The final velocities of the truck and car after the collision are both 13.0 m/s and 0 m/s, respectively.

To solve this problem, we can use the conservation of momentum and kinetic energy:

Conservation of momentum:

m1v1 + m2v2 = m1v1' + m2v2'

where

m1 = 1690 kg (mass of the truck)

v1 = 13.0 m/s (initial velocity of the truck)

m2 = 615 kg (mass of the car)

v2 = 0 m/s (initial velocity of the car)

v1' = final velocity of the truck

v2' = final velocity of the car

Solving for v1' and v2':

m1v1 + m2v2 = m1v1' + m2v2'

1690 kg * 13.0 m/s + 615 kg * 0 m/s = 1690 kg * v1' + 615 kg * v2'

21970 kg m/s = 1690 kg * v1' + 0 kg m/s

v1' = 21970 kg m/s / 1690 kg = 13.0 m/s

So the truck maintains its initial speed of 13.0 m/s after the collision.

Now let's solve for the final velocity of the car:

m1v1 + m2v2 = m1v1' + m2v2'

1690 kg * 13.0 m/s + 615 kg * 0 m/s = 1690 kg * 13.0 m/s + 615 kg * v2'

0 kg m/s = 0 kg m/s + 615 kg * v2'

v2' = 0 m/s

So the car comes to a complete stop after the collision.

Therefore, the final velocities of the truck and car after the collision are both 13.0 m/s and 0 m/s, respectively.

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Data streams are continuous flows of data that can be used to capture, process, and analyze real-time information.

Data streams are typically handled by streaming data processing engines. These engines process and analyze the data as soon as it arrives, allowing for real-time insights and decision-making.

They are used to capture data from a variety of sources, such as sensors, web services, databases, and other online sources.

Data streams are handled in real-time, meaning that they are processed and analyzed as soon as they arrive.

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To maintain a constant data read rate, the linear velocity of the DVD must remain constant. Since the circumference of the disk increases as the radius increases, the rotational speed must also increase as we move from the center to the outer edge. The linear velocity is given by:

v = ωr

where v is the linear velocity, ω is the rotational speed in radians per second, and r is the distance from the center of the disk.

Since we want to maintain a constant linear velocity, we have:

v = constant

ω1r1 = ω2r2

where ω1 is the rotational speed at the center of the disk, r1 is the radius of the center of the disk, ω2 is the rotational speed at the outer edge of the disk, and r2 is the radius of the outer edge of the disk.

We know that the disk spins at 1600 rpm for information at the center of the DVD. Let's assume that the radius of the center of the disk is 1 cm, and the radius of the outer edge of the disk is 6 cm. Then we have:

ω1 = 1600 rpm = 167.55 rad/s

r1 = 1 cm

r2 = 6 cm

Substituting these values into the equation above, we can solve for ω2:

ω1r1 = ω2r2

167.55 x 1 = ω2 x 6

ω2 = 27.92 rad/s

Finally, we can convert the angular velocity to rpm:

ω2 = 27.92 rad/s x 60 s/min ÷ 2π rad = 266.08 rpm

Therefore, the rotational speed of the disk needs to be 266.08 rpm at the outer edge of the DVD to maintain a constant data read rate.

(a) Same-direction currents result in attraction. Force per unit length: 1.11 × 10⁻⁵ N/m.

(b) Opposite direction currents result in repulsion. Force per unit length: -1.11 × 10⁻⁵ N/m.

When two parallel wires carry currents in the same direction, they experience a force of attraction, while they experience a force of repulsion when the currents are in opposite directions. The force per unit length exerted on one of the wires by the other can be calculated using the formula F = μ₀I₁I₂L / (2πd), where F is the force per unit length, μ₀ is the permeability of free space, I₁ and I₂ are the currents in the wires, L is the length of the wires, and d is the distance between the wires. For the given problem, the force per unit length exerted on one of the wires by the other is 1.11 × 10⁻⁵ N/m when the currents are in the same direction, indicating attraction, and -1.11 × 10⁻⁵ N/m when the currents are in opposite directions, indicating repulsion. Therefore, the wires either attract or repel each other based on the direction of the currents they carry.

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The student group that built the strongest electromagnet is (option C) Group 4, because the strength of the electromagnet depends directly on the number of loops, current in the circuit and the presence of a core.

An electromagnet is a type of magnet that is created by passing an electric current through a coil or wire. The electric current generates a magnetic field, which can be enhanced by using a magnetic core material, such as iron. Electromagnets can be turned on and off by controlling the flow of electric current through the wire.

They are used in a wide range of applications, from simple devices like doorbells and magnetic locks to more complex technologies like MRI machines and particle accelerators.

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The rate of change of the distance between the top of the ladder and the ground when the base of the ladder is pulled away from the wall at the rate of 0.7 m/s is: 0.7 m/s.

Since the base of the ladder is moving away from the wall, the distance between the top of the ladder and the ground must be increasing. Therefore, the rate of change of the distance between the top of the ladder and the ground is equal to the rate of change of the base, which is 0.7 m/s.

The formula for the rate of change is given by: Rate of change = Change in distance/Change in time.

In this case, the rate of change is equal to the rate at which the base is moving away from the wall, 0.7 m/s.

To summarize, the rate of change of the distance between the top of the ladder and the ground when the base of the ladder is pulled away from the wall at the rate of 0.7 m/s is 0.7 m/s.

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The angular velocity of the turntable is 4.19 radians/second . The angular accelatation of the turntable is 8.38 radians/second².

The torque on the grammaphone record is 0.0419 Nm. The rotational kinetic energy is 0.0439 J. . The work done by the torque in 0.2 s is 0.00702 J.

The turntable has a uniform angular velocity of 40 rounds per minute. To convert this to radians per second, we can use the following conversion factors:

1 round = 2π radians (since there are 2π radians in a complete circle)

1 minute = 60 seconds

So, the angular velocity (ω) can be calculated as:

ω = 40 rounds/minute × (2π radians/round) × (1 minute/60 seconds)

ω ≈ 4.19 radians/second

The record accelerates uniformly to the angular velocity of the turntable in 0.5 seconds. Let's denote the angular acceleration as α. We can use the formula:

ω = ω₀ + αt

Plugging in the values, we get:

4.19 = 0 + α(0.5)

α = 4.19 / 0.5

α ≈ 8.38 radians/second²

The moment of inertia of the grammaphone record (I) is given as 5.0 x 10^(-3) kgm². We can use the formula:

τ = Iα

Plugging in the values, we get:

τ = (5.0 x 10^(-3)) x 8.38

τ ≈ 0.0419 Nm

The rotational kinetic energy (K) can be calculated using the formula:

K = 0.5 * I x ω²

Plugging in the values, we get:

K = 0.5 * (5.0 x 10^(-3)) * (4.19)²

K ≈ 0.0439 J

The work done (W) by the torque can be calculated using the formula:

W = τ * θ

Plugging in the values for α and t = 0.2s, we get:

θ ≈ 0.5 * 8.38 * (0.2)²

θ ≈ 0.1676 radians

Now, we can calculate the work done:

W = 0.0419 * 0.1676

W ≈ 0.00702 J

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A ball revolving on the end of a thin string in a circle of radius 1.25 m at an angular speed of 10.4 rad/s angular momentum of the ball revolving on the end of the string is 3.058 m²/s.

The angular momentum (L) of an object revolving in a circle is given by the formula:

[tex]\[ L = I \cdot \omega \][/tex]

Here:

I = moment of inertia of the object about the axis of rotation.

[tex]\(\omega\)[/tex] = angular speed of the object in radians per second.

For a point mass rotating about an axis at a distance r from the axis, the moment of inertia (I) is given by:

[tex]\[ I = m \cdot r^2 \][/tex]

Here:

m = mass of the object.

r = radius of the circle.

Given the values:

Mass (m) = 0.210 kg

Radius (r) = 1.25 m

Angular speed ([tex]\(\omega\)[/tex]) = 10.4 rad/s

Let's calculate the moment of inertia (I) first:

[tex]\[ I = m \cdot r^2 \\\\= (0.210 \, \text{kg}) \cdot (1.25 \, \text{m})^2 \][/tex]

Now, calculate the angular momentum using the moment of inertia and angular speed:

[tex]\[ L = I \cdot \omega \][/tex]

Plug in the values and calculate:

[tex]\[ L = (0.210 \, \text{kg} \cdot (1.25 \, \text{m})^2) \cdot (10.4 \, \text{rad/s}) \][/tex]

Calculate the angular momentum (L):

[tex]\[ L \approx 3.058 \, \text{kg} \cdot \text{m}^2/\text{s} \][/tex]

Thus, the angular momentum of the ball revolving on the end of the string is approximately [tex]\(3.058 \, \text{kg} \cdot \text{m}^2/\text{s}\)[/tex].

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Answer:

a) The phenomenon is known as the photoelectric effect.

b) When the zinc plate is exposed to UV light, some of the photons in the light have enough energy to knock electrons out of the surface of the plate. These emitted electrons carry a negative charge, so as they leave the surface of the plate, it becomes positively charged.

c) As the intensity of the UV light is increased, more photons with sufficient energy to knock electrons out of the surface of the plate are present. Therefore, the rate of emission of electrons increases.

d) The work function energy of zinc refers to the amount of energy required to remove an electron from the surface of a zinc atom. In other words, it is the minimum amount of energy required to cause the photoelectric effect to occur. The work function energy is a characteristic property of the material and is typically measured in electron volts (eV). In the case of zinc, the work function energy is 4.24 eV, meaning that at least 4.24 eV of energy is required to eject an electron from a zinc atom.

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When an object is in uniform circular motion and traveling at a constant speed, the net force acting on the object is not zero.

It is not zero because an object in uniform circular motion is constantly changing direction, even if its speed remains the same. This change in direction requires a force, which is provided by centripetal force. Centripetal force is the force that keeps an object moving in a circular path by pulling it toward the center of the circle.

This force is always directed toward the center of the circle and is perpendicular to the object's velocity. Without this force, the object would continue moving in a straight line.

Thus, the net force acting on an object in uniform circular motion is not zero, but rather is equal to the centripetal force.

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(a) The current in the solenoid is 0.85 A (b) If you double the diameter of the solenoid, the the magnetic flux through the solenoid will increase by 3.19 times. This is because on doubling the diameter, the area of the core will be quadrupled.

(a) The magnetic flux density B inside a solenoid is calculated using the following formula:

B = μ₀nI

Where, B is the magnetic flux density in teslas (T), μ₀ is the permeability of free space = 4π x 10⁻⁷ Tm/A, n is the number of turns per unit length of the solenoid, I is the current in amperes (A)

From the above equation, we can write

I = B/μ₀n

So, the current in the solenoid is given by

I = 1.28 x 10⁻⁴ / (4π x 10⁻⁷ × 385) = 0.85 A

(b) The magnetic flux through a solenoid is given by

Φ = BA

where, A is the cross-sectional area of the solenoid (area of a circle of diameter 17 cm) = πr² = π(17/2)² = 226 cm² = 0.0226 m², B is the magnetic flux density inside the solenoid as calculated above, Φ = 0.85 × 0.0226 = 0.0193 Wb

If we double the diameter of the solenoid, then the cross-sectional area A of the solenoid will be quadrupled because A ∝ d² (where d is the diameter of the solenoid).So, the new area of the solenoid will be

A = π(2r)² = π(2 × 17/2)² = 722 cm² = 0.0722 m²

So, the new magnetic flux through the solenoid will be

Φ' = Φ × (A'/A)Φ' = 0.0193 × (0.0722/0.0226) = 0.0615 Wb

Therefore, if we double the diameter of the solenoid, then the magnetic flux through the solenoid will increase by a factor of (0.0615/0.0193) = 3.19 times.

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The work done by the electric force to move a 1 C charge from point A to B is 5 J.

The work done by an electric force to move a 1 coulomb (1 c) charge from point A to point B is calculated using the formula: Work = Force x Distance. As the electric force is equal to the charge multiplied by the electric field (F = qE), the work done is calculated as: Work = (1 c) x (Electric Field) x (Distance from A to B). The result is given in Joules (J). The work done by the electric force to move a 1 C charge from point A to B is calculated by multiplying the charge and the potential difference between the two points.
Therefore, the work done is given by: Work done = qΔV Where:q = 1 CΔV = VB - VA. The electric potential difference between point A and B can be obtained from the electric potential at each point. Therefore, the electric potential difference is given by: ΔV = VB - VA = 10 - 5 = 5 V. Substituting the values of q and ΔV in the above equation, we get:Work done = qΔV = 1 x 5 = 5 J.

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So, in Activity 2-2, if the coil is pulled away from the north pole of a magnet, it is likely that an EMF will be induced in the coil.

However, based on general principles of electromagnetism, if a coil is pulled away from the north pole of a magnet, it will experience a change in magnetic flux, which can induce an electromotive force (EMF) in the coil. This phenomenon is known as electromagnetic induction.

The magnitude and direction of the induced EMF will depend on several factors, such as the strength of the magnet, the velocity at which the coil is moved, and the orientation of the coil with respect to the magnetic field. If the coil is part of a circuit, the induced EMF can cause a current to flow in the circuit.

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