State and explain the effects on an electromagnet of: i) removing the core. ii) replacing the iron core with a steel core​

Answer:

3000 seconds

Explanation:

A hollow glass sphere has a density of 1.3g/cm at 20 C. Glycerine has a density of 1.26 g/cm at 20 C.

To determine the temperature at which the hollow glass sphere begins to float in glycerine, we need to calculate the density of glycerine at various temperatures and compare it to the density of the glass sphere.

The density of glycerine changes with temperature, so we need to use a density-temperature chart or equation to determine the density of glycerine at different temperatures.

Assuming the hollow glass sphere has a uniform wall thickness, we can calculate its volume by subtracting the volume of the hollow interior from the volume of the whole sphere

Volume of sphere = (4/3)π[tex]r^{3}[/tex]

Volume of hollow interior = (4/3)π[tex](r-t)^{3}[/tex]

Volume of glass wall = (4/3)π([tex]r^{3}[/tex] - [tex](r-t)^{3}[/tex]), where t is the thickness of the glass wall.

From the density and volume of the glass sphere, we can determine its mass

Mass of glass sphere = Density of glass sphere x Volume of glass sphere

Next, we can use Archimedes' principle to determine the volume of glycerine displaced by the glass sphere when it is submerged in the glycerine

Volume of glycerine displaced = Mass of glass sphere / Density of glycerine at the given temperature

When the glass sphere floats, the volume of glycerine displaced will be equal to the volume of the glass sphere. Thus, we can set the two volumes equal to each other and solve for the temperature at which the density of glycerine matches the density of the glass sphere

Volume of glass sphere = Volume of glycerine displaced

(4/3)π[tex]r^{3}[/tex] - (4/3)π[tex](r-t)^{3}[/tex] = Mass of glass sphere / Density of glycerine at the given temperature

Hence, for the temperature requires knowing the radius and thickness of the glass sphere and the mass of the sphere.

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Answer:4 m/s

Explanation:

Answer:0.65606 -> 0.66

Explanation:

Press 2nd on your calculator then hit sin, this will give you the inverse of sin. enter 0.61 in the ( ) and then enter.

To convert the equation  [tex]∆_{W}[/tex]= [tex]V_{ab}[/tex] I[tex]∆_t[/tex] to express power P for the circuit, we need to divide both sides of the equation by [tex]∆_t[/tex]. Option B is correct.

The power P for the circuit can be expressed using the equation;

P = [tex]∆_{W}[/tex]/[tex]∆_t[/tex]

where [tex]∆_{W}[/tex] is work done on the charge and [tex]∆_t[/tex] is time interval for which the work is done.

Starting with the equation [tex]∆_{W}[/tex]= [tex]V_{ab}[/tex] I[tex]∆_t[/tex], we can rearrange it as follows:

[tex]∆_{W}[/tex]/[tex]∆_t[/tex] = [tex]V_{ab}[/tex] I

Now, substituting the expression for power P, we get;

P = [tex]V_{ab}[/tex] I

Therefore, to convert the equation [tex]∆_{W}[/tex] = [tex]V_{ab}[/tex] I[tex]∆_t[/tex] to express power P for the circuit, we need to divide both sides of the equation by [tex]∆_t[/tex], as in option B;

[tex]∆_{W}[/tex]/[tex]∆_t[/tex] = [tex]V_{ab}[/tex] I

P = [tex]∆_{W}[/tex]/[tex]∆_t[/tex]

P = [tex]V_{ab}[/tex] I/[tex]∆_t[/tex]

Dividing both sides by [tex]∆_t[/tex], we get:

P =  [tex]V_{ab}[/tex]I

Hence, B. is the correct option.

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Answer:

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA

Initial diameter of the wire, d₁ = 4 mm = 0.004 m

Final diameter of the wire, d₂ = 1 mm = 0.001 m

Length of wire, L = 2.00 m

Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

The final area of the copper wire;

The initial drift velocity of the electrons is calculated as;

The final drift velocity of the electrons is calculated as;

The change in the mean drift velocity is calculated as;

The time of motion of electrons for the initial wire diameter is calculated as;

The time of motion of electrons for the final wire diameter is calculated as;

The average acceleration of the electrons is calculated as;

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

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Explanation:

The changing mean drift velocity of the electrons plays out at 3.506 x 10⁻⁷ m/s along with an average acceleration nearing 4.38 x 10⁻¹⁵ m/s².

As the electrons traverse one end of the wire to another, their mean drift velocity undergoes a shift of 3.506 x 10⁻⁷ m/s with an average acceleration of 4.38 x 10⁻¹⁵ m/s² in accordance with the following parameters:

- The current flowing through the wire is at 4.00 mA.

- The original diameter of the wire, d₁, measures at 4 mm or 0.004 m.

- Conversely, the final diameter, d₂, displays a measurement of 1 mm or 0.001 m.

- The length of the entire wire is consistent, measuring at 2 meters.

- Notably, the density of electrons present within copper reaches an estimated value of 8.5 x 10²⁸ /m³.

Calculations regarding both initial and final area coverage provided by copper must be explored along with numerical data involving the two varying drift velocities for accurate results.

Thus, we arrive at the change rate of the mean drift velocity between points in the wire as well as the plenitude of electron acceleration achieved after contemplation into the corresponding motion periods.

The conclusion reflects that our measurements find the changing mean drift velocity of the electrons plays out at 3.506 x 10⁻⁷ m/s along with an average acceleration nearing 4.38 x 10⁻¹⁵ m/s².

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Answer:

When you add 20 ml of vegetable oil to a beaker filled with half water and 20 ml of honey, the honey will settle at the bottom of the beaker, the vegetable oil will float on top of the water, and the honey will form a layer between the vegetable oil and the water. This happens because honey is denser than water, while vegetable oil is less dense than water. The difference in density causes the honey to sink to the bottom, while the vegetable oil floats to the top. Honey and vegetable oil are not miscible because they are not chemically compatible. They are both made up of different types of molecules, which do not mix together. Honey is made up of a complex mixture of sugars, while vegetable oil is made up of triglycerides. These different molecules do not dissolve into each other, which is why they separate into distinct layers in the beaker.

Hope this helps.

Answer:

46.5 N

Explanation:

attached is explanation

(i) The work done on the point charge by the electric force when moved from A to B is 2.1 x 10⁻⁶ J.

(ii) The work done on the point charge by the electric force when moved from A to C is -5.5 x 10⁻⁶ J.

The work done by an electric force is equal to the negative of the change in potential energy, which is given by the product of the charge and the change in potential. The change in potential between two points is equal to the potential difference between those points.

For (i), the potential difference between A and B is 6 V, so the work done is (3.9 x 10⁻⁷ C) x (-6 V) = -2.1 x 10⁻⁶ J (negative because the charge moves from higher to lower potential).

For (ii), the potential difference between A and C is -15 V, so the work done is (3.9 x 10⁻⁷ C) x (-(-15 V)) = -5.5 x 10⁻⁶ J (negative because the charge moves from lower to higher potential).

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The current I through the 4 kΩ resistor in the original circuit is 0.199 mA.

Thévenin's theorem states that any linear network of voltage and current sources and resistors can be replaced by an equivalent circuit consisting of a single voltage source and a single resistor. The equivalent circuit provides the same output voltage and current as the original circuit for any external load connected to it.

To find the current I in the circuit using Thévenin's theorem, we need to follow these steps:

Step 1: Find the Thévenin equivalent voltage (Vth) across the 4 kΩ resistor.

To find Vth, we need to first find the open circuit voltage (Voc) across the 4 kΩ resistor. We can do this by removing the 4 kΩ resistor and finding the voltage between its two terminals using a voltage divider:

Voc = 6 kΩ/(2 kΩ + 6 kΩ) x 2 mA = 1.2 V

Next, we need to find the Thévenin equivalent resistance (Rth) across the 4 kΩ resistor. To do this, we need to short-circuit all the independent voltage sources (in this case, there is only one) and find the equivalent resistance seen from the terminals of the 4 kΩ resistor. With the 2 mA current source shorted out, the 2 kΩ and 4 kΩ resistors are in parallel:

Rth = 2 kΩ || 4 kΩ = 1.33 kΩ

Step 2: Replace the original circuit with the Thévenin equivalent circuit.

We can now replace the original circuit with the Thévenin equivalent circuit, which consists of a voltage source Vth = 1.2 V in series with a resistor Rth = 1.33 kΩ.

Step 3: Find the current I through the 4 kΩ resistor in the Thévenin equivalent circuit.

To find the current I, we can use Ohm's law:

I = Vth/(Rth + 4 kΩ) = 1.2 V/(1.33 kΩ + 4 kΩ) = 0.199 mA

Therefore, the current I through the 4 kΩ resistor in the original circuit is 0.199 mA.

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Answer:

d

Explanation:

The arrows are drawn in the figure which shows gravitational forces on each person on earth.

Gravitational force is force of attraction between two masses. Gravitational force(F) between two bodies is directly proportion to the product of masses(m₁,m₂) of two bodies and inversely proportional to square of distance(r) between them. mathematically it is written as,

F∝ m₁.m₂

F ∝ 1/r²

F = G m₁,m₂÷r²

where G is gravitational constant, whose value is 6.6743 × 10⁻¹¹ m³ kg-1s⁻².

Force is expressed in Newton N in SI unit. its dimensions are [M¹L¹T⁻²].

This is analogous with coulomb's law which gives force between two charges.

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A student sets up four cups with 40 mL of water in each and adds different amounts of ice to each cup.

To determine the connection between the temperature change and the transfer of kinetic energy, the student can measure the initial and final temperatures of the water in each cup and the mass of the ice added to each cup. Then, the student can use the following equation to calculate the amount of heat transferred from the ice to the water

Q = m × c × ΔT

Where Q is the amount of heat transferred, m is the mass of the ice added, c is the specific heat capacity of water (4.184 J/g °C), and ΔT is the change in temperature of the water.

The student can then compare the amount of heat transferred from the ice to the water in each cup to the change in temperature of the water. If the temperature change is greater in a cup where more heat was transferred, this suggests a direct connection between the transfer of kinetic energy (as heat) from the ice to the water and the temperature change of the water.

Hence, This can be further supported by calculating the temperature change per unit of heat transferred, which should be approximately the same for each cup if there is a direct connection between the transfer of heat and the temperature change.

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The power being dissipated by the appliance is approximately 0.182 W.

To determine the power being dissipated by the appliance, we need to use Ohm's Law and the formula for power:

Ohm's Law: V = IR

where V is the voltage, I is the current, and R is the resistance.

Power formula: P = IV

where P is the power, I is the current, and V is the voltage.

First, let's find the current through the circuit:

V_battery = V_appliance + V_voltmeter

where V_battery is the voltage of the battery, V_appliance is the voltage across the appliance, and V_voltmeter is the voltage across the voltmeter.

Rearranging this equation to solve for V_appliance:

V_appliance = V_battery - V_voltmeter

V_appliance = 15.0 V - 11.3 V

V_appliance = 3.7 V

Now we can use Ohm's Law to find the current:

I = V_appliance / R

I = 3.7 V / 75.0 Ω

I = 0.0493 A

Finally, we can use the power formula to find the power:

P = IV

P = 0.0493 A x 3.7 V

P = 0.182 W

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--The complete question is, An idealized voltmeter is connected across the terminals of a 15.0 V battery. A 75.0 Ω appliance is also connected across the battery terminals. When the voltmeter measures 11.3 V, how much power is being dissipated by the appliance?--

There will be no change in the ration of Q/V. Therefore the correct answer is (a).

The ratio of charge Q to voltage V across a capacitor is given by:

Q/V = C

Where C is the capacitance of the capacitor.

If the capacitance is doubled, the new capacitance C' becomes 2C. Substituting into the equation, we get:

Q/V = C

Q/V = 2C/2 = C'

So the ratio of Q to V remains the same, and there is no change in the ratio Q/V when the capacitance is doubled.

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The wave speed is 840 cm/s and the fundamental frequency is 1120 Hz.

Frequency is the number of cycles of a periodic waveform that occur per unit of time. It is measured in Hertz (Hz).

We can use the equation λ=2L/n, where λ is the wavelength, L is the length of the string, and n is the harmonic number. Since the string has fixed ends, the harmonics must be odd-numbered, so we have n=1 for the fundamental frequency, n=3 for the second harmonic (315 Hz), and n=5 for the third harmonic (420 Hz).

Using n=1 and λ=2L/n, we get:

λ = 2L/1

λ = 2L

Using n=3 and λ=2L/n, we get:

λ = 2L/3

Using n=5 and λ=2L/n, we get:

λ = 2L/5

We can use the formula f=v/λ to relate the wave speed v, wavelength λ, and frequency f. For the two consecutive harmonics, we can write:

v/λ1 = f1

v/λ2 = f2

Since the two harmonics are consecutive, we can assume that they correspond to adjacent values of n, so we have:

λ1 = 2L/1 = 2L

λ2 = 2L/3

Substituting these values into the above equations and solving for v, we get:

v = f1λ1 = f2λ2 = (420 Hz)(2L) / (2L) = (315 Hz)(2L)/(2L/3) = 840 cm/s

To find the fundamental frequency, we use the formula f=v/λ1:

f = v/λ1 = 840 cm/s / 2L = (840 cm/s) / (0.75 m) = 1120 Hz

Therefore, the wave speed is 840 cm/s and the fundamental frequency is 1120 Hz.

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The structure of zinc telluride crystals is formed by a dense packing of anions, and cations occupy inter-nodes.

a) The structure of zinc telluride crystals is formed by a close packing of anions in a hexagonal close-packed (HCP) lattice. The stacking sequence of HCP lattice is ABABAB.

b) The cations occupy octahedral voids which are formed in between the closely packed anions.

c) In HCP lattice, there are 6 octahedral voids per unit cell. Each unit cell contains 2 zinc cations. Hence, the fraction of the available voids occupied by cations is 2/6 or 1/3.

d) Here is a depiction of two densely packed planes of anions stacked in the AB sequence with the voids filled with cations

 B           Cation in one octahedral void

  A       B       Cation in another octahedral void

        A           Anion

        A           Anion

  B       A       Cation in one octahedral void

        B           Cation in another octahedral void

The two densely packed planes of anions are labeled as A and B. Hence, The cations occupy the octahedral voids between these planes.

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The cuvature of the mirror is 26.67 cm.

The curvature of the mirror is calculated as follows;

1/f = 1/v + 1/u

Where;

The magnification of the mirror = 2

m = v/u

2 = v/u

v = 2u

The focal length of the mirror is calculated as;

1/f = 1/2u + 1/u

1/f = 3/2u

f = 2u/3

f = (2 x 20 cm )/3

f = 13.33 cm

Curvature = 2f = 26.67 cm

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Answer and Explanation:

A well-coated structure is defined as having a coating that meets a certain standard of quality. The answer to this particular question depends on the specific criteria being used to evaluate the coating. This would typically require a coating coverage of 90% or better, if not higher.

However, in general, a well-coated structure would typically refer to a surface that has been thoroughly and evenly covered with a coating material such as paint or varnish. This ensures that the underlying material is protected from environmental factors such as moisture and UV radiation. In addition, a well-coated structure can also improve the overall appearance of the surface, making it more aesthetically pleasing. Regarding the options provided in the question, the answer would depend on the specific criteria being used to evaluate the coating. However, it is safe to say that a well-coated structure would require a high level of coating coverage, with minimal areas left uncovered or with an uneven application. This would typically require a coating coverage of 90% or better, if not higher. Ultimately, the specific answer would depend on the standards and expectations set by the evaluating body.

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A well-coated structure is defined as having a coating that meets a certain standard of quality. The answer to this particular question depends on the specific criteria being used to evaluate the coating. This would typically require a coating coverage of 90% or better, if not higher.

However, in general, a well-coated structure would typically refer to a surface that has been thoroughly and evenly covered with a coating material such as paint or varnish. This ensures that the underlying material is protected from environmental factors such as moisture and UV radiation. In addition, a well-coated structure can also improve the overall appearance of the surface, making it more aesthetically pleasing.

Regarding the options provided in the question, the answer would depend on the specific criteria being used to evaluate the coating. However, it is safe to say that a well-coated structure would require a high level of coating coverage, with minimal areas left uncovered or with an uneven application. This would typically require a coating coverage of 90% or better, if not higher. Ultimately, the specific answer would depend on the standards and expectations set by the evaluating body

Answer:

The velocity of the box is related to the angular velocity of the spool, which is given by the equation:

v = r * ω

where r is the radius of the spool and ω is the angular velocity of the spool. The angular velocity of the spool, in turn, is related to the torque applied to the spool by the tension in the string, which is given by the equation:

τ = I * α

where τ is the torque, I is the moment of inertia of the spool, and α is the angular acceleration of the spool.

The tension in the string is equal to the weight of the box, which is given by:

T = m * g

Putting all of these equations together, we can solve for the time it takes for the box to reach the floor. Here's how:

First, we can find the angular acceleration of the spool using the torque equation:

τ = I * α

T = m * g = τ

m * g = I * α

α = (m * g) / I

α = (35.30 kg * 9.81 m/s^2) / 4.00 kg*m^2

α = 86.53 rad/s^2

Next, we can find the angular velocity of the spool using the kinematic equation:

ω^2 = ω_0^2 + 2 * α * θ

where ω_0 is the initial angular velocity (which is zero), θ is the angle through which the spool has turned (which is equal to the distance the box has fallen divided by the radius of the spool), and ω is the final angular velocity (which is what we want to find). Solving for ω, we get:

ω^2 = 2 * α * θ

ω = sqrt(2 * α * θ)

ω = sqrt(2 * 86.53 rad/s^2 * (3.50 m / 0.10 m))

ω = 166.6 rad/s

Finally, we can find the time it takes for the box to reach the floor using the equation:

v = r * ω

v = 0.10 m * 166.6 rad/s

v = 16.66 m/s

t = d / v

t = 3.50 m / 16.66 m/s

t = 0.21 s

The volume of the parallelepiped is 83 cubic units.

The volume of a parallelepiped with sides give as a = ( 7,2 ,4 ) , b = ( 4,7 ,6 ) and c = ( 3,4 ,7 ).

The volume of a parallelepiped with adjacent sides a, b, and c is given by the scalar triple product (a × b) · c.

First, need to calculate the cross product of vectors a and b

a × b =

[tex]\left[\begin{array}{ccc}i &j&k\\7&2&4\\4&7&6\end{array}\right][/tex]

= (2 × 6 - 4 × 7) i - (7 × 6 - 4 × 4) j + (7 × 7 - 2 × 4) k

= -8 i - 26 j + 45 k

Now, calculate the scalar triple product

(a × b) · c = (-8)(3) + (-26)(4) + (45)(7) = 83

Therefore, the volume of the parallelepiped is 83 cubic units.

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A student slides a block on a surface by applying a force of 11 newtons (N) toward the left. The friction force on the block is 4 N and which is in right direction. the Net force acting on the box is 11-4 = 7N.( towards left)

Force is responsible for the motion of an object. it produces acceleration in the body. According to newton's second law force is mass times acceleration i.e. F =ma. Its SI unit is N which is equivalent to kg.m/s². There are two types of forces, balanced force and unbalanced force. Balanced forces are those forces which are opposite in direction and equal in magnitude. When Net force acting on a body is zero then we call it as balanced force. Balanced force is not responsible for the motion of the body. ex. when two persons pulling rope on both end with equal magnitude which cause them to be balanced force have 0 net force. Unbalanced forces are those when resultant of all the forces is not equal to zero is called as unbalanced force. unbalanced force is responsible for the motion of the body.

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The mass of the helium sample is approximately 0.187 g.

To solve this problem, we can use following formula:

w = nR(T2 - T1)

We can rearrange this formula to solve for n:

n = w / (R * (T2 - T1))

To find the mass of the helium sample, we can use following formula:

m = n * M

where m is the mass of the sample, n is  number of moles of gas, and M is the molar mass of helium.

Substituting the given values into the first equation, we get:

70 J = n * 8.31451 J/mol*K * (358 K - 283 K)

Simplifying this equation, we get:

n = 0.0467 mol

Substituting this value into the second equation, we get:

m = 0.0467 mol * 4 g/mol = 0.187 g

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Paul Cezanne's Still Life with Apples in a Bowl (1879-83) represents a break with the tradition of using linear perspective in art.

One of the pioneers of modern art, Cezanne used a novel approach to painting at the time. In his still life paintings, Cezanne represented things utilizing a system of flattened planes and simplified forms rather than the conventional perspective techniques that provide the impression of depth and space.

Additionally, he played around with color, relying on color blocks rather than shading and modeling to convey a sense of volume and form.

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Letter F shows the ball when it has the maximum kinetic energy.

Letter A shows the ball when it has the maximum potential energy.

Letter G shows the ball when it has the least kinetic energy.

Letter C shows the ball when it has the least potential energy.

The kinetic energy of an object is  described as the form of energy that it possesses due to its motion.

potential energy  on the hand is described as  the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.

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