As a 6 x 102 kg object is pushed horizontally with a force of 1 x 102 N, it gains 5 x 102 J of kinetic energy. Through what distance did the force act?

A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected to a 4,5-kN axial load. Knowing that E= 70 GPa, determine the required diameter of the rod. The required diameter of the rod is 377 mm

To determine the required diameter of the aluminum rod, we can use the equation for normal stress in a prismatic bar under axial load:

σ = F / A

where σ is the normal stress, F is the axial load, and A is the cross-sectional area of the rod.

To ensure that the normal stress does not exceed 40 MPa, we can set the equation equal to 40 MPa and solve for the required cross-sectional area:

40 MPa = 4,500 N / A

Solving for A, we get:

A = 4,500 N / (40 MPa) = 0.1125 m^2

To ensure that the rod does not stretch more than 1 mm, we can use the equation for axial strain in a prismatic bar under axial load:

ε = ΔL / L

where ε is the axial strain, ΔL is the change in length, and L is the original length of the rod.

We can rearrange this equation to solve for the required change in length:

ΔL = ε * L = (1 mm) * (1.5 m)

= 0.0015 m

Using the equation for axial strain, we can also express the change in length in terms of the normal stress and the material's elastic modulus:

ε = σ / E

ΔL = ε * L = (σ / E) * L

Solving for the required cross-sectional area using this equation, we get:

A = (F * L) / (E * ΔL) = (4,500 N * 1.5 m) / (70 GPa * 0.0015 m) = 0.0893 m^2

Now that we have two equations for the required cross-sectional area, we can set them equal to each other and solve for the diameter:

A = π * d^2 / 4

d = sqrt(4A / π) = sqrt(4 * 0.1125 m^2 / π) = 0.377 m = 377 mm

Therefore, the required diameter of the aluminum rod is 377 mm.

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C is the duration of the heartbeat in seconds, whereas m is the kilos of the animal. The decrease in is C'(90) occurs at a rate of 0.13 sec kg1.

A known quantity of inertia, an essential characteristic of all matter, is known as mass in physics. The resistance a body of substance offers to either a modification in its movement or location as a result of the force that is applied is what it is in essence.

The amount of substance in a thing is measured by its mass. Its mass will remain constant regardless of where it is situated in the vast universe. The amount of gravitational force acting on an object is measured by its heaviness, on the other hand.

C(m) = 18 m0.23 sec

C(m = 90 kg) = 18 (90)0.23 sec

= 50.67 sec

C (m) = 18m0.23

C' m = dc/dm

= 18 (0.23) m0.23⁻¹

= 4.14 m⁻⁰·77

C' (90kg) = 0.13 sec kg⁻¹

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The stated statement indicates that Scales less than 1:24,000 are 1:900,000 and 1:24,000, respectively.

The correct option is B.

a graded object, particularly when used as a rule or measure: a set of markings or points spaced apart at defined intervals used to measure distances, such as: a relationship between the distances shown on a map and their corresponding actual distances (such as the height of the mercury in a thermometer)

To visually depict the scale of a map, maritime chart, engineering design, or architectural drawing, use a linear scale, also known as a bar scale, scale bar, or graphic scale.

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a) The length of the rope is π/4 meters. b) The speed of the waves on the rope is 6π m/s. c) The mass of the rope is 0.485 kg. d) For the third harmonic, the period of oscillation is π/6 seconds.

a) The displacement of the rope oscillates between -0.1 and 0.1 meters. The length of the rope is half of the wavelength of the standing wave pattern, which is given by λ = 2L/n, where L is the length of the rope and n is the harmonic number. For the second harmonic, n = 2, so we have:

λ = 2L/2 = L

The wavelength is π/2 meters, so the length of the rope is:

L = λ/2 = (π/2)/2 = π/4 meters.

b) The speed of the waves on the rope can be determined from the wave equation, v = fλ, where v is the speed of the waves, f is the frequency, and λ is the wavelength. The frequency of the wave is 12 Hz (since it is the second harmonic), so we have:

v = fλ = 12 × (π/2)/2 = 6π m/s.

c) The mass of the rope can be calculated using the formula for the linear density of a string, μ = m/L, where μ is the mass per unit length, m is the total mass of the rope, and L is the length of the rope. The tension in the rope is 232 N, so we have:

v = √(T/μ)

μ = T/v^2 = 232/(6π)^2 = 0.616 kg/m

The mass of the rope is:

m = μL = 0.616 × π/4 = 0.485 kg.

d) For the third harmonic, n = 3, so the wavelength is:

λ = 2L/3

The period of oscillation is given by T = 1/f, where f is the frequency. The frequency for the third harmonic is:

f = v/λ = v/(2L/3) = 3v/2L = 9f0/2,

where f0 is the fundamental frequency (f0 = v/2L). Therefore, the period is:

T = 1/f = 2/(9f0) = 2L/9v = π/(9f0) = π/6 seconds.

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Complete question:

A rope, under a tension of 232n and fixed at both ends, oscillates in a second-harmonics standing wave pattern. the displacement of the rope is given by
y=0.1sin( πx/2 )sin12πt.
where x = 0 at one end of the rope, x is in meters, and t is in seconds.
a) What is the length of the rope?
b) What is the speed of the waves on the rope?
c) What is the mass of the rope?
d) If the rope oscillates in a third harmonic standing wave pattern, what will be the period of oscillations?

The electric field at x = 15 cm due to the given electric dipole of two charges ±2.5 µC separated by 1.0*10^-4m and centred on the origin, oriented along the x axis, is approximately 8.8×10^4 N/C, directed along the negative x-axis.

An electric dipole is a pair of electric charges of equal magnitude but opposite sign that are separated by a small distance. The dipole has a net electric charge of zero because the charges are equal and opposite, but it has a permanent electric dipole moment due to the separation between the charges.

The dipole moment is a vector that points from the negative charge to the positive charge and has a magnitude equal to the product of the charge magnitude and the distance between the charges. The dipole moment is often denoted by the symbol "p" and has units of coulomb-meters (c.m).

Electric dipoles are important in many areas of physics and engineering, as they play a role in a wide range of phenomena, such as the interaction of molecules with electric fields, the behaviour of capacitors, and the generation and propagation of electromagnetic waves. They can also be used to create and control electric fields in various applications, including in electrical circuits, in sensors, and in medical devices.

To calculate the electric field at a distance of x = 15 cm from the origin, we need to use the formula for the electric field due to an electric dipole, which is:

E = (1/4πε₀) [2p cosθ / r³]

where ε₀ is the permittivity of free space, p is the dipole moment, θ is the angle between the dipole moment and the position vector, and r is the distance from the dipole to the point where we want to find the electric field.

In this problem, the electric dipole is oriented along the x-axis, which means that the angle θ between the dipole moment and the position vector is zero. We can therefore simplify the formula to:

E = (1/4πε₀) [2p / r³]

where r is the distance from the origin to the point where we want to find the electric field.

The dipole moment is given by:

[tex]p = qd[/tex]

where q is the magnitude of each charge in the dipole and d is the separation between the charges. In this case, q = ±2.5 µC and d = 1.0×10^-4 m, so:

p = (2.5 µC)(2d) = 5 µC·m

Now, substitute the values of p and r into the formula to find the electric field at x = 15 cm:

E = (1/4πε₀) [2p / r³] = (1/4πε₀) [2(5 µC·m) / (0.15 m)³]

Using the value of ε₀ = 8.85×10^-12 C²/N·m², we can evaluate this expression to get:

E = (9×10^9 N·m²/C²) [10 / (0.15 m)³] ≈ 8.8×10^4 N/C

Therefore, the electric field at x = 15 cm due to the given electric dipole is approximately 8.8×10^4 N/C, directed along the negative x-axis.

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The strength of the magnetic field will be increased by increasing the number of coils. The correct option is D.

A magnetic field is a vector field that describes the magnetic influence on moving charges, currents, and magnetic materials. A moving charge in a magnetic field is subjected to a force that is perpendicular to both its own velocity and the magnetic field.

An electromagnet is a type of magnet in which the magnetic field is produced by an electric current. Electromagnets usually consist of wire wound into a coil.

By increasing the number of coils there will be more current around the space resulting in a more magnetic field. In this way the strength of the magnetic field is increased.

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(a) Work done by the cord's force on the block [tex](WF) = Mg/4\times d[/tex]

(b) Work done by the weight of the block [tex](Wg) = Mg/4 \times d[/tex]

(c) Kinetic energy of the block [tex](K) = Mg/2\times d[/tex]

(d) Speed of the block [tex](v) = sqrt(gd/2)[/tex] where g is the acceleration due to gravity and d is the distance the block has fallen.

Kinetic energy is the energy that an object possesses due to its motion. In other words, if an object is moving, it has kinetic energy. The amount of kinetic energy an object has depends on its mass and velocity. The formula for calculating kinetic energy, K, is:

K = 1/2 mv^2

where m is the mass of the object and v is its velocity. The factor of 1/2 in the formula is due to the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. If a force is applied to an object to put it in motion, work is done on the object, and its kinetic energy increases.

The units of kinetic energy are joules (J). One joule is defined as the amount of work done when a force of one newton (N) is applied over a distance of one meter (m). Since work and energy are measured in the same units, the unit of kinetic energy is the same as the unit of work, which is the joule.

Kinetic energy is a scalar quantity, which means it has magnitude but no direction. The kinetic energy of an object increases as its mass and velocity increase.

(a) The work done by the cord's force on the block can be found using the formula:

[tex]WF = force\times distance[/tex]

The force applied by the cord is equal to the tension in the cord, which is equal in magnitude to the weight of the block. Therefore, the force applied by the cord is:

[tex]F = Mg[/tex]

where M is the mass of the block, and g is the acceleration due to gravity.

The distance over which the force is applied is d, the distance the block has fallen. Therefore, the work done by the cord's force on the block is:

[tex]WF = Fd = Mg \times d[/tex]

Substituting g/4 for g

[tex]WF = M(g/4) \times d = Mg/4 \times d[/tex]

Hence, the work done by the cord's force on the block is Mg/4 × d.

(b) The work done by the weight of the block can be found using the formula:

[tex]Wg = force \times distance[/tex]

The weight of the block is equal to its mass times the acceleration due to gravity, which is:

[tex]W = Mg[/tex]

The distance over which the weight is applied is again d. Therefore, the work done by the weight of the block is:

[tex]Wg = Wd = Mg\times d[/tex]

Substituting g/4 for g:

[tex]Wg = M(g/4) \times d = Mg/4 \times d[/tex]

Hence, the work done by the weight of the block is also [tex]Mg/4 \times d.[/tex]

(c) The change in the kinetic energy of the block can be found using the work-energy principle:

Wnet = ΔK

where Wnet is the net work done on the block, and ΔK is the change in the kinetic energy of the block.

The net work done on the block is equal to the sum of the work done by the cord's force and the work done by the weight of the block:

[tex]Wnet = WF + Wg = Mg/4 \times d + Mg/4 \times d = Mg/2 \times d[/tex]

The change in the kinetic energy of the block is equal to the net work done on the block:

ΔK = Mg/2 × d

Since the block starts from rest, its initial kinetic energy is zero. Therefore, the final kinetic energy of the block is equal to the change in kinetic energy:

[tex]K = Mg/2\times d[/tex]

(d) The final speed of the block can be found using the equation for the final kinetic energy:

[tex]K = 1/2 mv^2[/tex]

where m is the mass of the block, and v is its final speed.

Substituting Mg/2 × d for K and M for m

[tex]Mg/2\times d = 1/2 Mv^2[/tex]

Simplifying and solving for v, we get:

[tex]v = sqrt(gd/2)[/tex]

Therefore, the speed of the block is[tex]sqrt(gd/2)[/tex], where g is the acceleration due to gravity and d is the distance the block has fallen.

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The plate is about 12.65 units away from the top of the bowl, with a contact area of 48 square units on the bottom of the bowl.

The plate is parallel to the rim of the bowl, it is not in contact with the curved surface of the bowl. The area of the plate in contact with the bottom of the bowl can be calculated as the product of its dimensions: 6 x 8 = 48 square units.

The distance between the bottom of the bowl and the top of the plate can be calculated using the Pythagorean theorem. The height of the bowl is the radius, which is 13 units. The width of the bowl can be calculated as the diameter, which is twice the radius, or 26 units. The width of the plate is 6 units, so the distance between the bottom of the bowl and the top of the plate can be calculated as,

sqrt(13^2 - 3^2) = sqrt(160) = 12.65 units

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The correct answer of the angular speed and linear speed respectively is option A) 0.1047,0.01047

To find the angular speed of the second hand of a clock, we use the formula:

Angular speed = 2π / time

Since the second hand of a clock completes one full rotation every 60 seconds, the time is 60 seconds. Therefore:

Angular speed = 2π / 60

Angular speed = 0.1047 rad/s

To find the linear speed of the tip of the second hand, we use the formula:

Linear speed = angular speed x radius (v = ωr)

Since the radius of the second hand is 10cm or 0.1m, we can plug in the values:

Linear speed = 0.1047 x 0.1

Linear speed = 0.01047 m/s

Therefore, the angular speed of the second hand of a clock is 0.1047 rad/s and the linear speed of its tip is 0.01047 m/s.

The correct answer to the angular speed and linear speed is therefore option A) 0.1047, 0.01047.

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The correct answer is option B.

In the diagram below, point P is located in the electric field between two oppositely charged parallel plates. Compared to the magnitude and direction of the electrostatic force on an electron placed at point P, the electrostatic force on a proton placed at point P has B) the same magnitude, but the opposite direction.

Electrostatic force is a type of force that occurs between two charged particles. The magnitude of this force is determined by the amount of charge on each particle and the distance between them. The direction of the force is determined by the sign of the charges.

If the charges are opposite, the force is attractive and if the charges are the same, the force is repulsive.

In the case of an electron and a proton placed at point P between two oppositely charged parallel plates, the magnitude of the electrostatic force will be the same because they both have the same amount of charge (albeit with opposite signs).

However, the direction of the force will be opposite because the electron and proton have opposite charges.

Therefore, the correct answer to, Compared to the magnitude and direction of the electrostatic force on an electron placed at point P, the electrostatic force on a proton placed at point P has B) the same magnitude, but the opposite direction.

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"If we draw magnetic field lines for a magnet, field lines loop around the magnet starting at the north pole and ending at the south pole." Correct option is B.

Magnetic field lines are the fictitious lines that surround a magnet and reveal the magnet's magnetic field pattern. These lines have no beginning or end. They create a closed circuit. Outside of the magnet, they move from North to South, but inside the magnet, they move from South to North.

North and South poles are the two poles of a magnet. The lines along which magnetic force is applied are known as magnetic field lines. Around the magnet, magnetic field lines create closed loops. These lines originate at the North Pole and come to an end at the South Pole. The magnet's strength is greatest close to its poles. The lines are distinct and do not intersect.

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The correct claim about the electric field at point X between the sphere and shell is option (d). The electric field is zero, because a Gaussian sphere concentric with the sphere with X on its surface encloses zero net charge.

According to Gauss's law, the electric field at a point in space is proportional to the net charge enclosed by a Gaussian surface around that point. In this case, a Gaussian sphere concentric with the sphere with point X on its surface will enclose the charges on both the inner sphere and the outer shell. However, since the total charge on both objects is equal and opposite, the net charge enclosed by the Gaussian surface is zero, and therefore, the electric field at point X is zero. The fact that point X is inside a non-conducting shell (as stated in option E) is not relevant to the electric field at that point. Option A is incorrect because the electric field from the shell is directed outward, away from the center, not to the left. Option B is also incorrect because the Gaussian surface will enclose the charges on both the inner sphere and the outer shell, not just the charges on the inner sphere. Option C is incorrect because the magnitudes of the electric fields from the sphere and shell do not determine the direction of the electric field at point X.

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The correct system that can be classified as a closed system in this scenario is (C). A system consisting of the block and spring.

A closed system is one that does not exchange matter or energy with its surroundings. In this scenario, the block and the spring are the only objects that interact with each other, and there is no energy or matter exchange with the surroundings. Therefore, the system consisting of the block and spring can be considered as a closed system. Options A and B are incorrect because they do not include the spring or the block, respectively, both of which are essential components of the system. Option D is incorrect because it includes the Earth, which is not part of the system and can exchange matter and energy with the block and spring.

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To answer the question: "A student made a model of isostasy by placing a block of wood in a beaker of water how can a student demonstrate isostaic rebound using her model?" If she "Press the block down and let go

The state of gravitational equilibrium between the Earth's crust and mantle known as isostasy or isostatic equilibrium causes the crust to "float" at an elevation that is dependent on its thickness and density.

This idea is used to explain how various topographic heights can occur on the surface of the Earth.

Hence, to test the model she has, she should press the block down and let go

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A student made a model of isostasy by placing a block of wood in a beaker of water how can a student demonstarte isostaic rebound using her model

A. Blow vigorously on the block

B. Press the block down and let go

C. Add water until the beaker overflow

D. Pour out all the water

Because kinetic energy is directly proportional to the square of the velocity.

The increase in kinetic energy of an object is proportional to the square of its velocity.

This relationship is described by the equation;

KE = ¹/₂mv²

where;

When an object accelerates from 20 mph to 40 mph, its velocity increases by a factor of 2. Squaring this factor of 2 results in an increase in kinetic energy by a factor of 4 (2² = 4).

Therefore, the kinetic energy of the object is quadrupled as it increases its velocity from 20 mph to 40 mph.

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The complete question is below:

Which statement best explains why accelerating a car from 20 mph to 40 mph quadruples its kinetic energy? Hint: look at the equation for KE.

A. Because kinetic energy is directly proportional to the cube of the velocity.

B. Because kinetic energy is directly proportional to the square of the velocity.

The statement best explains why accelerating a car from 20 mph to 40 mph quadruples its kinetic energy is:

Kinetic energy is directly proportional to the square of the velocity.

The correct option is A.

The energy that an object has as a result of motion is known as kinetic energy. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes.

Mathematically:

Kinetic energy = ¹/₂ mass * velocity²

Hence, when the velocity of a body is doubled, the kinetic energy of the body quadruples.

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Complete question:

Which statement best explains why accelerating a car from 20 mph to 40 mph quadruples its kinetic energy? Hint: look at the equation for KE.

Kinetic energy is directly proportional to the square of the velocity.

Kinetic energy is inversely proportional to the square of the velocity.

Kinetic energy is directly proportional to the fourth power of the velocity. Kinetic energy is inversely proportional to the fourth power of the velocity.

Kirchhoff's laws are fundamental principles in electrical circuit theory.

Your question is incomplete but you seem to want to know something about Kirchhoff's laws hence I will describe it generally.

Kirchhoff's laws are fundamental principles in electrical circuit theory. Kirchhoff's first law, also known as the law of conservation of current, states that the sum of currents entering any node or junction in a circuit must be equal to the sum of currents leaving that node.

Kirchhoff's second law, also known as the voltage law, states that the sum of voltage drops around any closed loop in a circuit must be equal to the sum of the voltage sources in that loop. These laws are used to analyze and design electrical circuits.

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a) The time of flight is 4.04 seconds.

b) The maximum height is 115 meters

c) The horizontal range is 63.3 meters.

d) velocity of hitting the ground is 35.35 m / s.

Projectile motion is a type of motion experienced by an object or particle that is launched in a gravitational field, such as from the Earth's surface, and moves along a curved path solely under the influence of gravity.

Given that a ball is thrown with an initial velocity of 25.2 m/s at 52.0° above the horizontal from the top of a cliff 95.0 m high.

a) The time of flight is calculated as:-

t = ( 2usinθ) / g

t = ( 2 x 25.2 x sin52 ) / 9.81

t = 4.04 sec

b) The maximum height is calculated as:-

hm = h + ( usinθ)²/ 2g

hm = 95 + ( 25.2 x sin52)² / ( 2 x 9.81 )

hm = 95 + 20

hm = 115 meters

c) The horizontal range will be,

d = u²sin2θ / g

d = ( 25.2² x sin( 52 x 2 ) / 9.81

d = 63.3 meters

d) The velocity of hitting the ground is calculated as:-

mgh + (1/2 ) mu² = 1/2 mv²

(9.81 x 95 ) + 1/2(25.2)² = v²

v = 35.35 m / s

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Car A's time to halt, tA = 0.33T. This is the right response to the question that was asked. The answer has two significant figures.

Provide an illustration of two vehicles with similar speeds but differing velocities. Example: Two 40 km/hr automobiles, one heading east and the other north.

Equal accelerations do not necessarily imply identical velocities, and vice versa. For instance, even if both of your cars have the same acceleration, if one of them accelerates first, it will certainly go more quickly than the other.

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The sun is not located at the center of an elliptical orbit, but is offset somewhat toward one of its ends - True.

The Sun is not relatively at the center of a earth's elliptical route. An cirque has a point a little bit down from the center called the" focus". The Sun is at the focus of the cirque. Because the Sun is at the focus, not the center, of the cirque, the earth moves near to and further down from the Sun every route. The close point in each route is called perihelion. The far down point is called aphelion.

The Sun isn't located at the center of an elliptical route, but is neutralize kindly toward one of its ends. The Greek champion Claudius Ptolemaeus ( Ptolemy) tutored that all heavenly bodies moved in perfect indirect routeways centered on the Earth.

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The maximum height achieved by the body is h = v√(1/2g).

The formula to find the potential energy near earth's surface is -

U = mgh

Given is to identify the maximum height to which the object will rise.

Using the energy conservation, we can write -

1/2mv² = mgh

1/2v² = gh

h = v√(1/2g)

Therefore, the maximum height achieved by the body is h = v√(1/2g).

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The net work is done just on crate as it is moved across the uneven part of the floor is W= 882 N.

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

The work W is equivalent to the force f twice a distance d, or W = fd, to represent this idea numerically. Work is defined as W = fd cos if the force is exerted at an angle to the displacement.

The resistive friction force (Fr) is divided by the normal or perpendicular force (N) pushing both objects together to produce the friction coefficient (fr), which is a numerical value. It can be described by the formula fr = Fr/N. Keep the block stationary in its horizontal position. the box was subjected to the floor's frictional force as a result is 110 N.

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The y component of the velocity vector is 10 m/s. Then the v is 12.86 m/s. Now the x component of this velocity is v sin θ.That is , x component is 8 m/s.

Velocity of an object is the measure of its distance travelled per unit time. Velocity is a vector quantity thus having both magnitude and acceleration. The magnitude is called speed.

For a velocity vector there are three translation components possible for three different axes.

here, x component = v sinθ

y component = v cos θ

Given that v cos θ = 10 m/s

θ = 39

Vy = v cos 39 = 10 m/s

then v = 12.8 m/s

Now, the x component is calculated as:

vx = 12.8 m/s sin 39 = 8 m/s.

Therefore, the x component of velocity here is 8 m/s.

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The value of KB when KA and KC are given along with their thickness is calculated to be 1.57 W/m.K.

The value of thermal conductivity kA is given as = 20 W/m.

The value of thermal conductivity kC is given as = 50 W/m.

The thickness LA is given as = 0.30 m.

The thickness LB is given as = 0.15 m.

The thickness LC is given as = 0.15 m.

Temperature on the inner surface Ts,0 = 20 degrees

Temperature on the outer surface Ts,i = 600 degrees

Temperature at infinity = 800 degrees

Convection coeffiecient h = 25W/m²

Thermal resistance due to conductivity,

R = L/KA

Thermal resistance due to heat transfer coefficient

R' = L/KA

We know that heat transfer Q

Q = ΔT /Rth

By equating heat,

(600 - 20)/[LA/A KA + LB/B KB + LC/C KC] = (800 - 20)/[LA/A KA + LB/B KB + LC/C KC + 1/Ah]

(600 - 20)/[0.3/20 + 0.15/B KB + 0.15/50] = (800 - 20)/[0.3/20 + 0.15/B KB + 0.15/50 + 1/25]

580/[0.015 + 0.15/ KB + 0.003] = 780/[0.015 + 0.15/ KB +0.003 + 0.04]

580/(0.018 + 0.15/ KB) = 780/(0.058 + 0.15/ KB)

580(0.058 + 0.15/ KB) = 780(0.018 + 0.15/ KB)

33.64 + 87/KB = 14.04 + 117/KB

33.64 - 14.04 = 117/KB - 87/KB

19 = 1/KB (117 - 87)

19 KB = 30

KB = 30/19 = 1.57 W/m.K

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Answer:

Explanation:

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

Work done = 335-125=110

Power=work done/time = 110/2.3 = 47.83 watts

Water's high specific heat capacity helps it resist changes in temperature, making it a useful tool for organisms to regulate their body temperature.

Water helps regulate body temperature through several mechanisms.

In addition, water's high specific heat capacity allows it to resist changes in temperature, which helps the body maintain a constant internal temperature even in the face of external temperature fluctuations.

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Sally's observation that the water bottle appeared untouched while the acetone bottle was empty can be explained by the stronger cohesive forces between the water molecules compared to the acetone molecules. The water molecules stick together more tightly, preventing them from evaporating as quickly as the acetone molecules

Sally's statement that particles in water stick together more is partially correct. The cohesive forces between the particles in water, also known as hydrogen bonds, are indeed stronger than the intermolecular forces between the particles in acetone, which are dipole-dipole interactions.

The strength of the intermolecular forces between particles is determined by the type of chemical bonding and the molecular structure of the substance. In the case of water, the oxygen atom in each molecule attracts electrons more strongly than the two hydrogen atoms, giving rise to partial negative charges around the oxygen and partial positive charges around the hydrogen. This creates an electrostatic attraction between neighboring water molecules, resulting in hydrogen bonding.

On the other hand, acetone molecules have a polar covalent bond between the carbon and oxygen atoms, resulting in partial positive and negative charges. The intermolecular forces between acetone molecules are weaker than those between water molecules, because the dipole moments in acetone are not as large as the dipole moments in water.

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Answer: Acceleration, a= -1.11*10^-5

Explanation:

Using a= (V(final speed)- U(Initial speed) )/ TIME IN SECONDS

5hr to seconds is 5*60*60=18,000s

(2.8m/s-3m/s)/ 18,000s = -1.11*10^-5

I'm sorry I wasn't able to explain well, but I hope this helps.  

a) The multiplicative factor by which the sound intensity decreases as you go from the first to the second site is 64.

b) The additive amount by which the sound level intensity decreases as you go from the first to the second site is -15.56 dB.

a) The intensity of the sound is measured in decibels and the sound intensity level with the intensity I is calculated by the expression,

β(dB) = 10 log(I/I₀)

where,

β/dB is the intensity level

I is sound intensity

The intensity is given by the relation,

Intensity = Power/Area

I = (P/πr²)

Intensity is inversely related to the square of radius.

I₁/I₂ = (r₂/r₁)² = (8/1)² = 64

I₁ = 64 I₂

So, the multiplicative factor is 64.

b) The formula for the intensity level is given by,

β(dB) = 10 log(I/I₀)

β₁ = 10 dB log(I₁/I₀)

β₂ = 10 dB log(I₂/I₀)

β₂ - β₁ = 10 dB[log(I₂/I₀) - log(I₁/I₀)]

β₂ - β₁ = 10 dB log(I₂/I₁)

β₂ - β₁ = 10 dB log(1/36) = -15.56 dB

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Yes, it is true, the longer a pendulum is, the longer it takes to swing. Therefore, when the pendulum of a clock is lengthened, the clock slows down.

The length of a pendulum influences the time it takes to complete one swing, which is referred to as the period. The pendulum swings more slowly and has a longer period as it gets longer. A clock slows down when the pendulum is stretched because it takes longer for each swing to be completed. Similar to how the clock speeds up when the pendulum is shorter, it swings quicker as well. The operation of pendulum clocks and other timekeeping instruments that employ pendulums to measure time is based on this idea.

A pendulum's period, or how long it takes to complete one full swing back and forth, can be used to define how it moves. The length, gravitational acceleration, and angle of displacement are some of the variables that affect a pendulum's period. However, the length of a particular pendulum is what essentially determines its period.

The period of a basic pendulum is described mathematically as

T = 2(L/g), where T is the period, L is the length of the pendulum, and g is the gravitational acceleration.

As you can see, the square root of the length and the period are precisely proportional. This implies that the period likewise grows as the pendulum's length does.

The pendulum offers a means of controlling the clock's timekeeping mechanism. The clock's hands are moved by a gear train that is moved back and forth by the pendulum. The pendulum's length can be changed by the clockmaker to alter the clock's rate of movement. The pendulum can be shortened to make it swing more quickly and speed up the clock if it is running quickly. The pendulum can be stretched to make it swing more slowly and slow down the clock if it is operating slowly.

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The required velocity and acceleration of this particle when position of the particle is given are never zero. Correct option is D.

The position of the particle is given as x = 3t³ - 6t, y = t² - 2t

The particle's velocity is determined by,

v = dx/dt i + dy/dt j = (9t² - 6) i +  (2t - 2) j

The velocity at the point t = 0 is,

v = (9t² - 6) i +  (2t - 2) j = - 6 i - 2 j

The velocity at the point t = 1 is,

v = (9t² - 6) i +  (2t - 2) j = 3 i

The acceleration of the particle is,

a = d²x/dt² i + d²y/dt² j = 18t i + 2 j

The particle's acceleration at time zero is,

a = 2 j

Thus, the velocity and acceleration of this particle are never zero.

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