how the kinetic frictional force on the wooden block depends on the normal force on the wooden block

Currents can be created by magnetic fields. A current is capable of creating a magnetic field.

Magnetic fields are constantly shifting, pushing and pulling electrons. Metals like copper and aluminium have slackly held electrons. As a magnet is moved around a wire or a coil of wire, the electrons in the wire are pushed, creating an electrical current.

Sound energy can be made up of both kinetic and potential energy. A musical instrument is one possible illustration. When the instrument is performed, sound waves are produced that have kinetic energy. Nevertheless, potential energy is all that is present when that same musical instrument is at rest.

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Answer: maybe read your book

Explanation:

When you place a thumbtack 54.0 cm in front of a lens, the resulting image is real, inverted, and the same size as the thumbtack, then the lens in this case is a converging lens.

Converging lenses are thick in the middle and thin at the edges. They are convex in shape, which means they bulge outwards. Converging lenses can converge light rays to a point on the other side of the lens. They are also known as convex lenses, and they have a positive focal length.

The image produced by a converging lens can be real or virtual depending on the position of the object relative to the lens. The image of the object is real when the object is placed beyond the focal point of the lens, as in this situation.

Therefore, when you place a thumbtack in front of a lens, the resulting image is real, inverted, and the same size as the thumbtack. that means the lens is a converging lens.

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Answer:

It would be 2850 kg m/s

Explanation:

The bolts on the cylinder head of certain engines require tightening to a torque of 90 nm. The force perpendicular to the wrench that the mechanic must exert at its end is 450 N.

Torque is a measure of how much a force acting on an object causes the object to rotate. One component of torque is force, the other is distance. If the distance between the point where the force is applied and the point where the object rotates is shorter, a higher torque can be produced using less force.The torque formula is given by τ = r × F, where τ is the torque, r is the lever arm, and F is the force.

The formula for torque is as follows:

Torque = Force x Distance perpendicular to the applied force

Where:

Distance perpendicular to the applied force is the distance between the applied force and the pivot point.

Given that:

To convert cm to m, divide by 100. So, the distance perpendicular to the applied force is 0.2 m.

Thus, the force perpendicular to the wrench that the mechanic must exert at its end is 450 N.

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If the winner started the acceleration first, he will be 31.24 m ahead of the loser when the loser finishes.

Let the winner's initial position be at x = 0. Assume that the loser is a distance d away from the winner's initial position. Therefore, the loser's initial position is at x = -d.The winner starts to accelerate when the other racer is 6.5 m ahead of him. When the winner reaches a speed of 12.2 m/s, the other racer has traveled a distance of 6.5 m.

The time it takes the winner to catch up to the other racer can be found using the equation:

x = vt + 1/2at^2

where x = 6.5 m, v = 12.2 m/s, and a = 0.

The time t is: 6.5 m = 12.2 m/s × t

Thus, t = 0.533 s.

From that time on, the winner continues to cycle at a constant speed of 12.2 m/s until the finish line. The distance the winner travels between catching up to the other racer and crossing the finish line is: Distance traveled by the winner = 12.2 m/s × (18.3 s – 0.533 s) = 224.43 m. Distance between the winner and the other racer after crossing the finish line is: 12.2 m/s × (18.3 s – 15.0 s) = 40.26 m.

Hence, the distance between the loser and the finish line is (65.0 m – 40.26 m) = 24.74 m. Distance the winner will travel before the loser finishes: 24.74 m + 6.5 m = 31.24 m.

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An acrobat with a mass of 49-kg jumping on a trampoline, the force that the trampoline has to apply to accelerate her straight up at [tex]7.9m/s^2[/tex] in newtons is 867.3 N.

How to calculate force? Force is the product of mass and acceleration.

Hence, F = ma

Where, F = Net force applied

            m = mass of an object

            a = net acceleration

Force can be calculated by multiplying the mass of the object with the acceleration experienced by it.

In this case, m = 49kg and a = [tex]7.9 m/s^2[/tex]

Force applied by earth gravity on the acrobat is mg, which equals 49*9.8 N i.e., 480.2 N.

[tex]F - F_{gravity}=ma \\F-F_{gravity} = ma\\F = F_{gravity}+ma\\F=480.2+49*7.9 N \\F=867.3 N\\[/tex]

So, F= 867.3 N is required.

Therefore, the trampoline has to apply a force of 867.3 N to accelerate the acrobat straight up at [tex]7.9 m/s^2[/tex].

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The frequency of the wave is 2.66Hz.

It is given that,

Wave velocity = Vw = 4m/s

Wavelength = λ = 1.5m

Frequency = f = ?

The propagation rate The wave travels one wavelength in one period of time, or Vw, the distance it covers in a given amount of time. It is represented by the equation,

Vw = f λ

f = Vw/λ

f = (4m/s)/1.5m

f = 2.66Hz

Therefore, the frequency of the wave would be 2.66Hz.

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The anthracite coal price per MMBtu is $3.00/MMBtu. Therefore, the correct option is A.

The heat content of anthracite coal is 15,000 BTU/pound. The question requires the conversion of an anthracite coal price of $90/ton to $/MMBtu. Let us use the given data to solve the problem. 1 ton equals 2000 pounds (lb). Hence,

2000 lb of anthracite coal has a heat content of 2000 lb x 15,000 BTU/lb = 30,000,000 BTU

Thus, the price per MMBtu can be calculated by the given formula;

Price per MMBtu = Price per ton / (BTUs per ton / MMBTUs per ton)

Price per MMBtu = $90 / (30,000,000 / 1,000,000)

Price per MMBtu = $90 / 30

Price per MMBtu = $3.00/MMBtu

Thus, the price per MMBtu of anthracite coal is $3.00/MMBtu. Therefore, the correct answer is option A: $3.00/MMBtu.

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The radial acceleration of a point on the second hand of a large clock is 0.14 cm/s². Using the formula for radial acceleration, the distance of the point from the axis of rotation is approximately 11.3 cm.

The radial acceleration of a point on a rotating object is the acceleration that points towards the center of rotation. This acceleration arises due to the object's circular motion and is given by the formula a = v²/r, where a is the radial acceleration, v is the tangential velocity of the object, and r is the radius of the circular path. In this case, we know that the radial acceleration of the point on the second hand of the clock is 0.14 cm/s². Since the second hand completes one full revolution in 60 seconds, we can find its tangential velocity using the formula v = 2πr/T, where T is the time taken for one revolution. Substituting T = 60 seconds, we get v = 0.1047 cm/s. We can now use the formula for radial acceleration to find the radius of the circular path followed by the point on the second hand. Rearranging the formula, we get r = v²/a, which gives us a value of approximately 11.3 cm for the distance of the point from the axis of rotation.

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Answer:

13. a) In Einstein's photoelectric equation, hf represents the energy of a single photon of the electromagnetic radiation, and mvmax² represents the maximum kinetic energy of the emitted photoelectrons.

b) i. To calculate the threshold frequency, we can use the equation c = fλ, where c is the speed of light, f is the frequency of the electromagnetic radiation, and λ is the wavelength:

c = fλ

f = c/λ

f = (3.00 x 10^8 m/s)/(6.5 x 10^-7 m)

f = 4.62 x 10^14 Hz

Therefore, the threshold frequency is 4.62 x 10^14 Hz.

ii. We can use the equation hf = Φ + KEmax, where hf is the energy of a single photon, Φ is the work function energy of the metal, and KEmax is the maximum kinetic energy of the emitted photoelectrons. We know that the wavelength of the incident electromagnetic radiation is 6.5 x 10^-7 m, which corresponds to a frequency of f = 4.62 x 10^14 Hz (as calculated in part i). We also know that this wavelength is the maximum for which photoelectrons are released, which means that the energy of the photons is equal to the work function energy:

hf = Φ

Substituting the values for h and f, we get:

(6.63 x 10^-34 J s)(4.62 x 10^14 Hz) = Φ

Φ = 1.93 x 10^-19 J

Converting this to electronvolts (eV), we get:

Φ = (1.93 x 10^-19 J)/(1.60 x 10^-19 J/eV)

Φ = 1.21 eV

Therefore, the work function energy of the metal is 1.9 eV.

c) If the intensity of the incident light is doubled, the rate of release of photoelectrons will also double. This is because the intensity of light is directly proportional to the number of photons incident on the metal surface per unit time. Each photon can cause the emission of one photoelectron, so doubling the number of photons will double the number of emitted photoelectrons per unit time. However, the kinetic energy of the emitted photoelectrons will not change, since this is determined only by the frequency (and therefore the energy) of the incident photons, and not by their intensity.

Answer:

Approximately [tex]15.5\; {\rm N}[/tex], assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].

Explanation:

To find the tension in the rope on the right side of the pulley, apply the following steps:

The net force on the block is:

[tex]F_{\text{net}} = m_{\text{b}} \, a[/tex], where

At the same time, the net force on the block can also be expressed as:

[tex]\begin{aligned}F_{\text{net}} &= T_{\text{left}} - (\text{weight}) \\ &= T_{\text{left}} - m_{\text{b}}\, g \end{aligned}[/tex], where

Rearrange and solve for [tex]T_{\text{left}}[/tex]:

[tex]T_{\text{left}} - m_{\text{b}}\, g = F_{\text{net}} = m_{\text{b}}\, a[/tex].

[tex]\begin{aligned}T_{\text{left}} &= m_{\text{b}}\, a + m_{\text{b}}\, g \\ &= m_{\text{b}}\, (a + g) \\ &= 1.3\, (1.2 + 9.81)\; {\rm N} \\ &= 14.313\; {\rm N}\end{aligned}[/tex].

Let [tex]r[/tex] denote the radius of the pulley. It is given that the diameter of the pulley is [tex]50\; {\rm cm}[/tex]. In standard units, the radius of the pulley would be [tex]r = 25\; {\rm cm} = 0.25\; {\rm m}[/tex].

On the left side of the pulley, tension in the rope exerts a torque of [tex]\tau_{\text{left}} = T_{\text{left}}\, r[/tex] on the pulley:

[tex]\begin{aligned}\tau_{\text{left}} &= T_{\text{left}}\, r \\ &= (14.313)\, (0.25)\; {\rm N\cdot m} \\ &= 3.57825\; {\rm N\cdot m} \end{aligned}[/tex].

Under the assumptions, the moment of inertia [tex]I[/tex] of this cylindrical pulley would be:

[tex]\begin{aligned} I &= \frac{1}{2}\, m\, r^{2} \end{aligned}[/tex], where

[tex]\begin{aligned} I &= \frac{1}{2}\, m\, r^{2} \\ &= \frac{1}{2}\, (2.0)\, (0.25)^{2}\; {\rm kg \cdot m^{2}} \\ &= 0.0625\; {\rm kg\cdot m^{2}} \end{aligned}[/tex].

Since the rope doesn't slip on the pulley, linear acceleration of the pulley would be equal to that of the rope, [tex]a = 1.2\; {\rm m\cdot s^{-2}}[/tex]. Divide this linear acceleration by the radius of the pulley to find the angular acceleration [tex]\alpha[/tex] of the pulley:

[tex]\begin{aligned}\alpha &= \frac{a}{r} \\ &= \frac{1.2}{0.25}\; {\rm s^{-2}} \\ &= 4.8\; {\rm s^{-2}}\end{aligned}[/tex].

Multiply angular acceleration by the moment of inertia to find the net torque [tex]\tau_{\text{net}}[/tex] on the pulley cylinder:

[tex]\begin{aligned}\tau_{\text{net}} &= I\, \alpha \\ &= (0.0625)\, (4.8)\; {\rm kg \cdot m^{2}\cdot s^{-2}}\\ &= 0.3\; {\rm kg \cdot m^{2} \cdot s^{-2}} \end{aligned}[/tex].

Note that the net torque of the pulley [tex]\tau_{\text{net}}[/tex] is in the same direction as [tex]\tau_{\text{right}}[/tex], but the opposite of [tex]\tau_{\text{left}}[/tex]. Hence:

[tex]\begin{aligned}\tau_{\text{right}} &= \tau_{\text{net}} + \tau_{\text{left}} \\ &= 0.3\; {\rm N\cdot m} + 3.57825\; {\rm N\cdot m} \\ &= 3.87825\; {\rm N\cdot m}\end{aligned}[/tex].

Divide the torque on the right [tex]\tau_{\text{right}}[/tex] by radius [tex]r[/tex] to find the tension in the string on the right [tex]T_{\text{right}}[/tex]:

[tex]\begin{aligned}T_{\text{right}} &= \frac{\tau_{\text{right}}}{r} \\ &= \frac{3.87825}{0.25}\; {\rm N} \\ &= 15.513\; {\rm N}\end{aligned}[/tex].

When the magnitude of the electric field in an EM wave is doubled, the intensity of the wave increases by a factor of 4. The correct option is c. It quadruples.

The intensity (I) of an electromagnetic wave is given by:

I = (1/2)ε0cE^2

where ε0 is the electric constant, c is the speed of light, and E is the magnitude of the electric field.

If the magnitude of the electric field in an electromagnetic wave is doubled, the intensity of the wave will increase by a factor of four (4), because:

I' = (1/2)ε0c(2E)^2 = 4(1/2)ε0cE^2 = 4I

The correct answer is (c) It quadruples.

Electromagnetic waves are waves that are produced by oscillating electric and magnetic fields. An electromagnetic wave is composed of electric and magnetic fields that oscillate perpendicularly to each other and to the direction of wave propagation. Light, microwaves, X-rays, and radio waves are all examples of electromagnetic waves.

The power transferred per unit area by an electromagnetic wave is known as the intensity of the wave. The magnitude of the electric field in an EM wave is related to its intensity. When the magnitude of the electric field in an EM wave is doubled, the intensity of the wave quadruples.

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Answer:

We dont see a pattern of light and dark fringes on a wall that is illuminated by two ordinary light bulbs beca.

Answer:

pls mrk me brainliest

Explanation:

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at any point of time due to the force of gravity. Hydrostatic pressure is proportional to the depth measured from the surface as the weight of the fluid increases when a downward force is applied. The hydrostatic pressure at any point in a fluid can be calculated by using the formula:

P = P0 + ρgh

where P is the hydrostatic pressure, P0 is the atmospheric pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the point from the surface.

a) In which fluid is pressure of 610 kPa first achieved?

It can be found out by adding up the hydrostatic pressures of each layer of fluid until we reach 610 kPa. Starting from gas layer:

Pgas = 60 kPa + (1 kg/m3)(9.81 m/s2)(1 m) = 60.00981 kPa

Palcohol = Pgas + (780 kg/m3)(9.81 m/s2)(60 m) = 460.00981 kPa

Poil = Palcohol + (840 kg/m3)(9.81 m/s2)(10 m) = 542.40981 kPa

Pwater = Poil + (990 kg/m3)(9.81 m/s2)(2 m) = 561.60981 kPa

Pglycerine = Pwater + (1236 kg/m3)(9.81 m/s2)(3 m) = 605.46981 kPa

Pmolasses = Pglycerine + (1500 kg/m3)(9.81 m/s2)(10 m) = 752.96981 kPa

The pressure of 610 kPa is first achieved in glycerine layer.

b) If the bottom of the pipe is at zero elevation what is

b) If the bottom of the pipe is at zero elevation what isthe pressure at bottom in kpa?

The bottom of pipe corresponds to molasses layer so use it to calculate hydrostatic pressure as calculated above:

Pbottom = Pmolasses = 752.96981 kPa

c) At what elevation is pressure of 640kpa?

It can be found out by subtracting hydrostatic pressures from each layer until it reach below 640kpa and then use interpolation to find exact elevation.

Starting from molasses layer:

Pmolasses - Pglycerine= (752.96981 - 605.46981)kpa=147.5kpa

This means that somewhere between glycerine and molasses layers there is a point with pressure of 640kpa.

Let x be distance from top surface of molasses layer to this point then:

640kpa=605.4698+1500(9.8)x

x=0.023m

Therefore elevation from bottom surface of pipe to this point is:

10-0-0-023=9-977m

d) If an open manometer attached to side pipe anywhere on oily portion determine height liquid level manometer.

An open manometer measures difference between atmospheric pressure and fluid pressure inside pipe.

Let y be height liquid level manometer above oil level then:

Patm-Poil=yρg

y=(Poil-Patm)/ρg

y=(542-4098-101325)/(1000*9-8)

y=-44-6m

This means that liquid level manometer will be below oil level by -44-6m or oil level will be above liquid level manometer by +44-6m.

The work done by the spring is -(mg)2/(2ks). Therefore, the correct option is A.

The work done by the spring, Ws, is determined by the force of the spring and the displacement of the block. The spring has a stiffness constant, ks, so the force exerted by the spring is ks multiplied by the displacement, s.

The block has a mass, m, and is in the presence of acceleration due to gravity, g. The work done by the spring is therefore

Ws = -(1/2) x ks x s² = -(1/2) x ks x (mg/ks)² = -(mg)²/(2ks).

Therefore, the answer is A: -(mg)2/(2ks)

The work done by the spring can be expressed in terms of the force of the spring and the displacement of the block. The force of the spring is proportional to the stiffness constant, ks, and the displacement of the block is dependent on the mass, m, and acceleration due to gravity, g.

The work done by the spring is equal to the negative of one-half of the stiffness constant multiplied by the displacement squared. Therefore, the answer is -(mg)2/(2ks).

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The statement that  best describes the result of moving the charge to  the point marked X is: B. Its electric potential energy increases because the electric field increases.

Moving the charge to the point marked X will change its electric potential energy because the electric field at that point is different from the electric field at its initial position.

At the initial position of the charge, the electric potential energy is determined by the electric potential at that point and the charge of the object. When the charge is moved to point X, the electric potential at that point changes, which means that the electric potential energy of the charge also changes. This is because the electric potential at a point is directly proportional to the electric field at that point.

Therefore, since the electric field is different at point X than at the initial position, the electric potential energy of the charge increases as it moves to point X.

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The final speed of the second object after the collision is 27.91 m/s.

To solve this problem, we need to use the conservation of momentum principle, Before the collision, the total momentum of the system is:

[tex]P = m1v1 + m2v2\\\P = (4.71 kg)(27.2 m/s) + (7.86 kg)(13.7 m/s)\\\P = 236.56 kgm/s + 107.82 kgm/s\\\P = 344.38 kg*m/s[/tex]

After the collision, the two objects stick together and move with a common velocity, which we will call v. Therefore, the total momentum of the system after the collision is:

[tex]P' = (m1 + m2)*v\\\\\P' = (4.71 kg + 7.86 kg)v\\\P' = 12.57 kgv[/tex]

Since the total momentum of the system is conserved, we can equate the two expressions for the momentum:

[tex]P = P'\\\236.56 kgm/s + 107.82 kgm/s = 12.57 kgv\\\\\v = (236.56 kgm/s + 107.82 kg*m/s) / 12.57 kg\\\\\v = 27.91 m/s[/tex]

Therefore, the final speed of the second object after the collision is 27.91 m/s.

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Answer:

Explanation:

An earthquake, with its sudden and cataclysmic rupturing of the earth's crust, is a sublime exhibition of the extraordinary might of nature. This geological upheaval arises from the movements of tectonic plates - vast and ponderous slabs of rock comprising the earth's outermost layer. These plates glide past each other, collide, or diverge, accumulating seismic energy that they discharge as seismic waves, shaking the earth to its core.

The factors that underlie an earthquake are complex and multifarious, spanning the entire spectrum of geological events, from volcanic activity to anthropogenic interventions such as drilling and mining. Nevertheless, the dominant causal factor is the intricate interplay of tectonic plates, ceaselessly engaged in a dynamic dance of motion, friction, and release, culminating in an earthquake's fearsome display of raw power.

Despite their potential for devastation, comprehending the causes of earthquakes can inform and enhance our preparedness for these events, and equip us with the knowledge to mitigate their impact on human society. It may also allow us to deepen our appreciation of the natural world's enigmatic processes, and our place within it, engendering a richer understanding of our planet and its precarious balance.

The speed required to push the box is approximately 11.27 m/s.

Given the angle of inclination of the slope, θ = 18°. The distance covered is, d = 15 meters.

The component of gravity parallel to the slope is, g|| = g.sinθ

The force required to push the box up the slope is, F = m.g, The acceleration due to gravity is, g = 9.81 m/s².

The weight of the box is, m.g. The component of gravity parallel to the slope is, g|| = g.sinθ.

The force required to push the box up the slope is, F = m.g||

The force required to push the box up the slope is,F = m.g.sinθ.

The frictionless surface has no frictional force acting upon the object moving upon it. Therefore, no work is done by friction on the object. Hence, the energy of the object in the form of kinetic energy is conserved. This is as given below:

K.E. initial + work done = K.E. finfinalitial kinetic energy is zero. This is because the box is initially at rest. The final kinetic energy is K. This is because we need to push the box up the slope with the required force. The work done is equal to the force exerted multiplied by the distance moved in the direction of the force.

Hence,W = F.dThe final equation becomes,0 + F.d = ½.m.K²The final speed required is,K = √(2.F.d/m)Substituting the values in the above equation, we get the required final speed,K = √(2.m.g.sinθ.d/m)K = √(2.g.sinθ.d)Putting the values of d, g, and θ in the equation, we get,K = √(2.9.81.sin18°.15)≈ 11.27 m/s.

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The size of the image is 20 times smaller than the size of the car, and the image is located 120 cm behind the mirror.

Length is a measure of the magnitude of a line segment, straight or curved, between two points in space. It is a fundamental physical quantity in many fields of mathematics, physics, and engineering. Length is usually measured in units of distance such as inches, feet, meters, and kilometers. Length is an important concept in geometry, which is the study of shapes and their properties. In geometry, length is used to describe the size of lines, angles, and other shapes. Length can also be used to measure the size and distance between two points in space.

The size and location of the image of the car in the trucker’s diverging rear view mirror is determined by the focal length of the mirror. The focal length of the mirror is -60 cm. Using the thin lens equation, the image distance can be calculated as:

1/f = 1/di + 1/do

1/-60 = 1/di + 1/6

di = -120 cm

The height of the image can be calculated using the magnification formula:

m = -di/do

m = -120/6

m = -20

Therefore, the size of the image is 20 times smaller than the size of the car, and the image is located 120 cm behind the mirror.

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The direction of the force of static friction on an incline is perpendicular to the surface, while on a flat surface, it is parallel to the surface. This is because the incline introduces a component of the gravitational force that acts perpendicular to the surface.

On a flat surface, the force of static friction is parallel to the surface and opposite in direction to the applied force. However, on an incline, the component of gravitational force perpendicular to the surface creates a normal force that is also perpendicular to the surface. The force of static friction acts perpendicular to both the normal force and the inclined surface since it's always perpendicular to the normal force. This change in direction occurs because the force of static friction is a reactionary force that opposes the motion, and its direction depends on the forces acting on the object. Therefore, the direction of the force of static friction changes based on the orientation of the surface on which the object is placed.

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The required work done in stretching the spring from 15 cm to 20 cm is calculated to be 11.2 j.

The original length of the spring is given as 10 cm.

The work done to stretch the spring to 15 cm is 8 j.

Let us find out the value of k spring constant.

W = 1/2 k x²

where,

W is work

k is spring constant

x is elongation of spring

Putting in the values,

8 = 1/2 k (15² - 10²)

8 = 1/2 k (225 - 100)

8 = 1/2 k(125)

k = 8×2/125 = 0.128

Now, let us calculate the work done in stretching the spring from 15 cm to 20 cm.

W = 1/2 k x² = 1/2 (0.128) [(20² - 10²)-(15² - 10²)] = 0.064 [300-125] = 11.2 j

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The amount of time the baseball was in air for at a momentum of 0.85kgm/s is 0.399s

Momentum is the product of its mass and velocity, or the vector sum of the products of its masses and velocities. It can be calculated as follows:

Momentum = mass (m) × velocity (v)

However, the momentum of an object can be estimated using the following;

ΔM = F × ΔT

Where;

F = mass × acceleration

F = 0.213kg × 10m/s² = 2.13N

0.85 = 2.13 × t

t = 0.399s

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Answer:

no

Explanation:

because when he jumps into the swimming pool he will not be able to swim comfortably

The wavelength of a wave is the distance between two consecutive points on the wave that are in phase.

The wavelength is the distance between two consecutive nodes or two consecutive antinodes.

The first harmonious wavelength on a rope is given by the equation

λ = 2L/ n

where λ is the wavelength, L is the length of the rope, and n is the harmonious number. For the first harmonious, n = 1.

In this case, the length of the rope is L = 4m. Substituting into the equation over, we get

λ = 2( 4m)/ 1

λ = 8m

thus, the first harmonious wavelength on the rope is 8 measures.

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In a uniform field, a beam has no magnetic poles and therefore experiences a force equal to the product of its length and the field intensity. whereas bar magnet is influenced by the interaction between its magnetic poles and the field intensity.

When placed in a uniform field, the bar magnet experiences a torque, which causes it to align itself with the field direction. In contrast, a beam placed in a uniform field experiences a uniform force, which acts parallel to the field direction and perpendicular to the beam length. This force tends to cause the beam to move in the direction of the field intensity. The magnitude of the force on the beam is proportional to the length and field intensity, as well as the product of its magnetic permeability and the field intensity.

The force on the beam also tends to cause it to rotate about its center, in the direction of the field intensity. This rotation will result in the beam becoming oriented along the field direction, but it will not cause the beam to align itself with the field direction as with a bar magnet. Instead, the beam will experience a net force, which tends to move it in the direction of the field intensity.

In conclusion, the behavior of a beam in a uniform magnetic field differs from the behavior of a bar magnet in several ways. A beam placed in a uniform field experiences a uniform force, which acts parallel to the field direction and perpendicular to the beam length. This force tends to cause the beam to move in the direction of the field intensity and rotate about its center, in the direction of the field intensity.

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Answer: False

Explanation: It will move at a constant speed

The ammeter and voltmeter should be read simultaneously to determine the resistance of a circuit component because the resistance depends on the current passing through the component and the voltage drop across it. By measuring both the current and voltage, Ohm's law (V=IR) can be applied to calculate the resistance of the component.

The resistance of a circuit component represents how much it impedes the flow of current through it. According to Ohm's law, the voltage drop across a component is directly proportional to the current flowing through it, and the constant of proportionality is the resistance of the component. Therefore, to determine the resistance of a component, we need to know both the current passing through it and the voltage drop across it.

An ammeter is used to measure the current flowing through a circuit, while a voltmeter is used to measure the voltage difference between two points in the circuit. By reading both instruments simultaneously, we can use Ohm's law to calculate the resistance of the component being measured.

It is important to note that the readings must be taken simultaneously because the current and voltage can change over time, and any delay between taking the readings can result in inaccurate measurements. Therefore, it is necessary to read the ammeter and voltmeter at the same time to obtain an accurate measurement of the resistance of the component.

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