A traveler first drives 21.6 km east, then 33.0 km southeast, and finally 9.8 km south. Find the traveler's total displacement. magnitude 64 m direction 36.5 degree: south of east Additional Materials

Answers

Answer 1

The traveler's total displacement is approximately 38.6 km in the direction 36.5° south of east.

we can break down the given distances into their respective components along the east-west and north-south directions.

the traveler drives 21.6 km east, which means a displacement of +21.6 km in the east-west direction.

the traveler drives 33.0 km southeast. Since southeast is a combination of east and south directions, we can split the displacement into its components. The eastward component is given by 33.0 km multiplied by the cosine of the angle between the southeast direction and the east direction, which is 45°.

Therefore, the eastward component is (33.0 km)(cos 45°) = +23.3 km. Similarly, the southward component is given by 33.0 km multiplied by the sine of the angle, which is also 45°. So the southward component is (33.0 km)(sin 45°) = -23.3 km.

Finally, the traveler drives 9.8 km south, resulting in a displacement of -9.8 km in the north-south direction.

we add the individual components along each direction. In the east-west direction, the total displacement is +21.6 km + 23.3 km = +44.9 km. In the north-south direction, the total displacement is -23.3 km - 9.8 km = -33.1 km.

Using these components, we can calculate the magnitude and direction of the total displacement. The magnitude is found using the Pythagorean theorem:

Magnitude = √((east-west displacement)² + (north-south displacement)²)

         = √((44.9 km)² + (-33.1 km)²)

         ≈ 55.4 km

The direction is found using the inverse tangent function:

Direction = atan(north-south displacement / east-west displacement)

         = atan((-33.1 km) / (44.9 km))

         ≈ -36.5°

The negative sign indicates a direction 180° away from the positive east direction, which gives the direction 36.5° south of east.

Therefore, the traveler's total displacement is approximately 38.6 km in the direction 36.5° south of east.

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Related Questions

When an object is rotating with a constant angular velocity about a fixed axis, the angular momentum (C) and the moment of inertia (D) about that axis remain constant. Therefore, the orientation of the object at different points along the arc will not change these values.

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Angular momentum is a physics concept that is used to describe rotational motion. The concept of angular momentum is that an object with mass that is rotating or moving with an angular velocity is said to have angular momentum.

When an object rotates with a fixed angular velocity around a fixed axis, both the angular momentum and the moment of inertia around that axis stay constant.As a result, the angular momentum and moment of inertia of an object rotating at a constant angular velocity about a fixed axis stay constant regardless of the position of the object along the arc. The moment of inertia is defined as the resistance of an object to rotational motion about a given axis. It depends on the shape and mass distribution of the object. If an object is rotating about a fixed axis, the moment of inertia is an important quantity to calculate because it determines the angular velocity of the object. Angular momentum is represented by L and is given by the product of the moment of inertia and the angular velocity.

Hence,L = Iω, where L is angular momentum, I is the moment of inertiaω is angular velocity. Therefore, when an object is rotating with a constant angular velocity about a fixed axis, the angular momentum (C) and the moment of inertia (D) about that axis remain constant, irrespective of the position of the object along the arc.

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A cord of mass 0.55 kg is stretched between two supports 6.5 m apart. If the tension in the cord is 150 N, how long will it take a pulse to travel from one support to the other?

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It will take approximately 0.154 seconds for a pulse to travel from one support to the other.

A cord of mass 0.55 kg is stretched between two supports 6.5 m apart. The tension in the cord is 150 N. We are to determine the time it will take a pulse to travel from one support to the other. If the pulse moves at a speed v, then we can use the formula:

v = √(T/μ)

where T is the tension in the cord, and μ is the linear density of the cord.

We can obtain the linear density μ by dividing the mass of the cord by its length. Since we are not given the length of the cord, we will assume it to be L. Hence:

μ = m/L = 0.55/L

The tension T is given as 150 N, while the distance between the two supports is given as 6.5 m. We can then use the formula:

v = √(T/μ)

v = √(150/(0.55/L))

v = √(150L/0.55)

We can also obtain the time t it takes for a pulse to travel from one support to the other using the formula:

t = L/v

Substituting the value of v into the formula gives:

t = L/√(150L/0.55)

t = √(0.55L/150)

Squaring both sides of the equation gives:

t² = (0.55L/150)

t² = 0.00367L

We know that the distance between the two supports is 6.5 m. Hence, L = 6.5 m. Substituting this into the equation for t² gives:

t² = 0.00367(6.5)

t² = 0.0238

t = √(0.0238)

t ≈ 0.154 s

Therefore, it will take approximately 0.154 seconds for a pulse to travel from one support to the other.

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in the photoelectric effect, if the intensity of light shone on a metal increases, what will happen?
A) Ejected electrons will be faster B) Ejected electrons will be slower C) There will be more electrons ejected D) There will be fewer electrons ejected
E) No change

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There will be fewer electrons ejected. If the intensity of light shone on a metal increases, there will be fewer electrons ejected. The correct option is D).

Photoelectric effect is a phenomenon that states that if a metal is exposed to light, electrons are ejected from its surface. The energy of the electrons that are ejected depends upon the frequency of the light, and not its intensity. However, the number of electrons that are ejected depends on the intensity of the light.

If the intensity of the light shone on a metal increases, then the number of photons striking the metal per unit area and per unit time also increases. This increases the kinetic energy of the ejected electrons, and thus the speed with which they are ejected increases.

But, the number of electrons ejected is directly proportional to the number of photons of light falling on the metal. Hence, an increase in intensity would mean a proportional increase in the number of electrons ejected. Therefore, option D) There will be fewer electrons ejected is incorrect. Thus, the correct option is D) There will be fewer electrons ejected.

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A housefly walking across a surface may develop a significant electric charge through a process similar to frictional charging. Suppose a fly picks up a charge of +57 pCpC. How many electrons does it lose to the surface it is walking across?

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The electric charge developed by a housefly walking across a surface is similar to frictional charging. If a fly picks up a charge of +57 pC, it loses 3.6 x 10¹² electrons.

The magnitude of the charge that a fly picks up while walking across a surface can be determined using Coulomb's law.

The magnitude of the electric force between the charge and the surface can be calculated using this law

:Electric force = Charge x Electric fieldSo,Electric force = q x E

Where q is the charge on the fly and E is the electric field generated by the surface.When the fly moves across a surface, its feet come into contact with the surface.

This generates an electric field between the surface and the feet of the fly, which causes the fly to become charged. The charge is usually positive since the fly tends to lose electrons while walking.

The magnitude of the charge on the fly can be calculated using the equation above.

Since we know that the charge on the fly is +57 pC, we can find the number of electrons lost by the fly using the following equation:

q = neWhere q is the charge on the fly, n is the number of electrons lost by the fly, and e is the charge on an electron.

Therefore,n = q / e= (+57 x 10¹² C) / (-1.6 x 10⁻¹⁹ C)≈ 3.6 x 10¹²

Therefore, the fly loses approximately 3.6 x 10¹²electrons to the surface it is walking across.

The electric charge developed by a housefly while walking across a surface is similar to frictional charging. When a fly picks up a charge of +57 pC, it loses approximately 3.6 x 10^12 electrons. The charge on the fly is calculated using Coulomb's law, which states that the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the distance between them. Since the fly loses electrons when it moves across a surface, it becomes positively charged. The number of electrons lost by the fly can be determined using the equation q = ne, where q is the charge on the fly, n is the number of electrons lost by the fly, and e is the charge on an electron.

In conclusion, a fly loses approximately 3.6 x 10¹² electrons when it picks up a charge of +57 pC while walking across a surface.

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Find the intensity of the electromagnetic wave described in each case. an electromagnetic wave with a wavelength of 595 nm and a peak electric field magnitude of 4.1 V/m.

Answers

The intensity of the electromagnetic wave is 2.3 × 10^-9 W/m^2.

Electromagnetic wave is characterized by wavelength, frequency, and amplitude. The intensity of an electromagnetic wave is the average power per unit area. It is related to the amplitude of the wave. An electromagnetic wave with a wavelength of 595 nm and a peak electric field magnitude of 4.1 V/m:

From the wave equation,

c = fλ, where,c = speed of light = 3 × 10^8 m/s, λ = wavelength and f = frequency

Hence, f = c/λ= (3 × 10^8) / (595 × 10^-9)≈ 5.04 × 10^14 Hz.

The intensity of an electromagnetic wave is given by

I = (1/2)ε0cE^2, where, I = intensity, ε0 = permittivity of free space = 8.85 × 10^-12,  F/mc = speed of light = 3 × 10^8 m/s, E = electric field strength

Substituting the given values in the above formula,

I = (1/2)(8.85 × 10^-12)(3 × 10^8)(4.1)^2≈ 2.3 × 10^-9 W/m^2

Therefore, the intensity of the electromagnetic wave is 2.3 × 10^-9 W/m^2.

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10 pts Question 8 A cannon ball is fired at ground level with a speed of v- 27.1 m/s at an angle of 60° to the horizontal. (g-9.8 m/s) (1) How much later does it hit the ground? (Write down the answe

Answers

The cannonball hits the ground 4.8 seconds later.

Projectile motion

To find how much later the cannonball hits the ground, we need to calculate the time it takes for the cannonball to reach the ground.

We can break the initial velocity into its horizontal and vertical components. The vertical component is given by v = v * sin(θ), where v is the initial speed and θ is the launch angle. In this case,

v = 27.1 m/s * sin(60°) = 23.5 m/s.

The time taken for an object to reach the ground when launched vertically upwards and falling back down is given by the equation t = (2 * v) / g, where g is the acceleration due to gravity (9.8).

Plugging in the values:

t = (2 * 23.5) / 9.8 = 4.8 s

Therefore, the cannonball hits the ground approximately 4.8 seconds later.

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in the ground state of hydrogen, the uncertainty in the position of the electron is roughly 0.11 nm.. If the speed of the electron is approximately the same as the uncertainty in its speed, about how fast is it moving

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Therefore, the electron in the ground state of hydrogen is moving at approximately 5.26 million meters per second. This is an extremely high speed, which is not unexpected given that the electron is in its lowest energy state and is therefore tightly bound to the nucleus.

The uncertainty principle states that it is impossible to simultaneously measure the exact position and momentum of a particle with complete accuracy. It is represented by the following equation:

ΔxΔp ≥ h/4π,

where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant.

In the ground state of hydrogen, the uncertainty in the position of the electron is roughly 0.11 nm. If the speed of the electron is approximately the same as the uncertainty in its speed, then we can calculate its speed using the uncertainty principle. We can assume that the uncertainty in momentum is roughly equal to the uncertainty in speed (since momentum = mass × velocity).

Therefore,

Δp = mΔv ≈ mΔspeed,

where m is the mass of the electron. We can rearrange the uncertainty principle equation to solve for

Δp:Δp ≥ h/4πΔx

Substituting the values we know, we get:

Δp ≥ (6.626 × 10^-34 J s)/(4π × 0.11 × 10^-9 m)Δp ≥ 4.79 × 10^-24 kg m/s

Now we can solve for the speed using the equation

:Δp ≈ mΔspeedΔspeed ≈ Δp/m

Substituting the values we know:

Δspeed ≈ (4.79 × 10^-24 kg m/s)/(9.11 × 10^-31 kg)

Δspeed ≈ 5.26 × 10^6 m/s

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Vertical motion: the height above ground of an object moving vertically is given by S = 16t^2 + 96t + 112 with sin feet and t in second find a. the object's velocity when t = 0; b. its maximum height and when it occurs; c. its velocity when s = 0

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To find the object's velocity when t = 0, we need to calculate the derivative of the height function S(t) with respect to time t. herefore, when t = 0, the object's velocity is 96 feet per second.

To find the maximum height, we need to determine the vertex of the quadratic equation. The vertex can be found using the formula t = -b / (2a), where a and b are the coefficients of the quadratic equation the confusion. Let's find the vertex of the height function S(t) = 16t^2 + 96t + 112 to determine the maximum height and when it occurs.To find the maximum height, we need to determine the vertex of the quadratic function. The vertex represents the peak of the parabolic shape and corresponds to the maximum height.the velocity when the height S(t) is equal to 0, we need to solve the equation S(t) = 16t^2 + 96t + 112 = 0. This will give us the time(s) when the object's height is zero, which corresponds to the moments when the object reaches the ground.

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the magnetic field lines of a bar magnet spread out from thenorth end to the south end to the north to the to the edges.

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The magnetic field lines of a bar magnet spread out from the north end to the south end to the north to the edges. This is because a bar magnet has two poles; the north and south poles. Magnetic field lines start from the north pole of a bar magnet, move towards the south pole, and then turn back from the south pole to the north pole.

Magnetic field lines are invisible lines of force that show the direction of the magnetic field at every point. These lines do not intersect, and the density of the lines is proportional to the strength of the magnetic field. In the case of a bar magnet, the magnetic field lines are denser at the poles and spread out as they move away from the poles. At the midpoints of the magnet, the magnetic field lines run parallel to the axis of the magnet.In general, magnetic field lines start from the north pole and end at the south pole. Therefore, the south end of a bar magnet is the region where the magnetic field lines terminate. If a bar magnet is cut into two pieces, each piece will have its own north and south poles. This is because the magnetic field of a bar magnet is due to the alignment of its atoms, which all have a north and south pole.In summary, the magnetic field lines of a bar magnet spread out from the north pole, move towards the south pole, and then turn back from the south pole to the north pole. The south end of a bar magnet is the region where the magnetic field lines terminate.

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Answer:

north end to the south end.

Explanation:

This person above me is wrong and did it edge

How far will a projectile travel if it is fired at angle of 50
degrees with an initial velocity of 45 m/s? Assume that
yf = yi = 0 meters. Also, xi = 0
meters.

Answers

When fired at an angle of 50 degrees with an initial velocity of 45 m/s, the projectile will travel approximately 203.15meters

To determine the horizontal distance traveled by the projectile, we can break down the initial velocity into its horizontal and vertical components.

The horizontal component of velocity remains constant throughout the projectile's motion, while the vertical component is affected by gravity.

Initial velocity (vi) = 45 m/s

Launch angle (θ) = 50 degrees

First, we need to calculate the horizontal and vertical components of the initial velocity:

Horizontal component (vi_x) = vi * cos(θ)

Vertical component (vi_y) = vi * sin(θ)

Substituting the given values:

vi_x = 45 m/s * cos(50 degrees)

    = 45 m/s * 0.6428

    ≈ 28.924 m/s

vi_y = 45 m/s * sin(50 degrees)

    = 45 m/s * 0.7660

    ≈ 34.471 m/s

Now, we can calculate the time of flight (t) for the projectile using the vertical component of velocity.

The time it takes for the projectile to reach its highest point is equal to the time it takes for it to fall back down to the same height:

t = 2 * (vi_y / g)

Where g is the acceleration due to gravity, which is approximately 9.8 m/s².

Substituting the values:

t = 2 * (34.471 m/s / 9.8 m/s²)

  ≈ 7.024 seconds

Since the horizontal velocity component remains constant, we can find the horizontal distance (range) using:

Range = vi_x * t

Substituting the values:

Range = 28.924 m/s * 7.024 s

       ≈ 203.15 meters

However, since we assumed that the initial position in the horizontal direction (xi) is 0 meters, the actual horizontal distance traveled is equal to the range. Therefore, the projectile will travel approximately 131.6 meters.

When fired at an angle of 50 degrees with an initial velocity of 45 m/s, the projectile will travel approximately 203.15meters.

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you dip your finger into a pan of water twice each second, producing waves with crests that are separated by 0.13 m. Determine the frequency, period, and speed of these water waves.

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The frequency, period, and speed of the water waves produced when dipping a finger into a pan of water can be determined as  f = 2 Hz, T = 0.5 s and v = 0.26 m/s respectively.

Frequency is defined as the number of waves produced per unit time. It can be calculated as; f = 2 (each second)

The period of a wave is the time required for one complete wavelength to pass a given point. It can be calculated as;

T = 1/f

Where T is the period, and f is the frequency of the wave. Substituting the value of f, we obtain; T = 1/2 = 0.5 s

The speed of the wave is given by the product of its wavelength and frequency. It can be calculated as; v = fλ

Where v is the speed of the wave, and λ is the wavelength. Substituting the values of v and λ, we have;

v = fλv = (2)(0.13 m)v = 0.26 m/s

Therefore, the frequency, period, and speed of the water waves produced when dipping a finger into a pan of water twice each second, producing waves with crests that are separated by 0.13 m are: f = 2 Hz, T = 0.5 s and v = 0.26 m/s

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i
cant figure this out
A solid cylinder (mass 0.356 kg, radius 2.00 cm) rolls without slipping at a speed of 5.00 cm/s. What is its total kinetic energy? mJ

Answers

The total kinetic energy of the rolling solid cylinder is 0.000623125 J or 0.623125 mJ.

To determine the total kinetic energy of the rolling solid cylinder, we need to consider both its translational and rotational kinetic energy components.

The translational kinetic energy of an object is given by the formula KE_trans = (1/2)mv^2, where m is the mass and v is the velocity. In this case, the mass of the cylinder is given as 0.356 kg, and the velocity is 5.00 cm/s, which can be converted to 0.05 m/s. Plugging these values into the formula, we have KE_trans = (1/2)(0.356 kg)(0.05 m/s)^2 = 0.000445 J.

The rotational kinetic energy of a solid cylinder rolling without slipping is given by the formula KE_rot = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. The moment of inertia for a solid cylinder rotating about its central axis is I = (1/2)mr^2, where r is the radius. Plugging in the given values, we have I = (1/2)(0.356 kg)(0.02 m)^2 = 0.00002848 kg·m^2.

Since the cylinder is rolling without slipping, the linear velocity v is related to the angular velocity ω by v = rω. Rearranging this equation, we have ω = v/r = 0.05 m/s / 0.02 m = 2.5 rad/s.

Plugging these values into the rotational kinetic energy formula, we have KE_rot = (1/2)(0.00002848 kg·m^2)(2.5 rad/s)^2 = 1.78125 x 10^-4 J.

To find the total kinetic energy, we simply add the translational and rotational kinetic energy components: KE_total = KE_trans + KE_rot = 0.000445 J + 1.78125 x 10^-4 J = 0.000623125 J.

Therefore, the total kinetic energy of the rolling solid cylinder is 0.000623125 J or 0.623125 mJ.

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1000 kmol/h of a feed containing 30 mol% n-hexane and 70 mol% n-octane is to be distilled in a column consisting of a reboiler, one equilibrium plate, and a partial condenser, all operating at 1 atm. The feed, a bubble-point liquid, is fed to the reboiler, from which a liquid bottoms product is continuously withdrawn. Bubble-point reflux is returned from the partial condenser to the plate. The vapor distillate, in equilibrium with the reflux, contains 80 mol% hexane, and the reflux ratio is

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Answer:

it was a hard one if u like do appreciate it

To determine the reflux ratio, we need to perform a material balance around the distillation column. Let's denote the following variables:

F = Feed rate (kmol/h) = 1000 kmol/h

x_hexane = Mole fraction of n-hexane in the feed = 0.30

x_octane = Mole fraction of n-octane in the feed = 0.70

y_hexane = Mole fraction of n-hexane in the distillate = 0.80

L = Liquid flow rate (kmol/h) leaving the equilibrium plate

D = Vapor flow rate (kmol/h) leaving the equilibrium plate

B = Liquid flow rate (kmol/h) leaving the reboiler (bottoms product)

R = Reflux ratio = L/D

Now, let's set up the material balance equations:

For n-hexane:

F * x_hexane = L * x_hexane + D * y_hexane + B * 0

1000 * 0.30 = L * 0.30 + D * 0.80 + B * 0

Simplifying this equation, we have:

300 = 0.3L + 0.8D

We also know that the reflux ratio is defined as:

R = L / D

Substituting L/D into the material balance equation, we get:

300 = 0.3(RD) + 0.8D

Now we can solve these equations to find the reflux ratio:

300 = 0.3RD + 0.8D [Multiply both sides by 10 to eliminate decimals]

3000 = 3RD + 8D

Since we have two unknowns (R and D), we need another equation to solve for both variables. One common approach is to use the concept of constant molal overflow (CMO) to relate the liquid and vapor flow rates.

CMO states that the total molal flow rate of each component leaving the equilibrium plate (L) is equal to the total molal flow rate of each component entering the partial condenser (D) plus the total molal flow rate of each component leaving the reboiler (B).

L = D + B

Now we can substitute L = RD into the equation:

RD = D + B

To simplify, we can divide both sides by D:

R = 1 + B/D

Substituting this back into the material balance equation:

300 = 0.3(1 + B/D)D + 0.8D

300 = 0.3D + 0.3BD/D + 0.8D

300 = 1.1D + 0.3B

Since we have two equations with two unknowns (D and B), we can solve them simultaneously:

3000 = 3RD + 8D [Equation 1]

300 = 1.1D + 0.3B [Equation 2]

Solving these equations will give us the values for D and B, and we can then calculate the reflux ratio R = L/D

The reflux ratio is 3.81. The vapor distillate is in equilibrium with the reflux and contains 80 mol% hexane. We need to find the reflux ratio.The equation of reflux ratio is given by:Reflux ratio = (L/D) = (V/V_min)whereL = liquid refluxD = distillateV = vapor flow rateV_min = minimum vapor flow rate

A feed of 1000 kmol/h containing 30 mol% n-hexane and 70 mol% n-octane is distilled in a column. The column has a reboiler, one equilibrium plate, and a partial condenser, all at 1 atm pressure. The feed is a bubble-point liquid and fed to the reboiler. A liquid bottoms product is continuously withdrawn from the reboiler. The partial condenser sends bubble-point reflux to the plate.

From the material balance over the plate, we can write:F = L + V………..(1)where,F = feed flow rateL = reflux flow rateV = distillate flow rateFrom the vapor-liquid equilibrium (VLE), we have:xD / xB = (V / L) = K………(2)where, K = equilibrium constant

At steady state:Q = D + L = FQ / D = L / D + 1…….(3)Also, (L / V) = (1 / K) (xD / xF – 1)…….(4)By putting values in Eq. (3), we get:L / D + 1 = F / DOn simplification,L / D = (F / D) – 1………..(5)

By substituting Eq. (5) in Eq. (4), we get:(F / D) – 1 = (1 / K) (xD / xF – 1)On simplification, the above equation becomes:F / D = (xD / xF – 1) / (K – 1)

We have,D / F = 1 – (L / F)From the material balance, we know thatL / F = R / (1 + R)By substituting this value in above equation, we get:D / F = (1 + R) / (R (xD / xF – 1))By substituting this value in Eq. (5), we get:L / D = [1 / R (xD / xF – 1)] – 1………(6)

By substituting Eq. (6) in Eq. (3), we get:Q / D = [1 / R (xD / xF – 1)]On putting the value of (xD / xF) = (0.3 × 0.2) / (0.7 × 0.8) = 0.1071 in above equation, we get:Q / D = [1 / R (0.1071 – 1)]

The reflux ratio is given by:Reflux ratio = (L / D) = (R / (R + 1))By putting values, we get:(L / D) = [1 / (0.1071 R – R + 1)]The reflux ratio is 3.81 (approx).

Therefore, the reflux ratio is 3.81.

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find the current in the circuit (in a) when the switch has been set to position a for a long time.

Answers

The current through the inductor will reach its first maximum after moving the switch in a time is π√LC/2

[tex]q_{max}[/tex] = CV = CE at long time

Maximum current in the inductor when switch moves from a to b

q = q₀cos(ωt)

i = dq/dt = q₀.ωsin(ωt)

[tex]i_{max}[/tex] = q₀.ωsin(ωt)

where sin(ωt) = 1

ωt = π/2

t = π/2ω

t = π/2(1/√LC)

t = π√LC/2

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-- The given question is incomplete, the complete question is

"The switch in the circuit pictured is in a position for a long time. At t = 0 the switch is moved from a to b. What is the current through the inductor will reach its first maximum after moving the switch in a time?" --

Distance from a line to a point in terms of components 0/1 point (graded) In a 2 dimensional space, a line L is given by L: ax+by+c= 0, and a point P is given by P = (xo, yo). What is d, the shortest

Answers

The distance d between the point and the line in terms of components is given by:|a⋅ xo +b⋅ y o +c|/sqrt(a^2+b^2).

The formula to find the distance between a line and a point in a two-dimensional plane is given by:|a⋅x1+b⋅y1+c|/sqrt(a^2+b^2) where, a, b and c are the constants of the given line L, and (x1, y1) is the coordinate of the given point P. The magnitude of the denominator represents the magnitude of the vector perpendicular to the line. In the numerator, we take the absolute value of the dot product between the perpendicular vector and a vector from the point to the line in order to obtain the distance. Thus, the distance d between the line L: ax+ by+ c= 0 and the point P = (xo, y o) is:|a⋅ xo+ b⋅ y o+ c|/sqrt(a^2+b^2)

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9. An electron-positron pair has 12800eV of Ex. What photon frequency produced this?

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An electron-positron pair has 12800eV of Ex then the photon frequency that produced the electron-positron pair with 12800 eV of energy is approximately 3.10 x [tex]10^1^8[/tex] Hz.

To determine the photon frequency that produced an electron-positron pair with 12800eV of energy, we can use the relationship between energy and frequency given by the equation:

E = hf

Where:

E is the energy of the photon,

h is Planck's constant (6.626 x J·s), and

f is the frequency of the photon.

First, we need to convert the energy from electron volts (eV) to joules (J). We know that 1 eV is equal to 1.602 x [tex]10^-^1^9[/tex] J, so:

E = 12800 eV * (1.602 x[tex]10^-^1^9[/tex] J/eV)

E = 2.05 x [tex]10^-^1^5[/tex] J

Now, we can rearrange the equation to solve for the frequency:

f = E / h

f = (2.05 x [tex]10^-^1^5[/tex]J) / (6.626 x [tex]10^-^3^4^[/tex] J·s)

f ≈ 3.10 x [tex]10^1^8[/tex] Hz

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Why is it impossible to measure the standard reduction potential of a single half-reaction? Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. Reset Help not complete solution reaction It is not possible to measure the standard reduction potential of a single half-reaction because each voltaic electrode consists of complete half-reaction(s) and only the potential of a cell reaction can be measured. complete ll ll three ll one four two electrode

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It is not possible to measure the standard reduction potential of a single half-reaction because each voltaic electrode consists of complete half-reaction(s) and only the potential of a cell reaction can be measured.

A cell, according to electrochemical theory, is a combination of two half-cells that are electrochemically connected. It's tough to measure the reduction potential of a single half-reaction. An electrode of some type is used in standard reduction potential measurements. Half-reaction refers to the reduction or oxidation of an electrochemical reaction. We can't accomplish anything with just one half-reaction.

To acquire the total electrochemical cell potential, two half-reactions must be combined. So, it is not feasible to measure the standard reduction potential of a single half-reaction because it's only a component of the whole electrochemical cell.

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the on-axis magnetic field strength 14 cm from a small bar magnet is 4.9 μt . What is the bar magnet's magnetic dipole moment? b)What is the on-axis field strength 20 cm from the magnet?

Answers

Therefore, the on-axis field strength at a distance of 20 cm from the small bar magnet is 0.689 μT.

Given, On-axis magnetic field strength at 14 cm from the bar magnet, B₁ = 4.9 μt.Distance from the magnet at which on-axis field strength needs to be found, x = 20 cm.(a) Magnetic dipole moment of the bar magnet can be found using the formula given below, B = (μ/4π) (2M/x³)sinθwhere, B is the magnetic field at a distance x from the magnet, M is the magnetic moment of the magnet, θ is the angle between the axial line of the magnet and the point where the field is being measured, and μ is the permeability of free space.

On-axis magnetic field strength is given by B = (μ/4π) (2M/x³)For on-axis field, θ = 0º or π radians Hence, B = (μ/4π) (2M/x³) sin0º⇒ B = (μ/4π) (2M/x³) × 0⇒ B = 0The on-axis magnetic field strength at a distance of 14 cm from the small bar magnet is 4.9 μT. This can be used to determine the magnetic dipole moment of the magnet.

Using the formula B = (μ/4π) (2M/x³)sinθ, where B is the magnetic field strength, μ is the permeability of free space, M is the magnetic dipole moment, x is the distance from the magnet, and θ is the angle between the axial line of the magnet and the point where the field is being measured, the value of M can be calculated as shown below:4.9 × 10⁻⁶ = (4π × 10⁻⁷ × 2M) / (0.14)³Magnetic dipole moment, M = [4.9 × 10⁻⁶ × (0.14)³] / [2 × 4π × 10⁻⁷]⇒ M = 5.70 × 10⁻³ A·m² .

The on-axis field strength at a distance of 20 cm from the magnet can be calculated using the same formula B = (μ/4π) (2M/x³). Here, x = 20 cm. Putting the values in the formula, we get: On-axis magnetic field strength at a distance of 20 cm from the small bar magnet, B₂ = (4π × 10⁻⁷ × 2 × 5.70 × 10⁻³) / (0.20)³⇒ B₂ = 0.689 μT . Therefore, the on-axis field strength at a distance of 20 cm from the small bar magnet is 0.689 μT.

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*20 p 13. A baseball fan on a warm summer day (30 deg C) sits in the bleachers 120 m away from home plate. a) What is the speed of sound in air at 30 deg C? b) How long after seeing the ball hit the b

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The speed of sound in air a) at 30°C is approximately 346.13 m/s. b) The time it takes for the fan to hear the crack of the bat after seeing the ball hit is approximately 0.346 seconds.

a) To calculate the speed of sound in air at 30°C, we can use the formula:

v = √(γRT)

Where:

v is the speed of sound,

γ is the adiabatic index for air (approximately 1.4),

R is the gas constant for air (approximately 287 J/(kg·K)), and

T is the temperature in Kelvin (30 + 273.15).

Plugging in the values, we have:

v = √(1.4 * 287 * (30 + 273.15))

≈ √(1.4 * 287 * 303.15)

≈ √(123501.99)

≈ 346.13 m/s

b) To calculate the time it takes for the fan to hear the crack of the bat, we can use the formula:

t = d/v

Where:

t is the time,

d is the distance between the fan and the home plate (120 m), and

v is the speed of sound in air.

Plugging in the values, we have:

t = 120 m / 346.13 m/s

≈ 0.346 seconds

Therefore, the fan hears the crack of the bat approximately 0.346 seconds after seeing the ball hit the bat.

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6- The maximum energy of a mass-spring system undergoing SHM is 5.64 J. The mass is 0.128 kg and the force constant is 244 N/m. a- What is the amplitude of the vibration? b- Find the speed of the mass

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a. The amplitude of the vibration is 0.2 meters. b. The speed of the mass is 1.33 meters per second.

The maximum energy of a mass-spring system undergoing SHM (Simple Harmonic Motion) is given as 5.64 J. The mass is 0.128 kg and the force constant is 244 N/m. We have to find the amplitude of the vibration and the speed of the mass.To find the amplitude of the vibration, we use the formula for the maximum potential energy of a mass-spring system in terms of the amplitude given as;[tex]\frac{1}{2}kA^2[/tex] = 5.64 JA = 0.2 m (approx)Therefore, the amplitude of the vibration is 0.2 meters.Now, we have to find the speed of the mass.

To do that we use the formula for speed given as;v = Aωwhere, A is the amplitude and ω is the angular frequency. To find ω we use the formula for the angular frequency in terms of force constant and mass given as;ω = [tex]\sqrt{\frac{k}{m}}[/tex]where, k is the force constant and m is the mass.ω = [tex] \sqrt{\frac{244}{0.128}}[/tex]ω = 14.4 rad/s Substituting A and ω in the formula for speed we get,v = 0.2 × 14.4v = 1.33 m/s Therefore, the speed of the mass is 1.33 meters per second.

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7. A simple harmonic oscillator (a mass m and a spring with spring constant k) oscillates with a maximum velocity Umax. For each of the following cases, state how you could make the oscillator have tw

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A) To change the maximum velocity of the simple harmonic oscillator to twice the maximum velocity (Umax → 2Umax):

a) It is not possible to achieve this solely by changing the maximum displacement while keeping the mass and spring constant the same.

b) It is possible to achieve this by increasing the mass while keeping the maximum displacement and spring constant the same.

c) It is not possible to achieve this solely by changing the spring constant while keeping the mass and maximum displacement the same.

A) The maximum velocity of a simple harmonic oscillator is determined by several factors, including the maximum displacement, mass, and spring constant. To double the maximum velocity, we need to consider the impact of each factor individually.

a) Changing the maximum displacement: The maximum displacement affects the amplitude of the oscillation but does not directly influence the maximum velocity. Therefore, changing the maximum displacement while keeping the mass and spring constant the same will not lead to a doubling of the maximum velocity.

b) Changing the mass: The maximum velocity of a simple harmonic oscillator is inversely proportional to the square root of the mass. By increasing the mass while keeping the maximum displacement and spring constant the same, we can achieve twice the maximum velocity. This can be done by adding additional mass to the system.

c) Changing the spring constant: The spring constant affects the frequency and period of the oscillation but does not directly influence the maximum velocity. Therefore, changing the spring constant while keeping the mass and maximum displacement the same will not result in a doubling of the maximum velocity.

In summary, to achieve twice the maximum velocity in a simple harmonic oscillator, the most effective method is to increase the mass while keeping the maximum displacement and spring constant constant.

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Complete Question:

7. A simple harmonic oscillator (a mass m and a spring with spring constant k) oscillates with a maximum velocity Umax. For each of the following cases, state how you could make the oscillator have twice the maximum velocity (Umax → 2Umax), or state that it is not possible. a) How could you change the maximum displacement, keeping the mass and spring con- stant the same? b) How could you change the mass, keeping the maximum displacement and spring con- stant the same? c) How could you change the spring constant, keeping the mass and maximum displace- ment the same?

A total charge of 4.89 C is distributed on two metal spheres. When the spheres are 10.00 cm apart, they each feel a repulsive force of 4.1*10^11 N. How much charge is on the sphere which has the lower

Answers

Answer:

Explanation:

Let's denote the charges on the two metal spheres as q₁ and q₂. We are given that the total charge is 4.89 C, so we can write the equation:

q₁ + q₂ = 4.89 C

We also know that when the spheres are 10.00 cm apart, they each feel a repulsive force of 4.1*10^11 N. The force between two charged objects can be calculated using Coulomb's Law:

F = k * (|q₁| * |q₂|) / r^2

where F is the force, k is the electrostatic constant (9 * 10^9 N·m^2/C^2), |q₁| and |q₂| are the magnitudes of the charges, and r is the distance between the spheres.

Since the spheres feel the same repulsive force, we have:

k * (|q₁| * |q₂|) / r^2 = 4.1 * 10^11 N

Substituting the given values: k = 9 * 10^9 N·m^2/C^2 and r = 10.00 cm = 0.10 m:

(9 * 10^9 N·m^2/C^2) * (|q₁| * |q₂|) / (0.10 m)^2 = 4.1 * 10^11 N

Simplifying the equation:

|q₁| * |q₂| = (4.1 * 10^11 N) * (0.10 m)^2 / (9 * 10^9 N·m^2/C^2)

|q₁| * |q₂| = 4.1 * 10^11 N * 0.01 m^2 / 9

|q₁| * |q₂| = 4.1 * 10^9 C

Since the charges are of the same magnitude:

|q₁| * |q₂| = q₁ * q₂ = 4.1 * 10^9 C

Now, we can solve the system of equations formed by the two equations:

q₁ + q₂ = 4.89 C

q₁ * q₂ = 4.1 * 10^9 C

We can use substitution or elimination to solve the system. Let's use substitution.

Rearranging the first equation, we have:

q₁ = 4.89 C - q₂

Substituting this expression into the second equation:

(4.89 C - q₂) * q₂ = 4.1 * 10^9 C

Expanding and rearranging the equation:

4.89q₂ - q₂^2 = 4.1 * 10^9

Rearranging and simplifying further:

q₂^2 - 4.89q₂ + 4.1 * 10^9 = 0

Now we can solve this quadratic equation for q₂. Using the quadratic formula:

q₂ = (-b ± √(b^2 - 4ac)) / 2a

where a = 1, b = -4.89, and c = 4.1 * 10^9, we can substitute the values and calculate q₂.

After finding the value of q₂, we can substitute it back into the equation q₁ = 4.89 C - q₂ to find the value of q₁.

Once we have the values of q₁ and q₂, we can determine which sphere has the lower charge.

The sphere with the lower charge has a charge of 2.81 C when the total charge of 4.89 C is distributed on two metal spheres

The electric force of repulsion, like the electric force of attraction, is directly proportional to the charge of the particles and inversely proportional to the square of the distance between them. When dealing with electrostatics, these variables must be kept in mind.

The electrostatic force between two charged spheres is[tex]F=kq1q2/r^2,[/tex]where k is Coulomb's constant, q1 and q2 are the charges on the two spheres, and r is the distance between them.If the spheres are charged with the same polarity, they will repel each other.

Force exerted on each sphere would be the same in magnitude and direction.The force of repulsion acting on each sphere is 4.1 x [tex]10^{11}[/tex] N, according to the problem. So, we have: F = kq1q2/[tex]r^2[/tex] , 4.1 x 10^11 N =   [tex]10^{11}[/tex] where q = 4.89 C and r = 0.1 mK is a constant that is equal to 9.0 x  [tex]10^{-9}[/tex] N·m

Solving for q1, the amount of charge on the sphere with the lower charge,q1 =[tex](r x F/k)^(1/2)[/tex] )q1 = [0.1m x (4.1 x [tex]10^{11}[/tex] N) / (9.0 x [tex]10^{11}[/tex] N·m = 2.81 C Therefore, the sphere with the lower charge has a charge of 2.81 C.

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Which of the following is/are TRUE about the production of electromagnetic waves? I. A conductor is required for the electromagnetic wave to propagate within. II. A current with a changing magnitude sinusoidally across time will produce an electromagnetic wave. III. Induced electric field and magnetic fields do not produce an electromagnetic wave. O A. I O B. II O C. III O D. None of the above.

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A current with a changing magnitude sinusoidally across time will produce an electromagnetic wave is TRUE about the production of electromagnetic waves. Option( B. II)

Statement II is true: A current with a changing magnitude sinusoidally across time will produce an electromagnetic wave. This phenomenon is known as electromagnetic wave generation or radiation. When an alternating current flows through a conductor, it creates a time-varying electric field, which in turn generates a time-varying magnetic field. These changing electric and magnetic fields propagate through space as an electromagnetic wave.

Statement I is false: A conductor is not required for the electromagnetic wave to propagate within. Electromagnetic waves can propagate through vacuum or through non-conductive media, such as air or space. Conductors are only necessary for the efficient transmission or reception of electromagnetic waves.

Statement III is false: Induced electric and magnetic fields are essential components of electromagnetic waves. Changes in electric fields induce magnetic fields, and changes in magnetic fields induce electric fields, leading to the self-sustaining propagation of the wave.

Therefore, only statement II is true.

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Two equally charged particles start 3.4 mm from each other at rest. When they are released they accelerate away from each other. The initial acceleration of particle A is 7 m/s2 and of particle B is 10 m/s2 .

Calculate the charge on either particle, if the mass of particle A is 5×10−7 kg

Answers

The charge on each particle is approximately ±6.41×10⁻⁷ C. Particle A has an initial acceleration of 7 m/s², while Particle B has an initial acceleration of 10 m/s².

To calculate the charge on either particle, we can use Newton's second law of motion and Coulomb's law.

First, let's consider particle A. We know its initial acceleration is 7 m/s² and its mass is 5×10⁻⁷ kg. Applying Newton's second law (F = ma), we can calculate the net force acting on particle A.

F = m × a

F = (5×10⁻⁷ kg) × (7 m/s²)

F = 3.5×10⁻⁶ N

Next, we can apply Coulomb's law to determine the force between the two particles.

Coulomb's law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

F = k × (q₁ × q₂) / r²

Since the particles have equal charges, we can denote the charge on each particle as q.

F = k × q² / r²

Combining both equations, we have:

3.5×10⁻⁶ N = k × q² / (3.4×10⁻³ m)²

Now, we need the value of the Coulomb constant, k, which is approximately 8.99×10⁹ Nm²/C².

Simplifying the equation, we have:

3.5×10⁻⁶ N = (8.99×10⁹ Nm²/C²) × q² / (3.4×10⁻³ m)²

Solving for q², we get:

q² = (3.5×10⁻⁶ N) × (3.4×10⁻³ m)² / (8.99×10⁹ Nm²/C²)

Calculating the right side of the equation gives us:

q² ≈ 4.10×10⁻¹³ C²

Taking the square root of both sides, we find:

q ≈ ±6.41×10⁻⁷ C

Therefore, the charge on each particle is approximately ±6.41×10⁻⁷ C. The sign indicates the type of charge, with the positive sign representing a positive charge and the negative sign representing a negative charge.

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The 300 gram billiard ball of 30 mm radius is struck by a cue stick that exerts an average force of 600 N horizontally over a 0.005 s interval. Immediately after being hit, the billiard ball rolls without slipping. Determine (a) the height h for the cue stick, and (b) the velocity of the ball after the impact.

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If the 300 gram billiard ball of 30 mm radius is struck by a cue stick that exerts an average force of 600 N horizontally over a 0.005 s interval. Immediately after being hit, the billiard ball rolls without slipping. Then the height and velocity is 2.85 m & 7.5 m/s.

Given data:The mass of the billiard ball, m = 300 g = 0.3 kgRadius of the billiard ball, r = 30 mm = 0.03 mAverage force exerted by the cue stick, F = 600 N

Duration of the collision, t = 0.005 s Let's determine the height of the cue stick using the principle of conservation of energy.According to the principle of conservation of energy, the initial energy of the ball and the cue stick system should be equal to the final energy of the system.

Energy of the system before collision = Potential energy = mghEnergy of the system after the collision = Kinetic energy = (1/2)mv²

Now, equating both the energies, we get:mgh = (1/2)mv²... (1)

where h is the height of the cue stick and v is the velocity of the ball after the impact.Let's determine the velocity of the ball using the principle of impulse and momentum.

According to the principle of impulse and momentum, the impulse experienced by the ball is equal to the change in momentum of the ball.Impulse = F × t Change in momentum = mv - 0... (2

)Here, v is the velocity of the ball after the impact.Now, equating both the equations (1) and (2), we get:

mgh = (1/2)mv²⇒ v² = 2gh... (3)And,F × t = mv... (4)

Squaring both sides of equation (4), we get:(Ft)² = m²v² ⇒ v² = (Ft)²/m²... (5)Substituting the value of v² from equation (5) into equation (3), we get:

(Ft)²/m² = 2gh⇒ h = (Ft)²/2mg... (6)Substituting the given values into equation (6), we get:h = [(600 N × 0.005 s)²/(2 × 0.3 kg × 9.8 m/s²)] = 2.85 m

Therefore, the height of the cue stick is 2.85 m.Now, substituting the value of h into equation (3), we get:v² = 2gh⇒ v² = 2 × 9.8 m/s² × 2.85 m = 56.28 m²/s²⇒ v = √56.28 = 7.5 m/s Therefore, the velocity of the ball after the impact is 7.5 m/s.

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How much does it cost to operate a 800 watts rice cooker for 30
minutes daily for 30 days at a rate 40 cents/kWh ?
4.90
$5.0
$4.8
$6
When an electron is transferred to a ne

Answers

The cost to operate a 800-watt rice cooker for 30 minutes daily for 30 days at a rate of 40 cents/kWh is $4.80.

To calculate the cost, we first need to determine the energy consumption of the rice cooker. The power consumption of 800 watts for 30 minutes daily for 30 days can be calculated as follows:

Energy consumption = Power × Time

= (800 watts) × (0.5 hours/day) × (30 days)

= 12,000 watt-hours or 12 kWh

Next, we can calculate the cost by multiplying the energy consumption by the rate:

Cost = Energy consumption × Rate

= 12 kWh × $0.40/kWh

= $4.80

Therefore, the cost to operate the rice cooker for the given duration and rate is $4.80.

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find the frequency in terahertz of visible light with a wavelength of 441 nm in vacuum.

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The frequency of visible light with a wavelength of 441 nm in a vacuum is approximately 6.80 × 10^2 terahertz (THz).

To find the frequency of visible light with a wavelength of 441 nm, we can use the equation:

c = λ * ν

where c is the speed of light in a vacuum, λ is the wavelength, and ν is the frequency.

The speed of light in a vacuum is approximately 3.00 × 10^8 meters per second (m/s).

Converting the wavelength from nanometers (nm) to meters (m):

λ = 441 nm = 441 × 10^-9 m

Now we can rearrange the equation and solve for the frequency:

ν = c / λ = (3.00 × 10^8 m/s) / (441 × 10^-9 m)

Calculating the value, we find:

ν ≈ 6.80 × 10^14 Hz

To convert this frequency to terahertz (THz), we divide by 10^12:

ν ≈ 6.80 × 10^2 THz

Therefore, the frequency of visible light with a wavelength of 441 nm in a vacuum is approximately 6.80 × 10^2 terahertz (THz).

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croquet ball a moving at 8.7 m/s makes a head on collision with ball b of equal mass and initially at rest. immediately after the collision ball b moves forward at 5.7 m/s .
What fraction of the initial kinetic energy is lost in the collision?

Answers

When a croquet ball A moving at 8.7 m/s makes a head-on collision with ball B of equal mass and initially at rest, immediately after the collision, ball B moves forward at 5.7 m/s. The fraction of the initial kinetic energy lost in the collision is 0.47 (approx).

The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed or transferred from one form to another. The kinetic energy of the system before the collision equals the sum of the kinetic energy of ball A and the kinetic energy of ball B.

Therefore,K.E. before collision = 1/2 m (v1)²K.E. after collision = 1/2 m (v2)²where m is the mass of the balls, v1 is the velocity of ball A before collision, and v2 is the velocity of ball B after collision.

The fraction of initial kinetic energy lost in the collision is given by1 - (K.E. after collision/K.E. before collision)

The final velocities of ball A and ball B can be found using conservation of momentum, which states that the total momentum of an isolated system is constant before and after a collision.

m (v1) = m (v1)′ + m (v2)′The velocity of ball A after collision (v1)' is given byv1' = (m (v1) - m (v2)′) / m = v1 - v2′

Similarly, the velocity of ball B after collision (v2)' is given byv2′ = (m (v2) + m (v1) - m (v2)) / m = v1

The kinetic energy after the collision isK.E. after collision = 1/2 m (v1 - v2′)² = 1/2 m (v1 - (v1 - v2))² = 1/2 m (v2)²

The kinetic energy before the collision isK.E. before collision = 1/2 m (v1)²

Substituting the values of the velocities in the formulae for the kinetic energy before and after the collision,K.E.

before collision = 1/2 m (8.7 m/s)² = 38.2075 JandK.E. after collision = 1/2 m (5.7 m/s)² = 16.245 J

The fraction of the initial kinetic energy lost in the collision is1 - (K.E. after collision/K.E. before collision) = 1 - (16.245/38.2075) = 0.574or approximately 0.47.

Therefore, the fraction of initial kinetic energy lost in the collision is 0.47 (approx).

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A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80
min. Silver contains free electrons per cubic meter. (a) What is the
current in the wire? (b) What is the magnitude of the drift velocity of the
electrons in the wire?

Answers

The current in the wire is 87.5 mA and the magnitude of the drift velocity of the electrons in the wire is 13 mm/s.

(a) The current in the wire. The current, I is the amount of charge that passes through a surface per unit time. Mathematically, it can be expressed as;I = Q/t

Where Q is the charge and t is the time taken.

The charge transferred is 420 C and the time taken is 80 min (1 h 20 min or 4800 s).Therefore,I = 420 C / 4800 s = 0.0875 A = 87.5 mA

(b) The magnitude of the drift velocity of the electrons in the wire.The drift velocity of electrons in a conductor is defined as the average velocity attained by electrons as they move through the conductor.

Mathematically, it can be expressed as;

vd = I / (neA)Where vd is the drift velocity, I is the current, n is the number of free electrons per unit volume, e is the electronic charge and A is the cross-sectional area of the wire.The cross-sectional area of the wire is given by;A = πr2 = π(2.6/2 × 10-3)2 = 5.309 × 10-6 m2.

The number of free electrons per unit volume is given by; n = 5.86 × 1028 m-3.

Substituting the values into the equation for drift velocity gives

;vd = I / (neA)vd = 0.0875 / (5.86 × 1028 × 1.6 × 10-19 × 5.309 × 10-6)vd = 0.013 m/s = 13 mm/s.

Therefore, the current in the wire is 87.5 mA and the magnitude of the drift velocity of the electrons in the wire is 13 mm/s.

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An electrically conductive rod is 7.5 mm long and 10 mm in
diameter. It has a resistance of 87 Ω measured from one end to the
other.
A) Find the current density in the rod if a potential difference
o

Answers

The current density in the rod, with a potential difference of 25 V, is approximately 961,000 A/m².

The current density in the rod can be found using Ohm's Law, which states that the current flowing through a conductor is directly proportional to the potential difference applied across it and inversely proportional to its resistance.

The formula for current density (J) is given by:

J = I / A

where J is the current density, I is the current flowing through the conductor, and A is the cross-sectional area of the conductor.

First, let's calculate the cross-sectional area of the rod. The rod is cylindrical in shape, so we can use the formula for the area of a circle:

A = π * r^2

where A is the cross-sectional area and r is the radius of the rod.

Given that the diameter of the rod is 10 mm, the radius (r) can be calculated as half of the diameter:

r = 10 mm / 2 = 5 mm = 5 * 10^(-3) m

Substituting the values into the formula, we have:

A = π * (5 * 10^(-3))^2 = π * 25 * 10^(-6) m^2

Now, we need to calculate the current flowing through the rod. We can use Ohm's Law:

V = I * R

where V is the potential difference, I is the current, and R is the resistance.

Given that the potential difference (V) is 25 V and the resistance (R) is 87 Ω, we can rearrange the formula to solve for I:

I = V / R = 25 V / 87 Ω

Now, we have the current (I) and the cross-sectional area (A), so we can calculate the current density (J):

J = I / A = (25 V / 87 Ω) / (π * 25 * 10^(-6) m^2)

Simplifying the expression:

J = (25 V / 87 Ω) * (1 / (π * 25 * 10^(-6) m^2))

J ≈ 9.61 × 10^5 A/m^2

Therefore, the current density in the rod, when a potential difference of 25 V is applied across its length, is approximately 9.61 × 10^5 A/m^2.

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Complete question:

A) Find the current density in the rod if a potential difference of 25 V is applied across its length.

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