Three identical conducting spheres are charged as follows. Sphere A is positively charged, sphere B is negatively charged with a different magnitude of net charge than that of sphere A, and sphere C is uncharged. Spheres A and B are momentarily touched together and separated, then spheres B and C are briefly touched together and separated. After that series of processes is completed, which of the following interactions, if any, can be used as evidence to determine whether sphere A or sphere B had the initially larger magnitude of charge? A Sphere C is repelled from sphere A. B Sphere C is repelled from sphere B. Sphere A is repelled from sphere B. D It cannot be determined from observing whether the spheres repel, because they all have the same sign of charge.

Answers

Answer 1

The answer is C.  Sphere A is repelled from sphere B

Step by step explanation:

The question is asking which of the interactions between sphere A, B, and C can be used as evidence to determine which one had the initially larger magnitude of charge. This is because if sphere A has a larger magnitude of charge than sphere B, then when spheres A and B are touched and separated, the charge of sphere A would be transferred to sphere B, causing a conduction of charge.

This means that after the processes are completed, the charge of sphere A and B will have reversed - meaning that sphere A will now have the same, but opposite sign of charge as sphere B. As a result, when sphere A and B are close to each other, their charges will repel, so Sphere A is repelled from sphere B.

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Related Questions

You see the Moon on the meridian at sunrise. The phase of the Moon is a. waxing gibbous. b. full. c. first quarter. d. third quarter.

Answers

The phase of the Moon on the meridian at sunrise is third quarter. The correct answer is Option D.

When the moon is said to be on the meridian at sunrise, the moon is on the west horizon and it rises on the east horizon. The third quarter is when the Moon is on the meridian at sunrise. During the third quarter, the Moon appears as a half-circle with the right half illuminated. It is often seen in the morning sky because it rises at midnight and is visible through the morning hours. In conclusion, when you see the Moon on the meridian at sunrise, the phase of the Moon is the third quarter.

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true or false an action potential introduced at the neuromuscular junction is propagated along the sarcoplasmic reticulum.

Answers

The given statement an action potential introduced at the neuromuscular junction is propagated along the sarcoplasmic reticulum is false because the action potential is only propagated along the muscle cell membrane and not along the sarcoplasmic reticulum.

An action potential introduced at the neuromuscular junction (NMJ) triggers the release of calcium ions (Ca2+) from the sarcoplasmic reticulum (SR), but the action potential itself is not propagated along the SR. The SR is a specialized organelle found in muscle cells that stores and releases calcium ions, which are necessary for muscle contraction. When an action potential reaches the NMJ, it triggers the release of the neurotransmitter acetylcholine, which binds to receptors on the muscle cell membrane and causes an influx of sodium ions (Na+) and an efflux of potassium ions (K+), generating an action potential that spreads across the muscle cell membrane and into the SR. The release of Ca2+ from the SR then initiates the process of muscle contraction.

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4. Once the child in the sample problem reaches the bottom of the hill,
she continues sliding along flat; snow-covered ground until she comes
to a stop. If her acceleration during this time is -0.392 m/s², how long
does it take her to travel from the bottom of the hill to her stopping
point?

Answers

Answer:

8.04 seconds

Explanation:

Assuming that the child starts from rest at the bottom of the hill and travels until she comes to a stop, we can use the following kinematic equation:

v_f^2 = v_i^2 + 2ad

where v_f is the final velocity (which is zero since the child comes to a stop), v_i is the initial velocity (which is the velocity at the bottom of the hill), a is the acceleration (-0.392 m/s²), and d is the distance traveled.

We can solve for d:

d = (v_f^2 - v_i^2) / (2a)

= (0 - v_i^2) / (2-0.392)

= v_i^2 / 0.784

Since the child is sliding along flat snow-covered ground, there is no change in elevation, so we can use the distance traveled from the bottom of the hill to the stopping point as the distance d.

To find the time it takes for the child to travel this distance, we can use the following kinematic equation:

d = v_it + 0.5a*t^2

where t is the time and all other variables are as previously defined.

Substituting the expression for d obtained above, we get:

v_i^2 / 0.784 = v_it + 0.5(-0.392)*t^2

Solving for t, we get:

t = (2 * v_i) / 0.392

We still need to find the value of v_i, the initial velocity of the child at the bottom of the hill. To do so, we can use conservation of energy. The child starts at rest at the top of the hill, so all the initial energy is potential energy. At the bottom of the hill, all the potential energy has been converted to kinetic energy. Assuming no energy is lost to friction, we can equate these two energies:

mgh = 0.5mv_i^2

where m is the mass of the child, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the hill.

Solving for v_i, we get:

v_i = √(2gh)

Substituting this expression for v_i into the expression for t obtained earlier, we get:

t = (2 * √(2gh)) / 0.392

Plugging in the values of g, h, and a, we get:

t = (2 * √(29.820)) / 0.392 = 8.04 seconds

A loop of a wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius r= 0.24 m. The normal to the plane of the loop is parallel to a constant magnetic field of magnitude 0.77 T. What is the magnitude of the change in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through quarter of a revolution? x 27867 Wb X .27867 Wb B (into paper)

Answers

The magnitude of the change in the magnetic flux that passes through the loop when the semicircle is rotated through a quarter of a revolution is 0.27867 Wb.

The induced electromotive force (emf) in a coil is proportional to the rate of change of magnetic flux through it. This is also known as Faraday's law of electromagnetic induction.

The formula for magnetic flux is given by

Φ = BANcosθ

Where,

Φ = magnetic fluxB = magnetic field strengthA = area of the loopN = number of turnscosθ = angle between the magnetic field and the normal to the plane of the loop

If the angle between the magnetic field and the normal plane of the loop is , the maximum magnetic flux is achieved. If the angle is 90°, the flux is zero.

The area of the loop is given by

A = πr²

Therefore, the magnetic flux through the semicircular part of the loop is

Φ = (0.77)(πr²)cos0°

= (0.77)(π × 0.24²)

= 0.13636 Wb

When the semicircle is rotated through a quarter of a revolution, the angle changes from 0° to 90°. Therefore, the magnetic flux becomes zero. Hence, the change in the magnetic flux is given by

0 - 0.13636 = -0.13636

Wb = -136.36 m

Wb = -0.13636 × 10⁻³

Wb= -0.13636 mV

Therefore, the magnitude of the change in the magnetic flux that passes through the loop when the semicircle is rotated through a quarter of a revolution is 0.27867 Wb.

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17. a particle moves in simple harmonic motion with a frequency of 3.00 hz and an amplitude of 5.00 cm. (a) through what total distance does the particle move during one cycle of its motion? (b) what is its maxi- mum speed? where does this maximum speed occur? (c) find the maximum acceleration of the particle. where in the

Answers

A) Through one cycle of its motion, the particle will move a total distance of 10.00 cm (2π*amplitude).

B) The maximum speed of the particle will occur at the equilibrium point (amplitude/2). This speed can be calculated by multiplying the frequency and the amplitude is 94.25 cm/s.

C) The maximum acceleration of the particle will be [tex]1732 \frac{cm}{s^2}[/tex] .The maximum acceleration will occur at the extremes of the particle's motion (amplitude).

Given:

A=5.00 cm, f=3.00 Hz

(A) The distance travelled by the particle is equivalent to double the amplitude: 2 × 5.00 cm = 10.00 cm.

(B) The formula for the frequency of a particle in simple harmonic motion is:

[tex]f=\frac{v}{\lambda}[/tex] where v = velocity and λ = wavelength.

To find the maximum speed of the particle, we'll use the following formula:

[tex]v=A\sqrt{\omega^2-t^2}[/tex]

The maximum velocity occurs at the equilibrium point (i.e. at t = 0).

ω = 2πf = 2π(3.00 Hz) = 18.85 rad/s

v = Aω = 5.00 cm × 18.85 rad/s = 94.25 cm/s

Thus, the maximum velocity of the particle is 94.25 cm/s, and it occurs at the equilibrium point.

(C) The acceleration formula is: a = −Aω²sin(ωt).

We can obtain the maximum acceleration by putting t = 0.

a = Aω² = (5.00 cm)(18.85 rad/s)² = 1732 cm/s².

The maximum acceleration of the particle is 1732 cm/s², and it occurs at the ends of the motion.

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Five docks are being tested in a laboratory. Exactly at noon, as determined by the WWV

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Exactly at noon, as determined by the WWV time signal, on successive days of a week the clocks according to their relative value as good timekeepers, best to worst.

Time signals are also used in many everyday applications, such as GPS navigation, where precise timing is essential for calculating positions accurately.  A time signal refers to any signal that provides information about the passage of time. Time signals are often used in experiments to measure the duration of events or to synchronize the timing of multiple processes.

One common type of time signal is a periodic signal, which repeats itself at regular intervals. This can be used to measure the period or frequency of a phenomenon, such as the oscillation of a pendulum or the vibration of a guitar string. Another type of time signal is a pulse signal, which provides a brief burst of energy at a specific time. This can be used to trigger the start or stop of a process or to measure the time delay between different events.

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A column is fabricated by connecting the rolled-steel members shown by bolts of 3/4-in. diameter spaced longitudinally every 5 in. Determine the average shearing stress in the bolts caused by a shearing force of 30 kips parallel to the y axis.

Answers

The average shearing stress in the bolts caused by the given force is approximately 157 psi.

To determine the average shearing stress in the bolts, we can use the formula,

τ = F/A

where τ is the shearing stress, F is the force applied, and A is the cross-sectional area of the bolts.

First, we need to determine the cross-sectional area of one bolt. The area of a circle with a diameter of 3/4 inch is,

A = π/4 × (3/4 inch)^2 = 0.4418 square inches

Next, we need to determine the total number of bolts in the column. Since the bolts are spaced longitudinally every 5 inches, we can divide the length of the column (in the y direction) by 5 inches to find the number of bolt locations,

Number of bolt locations = (10 feet)/(5 inches/12 inches/foot) = 480

Since each bolt location has one bolt, the total number of bolts is 480.

Finally, we can calculate the shearing stress in one bolt using the formula above,

τ = F/A = 30,000 pounds / (480 bolts × 0.4418 square inches/bolt) ≈ 157 psi

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an incompressible substance with a density of 1000 kg/m3 is isothermally compressed from 100 to 1000 kpa. determine the change in enthalpy. multiple choice question. 0 kj/kg 0.9 kj/kg 10 kj/kg 900 kj/kg

Answers

The change in enthalpy of an incompressible substance with a density of 1000 kg/m³ that is isothermally compressed from 100 to 1000 kPa is 0 kJ/kg.

What is enthalpy?

Enthalpy is a measure of the total energy of a thermodynamic system. In addition, it incorporates the energy that is supplied to the system as heat, as well as any energy that is used as work. Enthalpy is represented by the symbol H and is usually calculated in units of joules (J).

What is an incompressible substance?

An incompressible substance is one that cannot be compressed or compressed to a significant degree. Liquids are examples of such materials. They are often described as having a constant density because, unlike gases, they do not easily change in volume in response to pressure or temperature changes. Therefore, the change in enthalpy is 0 kJ/kg.

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the orbital period of saturn is 29.46 years. determine the distance from the sun to the planet in km

Answers

The average distance from the Sun to Saturn is approximately 1,427,000,000 km. To calculate this, we can use the Third Kepler's Law of Planetary Motion, which states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of the orbit.

We can use Kepler's Third Law to relate the orbital period of a planet to its distance from the sun:

T^2 = (4π^2 / GM) * r^3

where T is the orbital period in years, G is the gravitational constant, M is the mass of the sun, and r is the average distance from the sun to the planet in astronomical units (AU).
Therefore, we can use the formula:

d^3 = (T^2 * 4π^2)/G*M

Where d is the distance, T is the orbital period, G is the gravitational constant, and M is the mass of the Sun.


Plugging in the values:

d^3 = (29.46^2 * 16π^2)/(6.67408 * 1.989 * 10^30)
d = 1,427,000,000 km

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a ball is dropped from rest from a tower and strikes the ground 122.5 m below. a) approximately how many seconds does it take the ball to strike the ground after being dropped? b) with what velocity does it strike the ground? neglect air resistance.

Answers

The approximate time taken by the ball to hit the ground after being dropped is: 5 seconds.

The velocity at which the ball hits the ground is approximately 49.05 m/s, and it moves in the downward direction (negative velocity).

A ball is dropped from rest from a tower and strikes the ground 122.5 m below.

We are asked to determine the time taken by the ball to hit the ground, and the velocity at which it hits the ground.

The formula to calculate the time taken by an object to fall from rest from a height h is given by: t = sqrt (2h/g)

Here, h = 122.5m; g = 9.81m/s² (acceleration due to gravity)

Using the given formula, t = sqrt (2h/g) = sqrt (2 × 122.5 / 9.81)≈ 5 seconds

We know that, `v = g.t`

Since the ball was dropped from rest, its initial velocity is 0.

So the final velocity `v` is equal to the velocity at which it hits the ground.

Since g is negative, the velocity `v` will be negative, which means it is moving in the downward direction.

Using `g = 9.81 m/s²`,`t = 5 seconds`, we have = g.t = 9.81 × 5 = 49.05 m/s

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a disturbance that transfers energy from place to place

Answers

A disturbance that transfers energy from place to place is called wave.

Waves can take many forms, including sound waves, light waves, water waves, seismic waves, and electromagnetic waves. Regardless of their type, all waves share certain characteristics, such as wavelength, frequency, amplitude, and speed. When a wave travels through a medium, it causes the particles in the medium to vibrate, but it does not transport the particles themselves. This means that waves can transfer energy over long distances without the transfer of matter. Waves are fundamental to many fields of science and technology, from communications and entertainment to medicine and engineering.

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--The complete question is, A disturbance that transfers energy from place to place is called _____.--

The volume of a sphere is increasing at the rate of 8cm3/s. Find the rate at which its surface area is increasing when the radius of the sphere is 12cm.

Answers

The rate at which the surface area of the sphere is increasing when the radius of the sphere is 12 cm is 1/226.5 cm/s.

What is the rate of surface area change?

The volume of a sphere is increasing at the rate of 8 cm³/s.

Radius of the sphere is 12 cm.

So, we need to find the rate at which its surface area is increasing.

Let, V be the volume of the sphere and r be the radius of the sphere. The volume of a sphere of radius r is given by:

V = (4/3)πr³

Differentiating with respect to time t, we get:

dV/dt = 4πr²(dr/dt) ...(1)

Also, the surface area of the sphere is given by:

A = 4πr²

Differentiating with respect to time t, we get:

dA/dt = 8πr(dr/dt) ...(2)

From equations (1) and (2), we can write:

dr/dt = dV/dt ÷ 4πr²

dr/dt = 8 / (4π × 12²)

dr/dt = 8 / 1808

dr/dt = 1 / 226.5 cm/s

Therefore, the rate at which the surface area of the sphere is increasing when the radius of the sphere is 12 cm is 1/226.5 cm/s.

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A block of massmis placed in a smooth-bored spring gun at the bottom of the inclineso that it compresses the spring by an amountx_c. The spring has spring constantk. The incline makes an anglethetawith the horizontal and the coefficient of kineticfriction between the block and the incline ismu. The block is released, exits the muzzle of the gun, andslides up an incline a total distancethe distance traveled along the incline by the block after it exitsthe gun. Ignore friction when the block is inside the gun. Also,assume that the uncompressed spring is just at the top of the gun(i.e., the block moves a distancex_cwhile inside of the gun). Usegfor the magnitude of acceleration due to gravity.

Answers

The block moves up the incline with a constant velocity, v² = 2gx.sin θ - 2μgd. The block will move up the incline as long as the numerator in the above equation is positive.

A block of mass m is placed in a smooth-bored spring gun at the bottom of the incline so that it compresses the spring by an amount x_c. The spring has spring constant k. The incline makes an angle theta with the horizontal and the coefficient of kinetic friction between the block and the incline is mu.

The block is released, exits the muzzle of the gun, and slides up an incline a total distance the distance traveled along the incline by the block after it exits the gun. Ignore friction when the block is inside the gun.

Also, assume that the uncompressed spring is just at the top of the gun (i.e., the block moves a distance x_c while inside of the gun). Use g for the magnitude of acceleration due to gravity. Determine the distance traveled along the incline by the block after it exits the gun.Given, Mass of the block = m Initial compression of the spring = xc, spring constant = k, Angle between incline and horizontal = θ, Coefficient of kinetic friction = μ, Distance traveled along the incline by the block = d.

Let us begin with the given problem,

the work done on the spring is

K = 1/2 k x_c²

As the spring is compressed, the potential energy of the spring increase. Thus, the work done on the block by the spring is -K.

This work is equal to the increase in kinetic energy of the block.

This kinetic energy is converted into potential energy as the block moves up the incline. Thus, work done by the block against the gravitational force is mgh where, h is the height the block reaches above its initial position. The work done against the friction is mgh.f where, f is the coefficient of friction between the block and the incline.

Then, K + mgh.f = 1/2mv²

where v is the velocity of the block after it exits the gun.

Determine the final velocity of the block,

v²= 2(k/m) x_c² - 2gh(f + sin θ).

The block moves up the incline with a constant velocity,

v² = 2gx.sin θ - 2μgd.

The above equation is obtained using the work-energy principle.

Then,

2gx.sin θ - 2μgd = 2(k/m) x_c² - 2gh(f + sin θ)

Here, solving for d, we get,

d = (1/2g) [x_c² (k/m) - μx_c² sin θ] / (μ + sin θ).

The distance traveled along the incline by the block after it exits the gun is

(1/2g) [x_c² (k/m) - μx_c² sin θ] / (μ + sin θ).

Thus, this is the required solution. The block will move up the incline as long as the numerator in the above equation is positive.

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A fire fighter is trying to shoot water straight to the window located at the second floor of a house 6 m above the ground: The distance between the fire fighter and the house is 8 m and he holds the fire hose 1.8 m above the ground: The water leaves the hose with a constant speed of 12.5 m/s. Initially, the fire fighter aims the hose at 53 above the horizontal and misses the window: (we can assume that the hose and the window are in the same vertical plane) How much time it will take for the water flow to reach the house? How far above the window does the water go? What is the magnitude of the velocity of water when it strikes the house? What must be the minimum angle and speed of the flow in order to get water right into the window?

Answers

The water flow takes 1.06 s to reach the house.

Water strikes 0.87 m above the window when the firefighter holds the hose at a [tex]53^o[/tex] angle from horizontal.

The magnitude of velocity when water hits the house is 7.52 m/s.

The minimum speed for water to enter into the window is[tex]v_0 = d / (v_0cos(\theta)) \times\sqrt{(8.4-2dtan\theta)/g}[/tex].

The question can be solved by applying the concept of projectile motion. When an object is projected into the air, it follows a curved path under the influence of gravity. The path followed by a projectile is called a parabolic path.

To solve this problem, we can break it down into a few parts.

First, let's find the time it takes for the water to reach the house:

We can use the horizontal distance between the firefighter and the house, which is 8 m, and the initial horizontal velocity of the water, which can be found using the initial speed and launch angle:

[tex]v_x = v_0 cos(53^\circ)[/tex]

[tex]v_x = 12.5 \ m/s \times cos(53^\circ)[/tex]

[tex]v_x = 7.5 \ m/s[/tex]

The time it takes for the water to travel the horizontal distance of 8 m can be found using the formula:

[tex]time = distance/velocity[/tex]

[tex]time = 8 \ m / 7.5 \ m/s[/tex]

[tex]time = 1.06\ s[/tex]

So it takes 1.06 seconds for the water to reach the house.

Next, let's find the height above the window that the water reaches:

We can use the vertical distance between the firefighter and the window, which is (6 - 1.8) m, and the initial vertical velocity of the water, which can be found using the initial speed and launch angle:

[tex]v_y = v_0 sin(53^\circ)[/tex]

[tex]v_y = 12.5 \ m/s \times sin(53^\circ)[/tex]

[tex]v_y = 9.98 \ m/s[/tex]

The time it takes for the water to reach the house is 1.06 s, so we can use this time and the initial vertical velocity to find the height above the window that the water reaches:

[tex]y = v_yt - 0.5gt^2[/tex]

[tex]y = 9.98 \ m/s \times 1.06 s - 0.5 \times 9.8 \ m/s^2 \times (1.06 \ s)^2[/tex]

[tex]y = 5.07\ m[/tex]

Since the firefighter is holding the fire hose 1.8 m above the ground, the total height reached by the water is

h = 1.8 + 5.07 = 6.87 m

Height above the window = 6.87 - 6 = 0.87 m

So the water reaches a height of 0.87 m above the window.

Next, let's find the magnitude of the velocity of the water when it strikes the house:

Vertical velocity of water when it stricks the house at t = 1.06 s.

[tex]v_{y(final)} = 9.98 - 9.81\times 1.06[/tex]

[tex]v_{y(final)} = 0.588 \ m/s[/tex] (downwards)

We can use the horizontal and vertical components of the velocity to find the total velocity using the Pythagorean theorem:

[tex]v = \sqrt{vx^2 + vy^2}[/tex]

[tex]v = \sqrt{(7.5\ m/s)^2 + (0.588\ m/s)^2}[/tex]

v = 7.52 m/s

So the magnitude of the velocity of the water when it strikes the house is 7.52 m/s.

Finally, let's find the minimum angle and speed of the flow in order to get water right into the window:

For the water to reach the window, its vertical displacement must be equal to the vertical distance between the firefighter and the window, which is 4.2 m. We can use this information to find the launch angle and speed using the equations of motion:

[tex]y = v_0 sin(\theta) t - 0.5 g t^2[/tex]

[tex]4.2 m = v_0 sin(\theta) t - 0.5 g t^2[/tex] ....(1)

[tex]v_x = v_0 cos(\theta)[/tex]

[tex]t = d / v_x[/tex]

[tex]t = {d}/{v_0 cos\theta}[/tex] .....(2)

Substituting the second equation into the first equation and solving for [tex]v_0[/tex] and θ, we get:

[tex]v_0 = d / (v_0cos(\theta) t)[/tex]

[tex]4.2 m = (\frac{d}{(cos(\theta) t)}) \times sin(\theta) t - 0.5 g t^2[/tex]

Solving for t and substituting into the equation for [tex]v_0[/tex], we get:

[tex]t = \sqrt{(8.4-2dtan\theta)/g}[/tex]

[tex]v_0 = d / (v_0cos(\theta) \times\sqrt{(8.4-2dtan\theta)/g)}[/tex]

Substituting the values given in the problem (d = 8 m, [tex]g = 9.8 m/s^2[/tex]), we can solve for θ and [tex]v_0[/tex]

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A student drops a rock in a pond and notices that her reflection in the water becomes less clear. Which statement best explains the student’s observation?(1 point)


The surface of the water is not shiny after the rock is dropped in the pond.

The surface of the water is no longer hard after the rock is dropped in the pond.

The water is not able to reflect waves after the rock is dropped in the pond.

The water is not a smooth surface after the rock is dropped in the pond.

Answers

Answer:

The best statement that explains the student's observation is: "The water is not a smooth surface after the rock is dropped in the pond." When the rock is dropped in the pond, it creates ripples and waves that disturb the smooth surface of the water. As a result, the reflection becomes less clear because the disturbed surface scatters the light and creates a distorted image. This is a common phenomenon observed when a disturbance is created on the surface of water, like when you throw a stone or object into it.

In the figure, a
4.4 kg
block is accelerated from rest by a compressed spring of spring constant
640 N/m
. The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction
μ k

=0.296
. The frictional force stops the block in distance
D=7.7 m
. What are (a) the increase in the thermal energy of the block-floor system, (b) the maximum kinetic energy of the block, and (c) the original compression distance of the spring? (a) Number Units (b) Number Units In the figure, a
4.4 kg
block is accelerated from rest by a compressed spring of spring constant
640 N/m
. The block leives the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction
μ 2

=0.296
. The frictional force stops the block in distance
D=7.7 m
. What are (a) the increase in the thermal energy of the block-floor system, (b) the maximum kineticenergy of the block, and (c) the original compression distance of the spring? (a) Number Units (b) Number Units

Answers

The (a) increase in the thermal energy of the block-floor system 139.3 J

(b), the maximum kinetic energy of the block 614.3 J

(c), and the original compression distance of the spring 0.625 m

(a) The increase in thermal energy of the block-floor system is equal to the work done by the friction force. This can be calculated using the equation

Work = Force × Distance,

where the friction force is equal to the coefficient of kinetic friction multiplied by the normal force, and the distance is equal to the stopping distance (7.7 m).

Therefore, the increase in thermal energy of the block-floor system is equal to

(0.296 x 4.4 kg x 9.8 m/s² x 7.7 m) = 139.3 J.

(b) The maximum kinetic energy of the block is equal to the kinetic energy of the block when it leaves the spring. This can be calculated using the equation

Kinetic Energy = ½ mv²,

where m is the mass of the block (4.4 kg) and v is the velocity of the block when it leaves the spring. This velocity can be found by using the equation

Force = Mass x Acceleration with the spring constant (640 N/m) and the mass of the block (4.4 kg).

Therefore, the maximum kinetic energy of the block is equal to

(0.5 x 4.4 kg x (640 N/m / 4.4 kg)²) = 614.3 J.

(c) The original compression distance of the spring can be found by using the equation

K.E (spring) 1/2 Kx² + Work done = 0

-1/2 * 640 N/m * x² + 99.93 J = 0

Solving for x, we get:

x = √(99.93 J / (1/2 * 640 N/m))

x = 0.625 m

Therefore, the original compression distance of the spring is 0.625 m.

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The colors on an oil slick are caused by reflection and (explain why)
a. Diffraction
b. Interference
c. Refraction
d. Polarization
e. Ionization

Answers

"The colours on an oil slick are caused by reflection and interference." Correct option is B.

Different bands of the oil slick create different colours as the oil film progressively thins from the centre to the edges.

Interference is what gives an oil slick drifting on water or a soap bubble in the sun their vibrant colours. The colours that interact most positively are the ones that are most vibrant. Thin film interference is the name given to the phenomenon because it occurs when light reflected from various thin film surfaces interferes with one another.

The most crucial interfering principle is the superposition principle.

This hair colour procedure primarily uses jewel tones and rainbow colours, including burgundy, royal blue, deep purple, green, and deep red. Alternating the colours that give your hair an oil spill appearance is the best method to make your skin tone and hair look good together. Best choice is B.

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in one method of measuring backlash, a bar is attached to the driven shaft and a dial indicator measures its movement. this method must be adjusted to account for the ___ of the bar.

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In one method of measuring backlash, a bar is attached to the driven shaft and a dial indicator measures its movement. this method must be adjusted to account for the flexion or bending of the bar.

It is because When using the bar method to measure backlash, a bar is attached to the driven shaft and a dial indicator measures its movement.

However, the bar may flex or bend due to its own weight or external forces, leading to inaccurate measurements of the backlash.

Therefore, to obtain accurate results, the method must be adjusted to account for the flexure or bending of the bar. This can be done by placing the bar in a support at some distance from the indicator or by using a more rigid and less flexible bar.

It is important to account for the flexure of the bar to ensure accurate measurements and proper functioning of the system.

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Learning Goal: To be able to calculate the tension in a string and the acceleration of each of two blocks in a two-pulley system. As shown, a block with mass mi is attached to a massless ideal string. The string wraps around a massless pulley and then wraps around a second massless pulley that is attached to a block with mass m2 and ultimately attaches to a wall. The whole system is frictionless.Part A - Tension in the string Given that a2 is the magnitude of the horizontal acceleration of the block with mass m2, what is T, the tension in the string? Express the tension in terms of m2 and a2. Part B - Acceleration of suspended block Given T, the tension in the string, calculate a1, the magnitude of the vertical acceleration of the block with mass mi. Express the acceleration's magnitude, a1, in terms of mi, g, and T. a1a_1 = ____ ?

Answers

The tension in the string is T = m2*a2.

The magnitude of the vertical acceleration of the block with mass m1, a1, is a1 = (T - m1*g)/m1.
In order to calculate the tension in the string, T, and the acceleration of the block with mass m1, a1, we must use Newton's second law of motion.

Part A - Tension in the string:

Since, the acceleration of the block with mass m2 is known, we can use the equation,

T = m2*a2 to calculate the tension in the string, T.

Tension= m2*a2

Part B - Acceleration of suspended block:

We can use the equation,

T = m1*a1 + m1*g to calculate the magnitude of the vertical acceleration of the block with mass m1, a1.

Rearranging this equation to solve for a1 gives us

a1 = (T - m1*g)/m1.

vertical acceleration= (T-m1*g)/m1

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A small source of sound waves emits uniformly in all directions. The total power output of the source is P. By what factor must P increase if the sound intensity level at a distance of 20. 0 m from the source is to increase 5. 00 dB?

Answers

The sound intensity must increase by a factor of 10(5.00/10) = 3.16 in order to increase the sound intensity level at a distance of 20.0 m from the source by 5.00 dB.

Because sound waves are uniformly released in all directions, a sphere with a radius of 20.0 m has an even distribution of power over its surface. Since r represents the distance from the source and P is the total power output, we can calculate the sound intensity at that distance as P/(4r2). The power output of the source must increase by the same factor, or by a factor of 3.162 = 10.0, in order to raise the sound intensity by a factor of 3.16. Hence, the total To increase the sound intensity level at a distance of 20.0 m from the source by 5.00 dB, the source's power output must rise by a factor of 10.0.

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Which of these devices are not based on magnetic effect of electric current? (Room heater, Magnetic crane, Electric bell, Loud speaker)

Answers

Room heater. The electric current's magnetic action is not the foundation of the space heater. Devices that depend on the magnetic effect of electric current to work include the magnetic crane, , and loudspeaker.

The magnetic crane creates a magnetic field that can lift large things using an electromagnet. The magnetic field produced by the electric bell's usage of an electromagnet forces a metal clapper to strike a bell, emitting a ringing sound. An electromagnet in the loudspeaker causes a diaphragm to vibrate, creating sound waves that can be perceived by the human ear. In contrast, the usual mechanism of a room heater is the use of a resistor or heating element to transform electrical energy into heat energy. After that, the heat is radiated or convected.

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Which lists the layers in a PNP transistor from the least negative to the most negative?

emitter, base, collector
base, emitter, collector
collector, base, emitter
collector, emitter, base

Answers

Answer:  

P-type layer (base)

N-type layer (emitter)

P-type layer (collector)

Explanation:

Consider a thin 30 m rod pivoted at one end. A uniform density spherical object (whose mass is 4 kg and radius is 3.3 m ) is attached to the free end of the rod and the moment of inertia of the rod about an end is I rod ​
= 3
1

mL 2
and the moment of inertia of the sphere about its center of mass is I sphere

= 5
2

mr 2
. What is the angular acceleration of the rod immediately after it is released from its initial position of 39 ∘
from the vertical? The acceleration of gravity g=9.8 m/s 2
. Answer in units of rad/s 2
.

Answers

The angular acceleration of the rod immediately after it is released from its initial position of 39 degrees from the vertical is - 0.022 rad/s²

Angular acceleration of the rod:We will use the law of conservation of energy. When the rod is released from the initial position, the gravitational potential energy will convert to kinetic energy of the rod and the sphere.Let the angular acceleration of the rod be α,The gravitational potential energy of the system when the rod is at the initial position is given by,PE = mgh where, m = mass of the sphere + mass of the rod = 4 kg,L = length of the rod = 30 mgr = radius of the sphere = 3.3 mθ = angle from the vertical = 39 degrees,h = vertical height = LcosθPE = mgLcosθ= 4 × 9.8 × 30 × cos39°PE = 1058.33 J

Now, when the rod falls, it will rotate about the pivot point. The kinetic energy of the system will be given by,K.E = 1/2 (Irod + Isphere) ω²where, ω = angular velocity of the rod + sphere after falling.The moment of inertia of the system about the pivot point is given by,I = Irod + Isphere. We can use the parallel axis theorem to calculate the moment of inertia of the sphere,I sphere = 2/5 mr² + mr² = 7/5 mr²So,I = Irod + Isphere= 3/12 mL² + 7/5 mr²= 3/12 × 4 × 30² + 7/5 × 4 × 3.3²= 126.48 kg.m²Now,K.E = 1/2 (Irod + Isphere) ω²= 1/2 I ω²

The initial velocity of the rod is 0, so the initial kinetic energy of the system is 0.The final velocity of the rod + sphere can be found using the conservation of energy equation,PE = K.EPE = K.E1/2 mv² = mgh1/2 I ω² = mghω² = 2mgh/Iω = sqrt (2mgh/I)Substitute the given values in the above equation,ω = sqrt (2 × 4 × 9.8 × 30 × cos39° / 126.48)ω = 1.479 rad/s

Now, we can use the torque equation to find the angular acceleration of the rod.The gravitational force acting on the sphere is mg.The torque due to the gravitational force about the pivot point is,τ = mgh sinθ= 4 × 9.8 × 30 × sin39°= 753.84 N.m The torque due to the weight of the rod is,T = Irod α= 3/12 mL² α= 3/12 × 4 × 30² × α= 90α N.m Using Newton's second law of motion,Net torque = Iαα = (mgh sinθ - T) / I= (mgh sinθ - Irod α) / I= (mgh sinθ) / I - (3/12 mL²) α= (4 × 9.8 × 30 × sin39°) / 126.48 - (3/12 × 4 × 30²) α= 1.98 - 90 αα = - 0.022 rad/s² (Negative sign indicates that the angular acceleration is in the opposite direction to the initial angular velocity of the rod)Therefore, the angular acceleration of the rod immediately after it is released from its initial position of 39 degrees from the vertical is - 0.022 rad/s².Answer: - 0.022 rad/s².

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How much force is required to accelerate a 5kg mass at 20m/s 2 ?

Answers

Нам не дано коэффициент трения, значит, можно не учесть силу трения. От этого, по второму закону Ньютона, F=ma=5×20=100 Н.

И это всё!

david walks 3 km north, and then turns east and walks 4 km. what is the distance?

Answers

David travelled a total of 5 kilometres.

To find the distance that David walked, we can use the Pythagorean theorem, which relates the sides of a right triangle. In this case, the two legs of the right triangle represent the distance that David walked north and east, respectively, and the hypotenuse represents the total distance that he walked.

If David walks 3 km north and then turns east and walks 4 km, we can draw a right triangle with legs of length 3 km and 4 km. Applying the Pythagorean theorem, we have:

distance²2 = (3 km)²+ (4 km)²

distance²2 = 9 km²+ 16 km²

distance = √(25) km

distance = 5 km

Therefore, the total distance that David walked is 5 km.

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which of the following describes the function and location of golgi tendon organs? which of the following describes the function and location of golgi tendon organs? monitor muscle length, situated in series with muscle fibers monitor muscle length, situated in parallel with muscle monitor muscle tension, situated in parallel with muscle fibers monitor muscle tension, situated in series with muscle fibers

Answers

The function and location of Golgi tendon organs are to monitor muscle tension, and they are situated in series with muscle fibers.

The Golgi tendon organ (GTO) is a sensory receptor found in the tendons of mammalian skeletal muscle. The GTO is positioned in series with the extrafusal muscle fibers in the tendons of mammalian skeletal muscle. It is situated where the muscle fibers blend with the tendon fibers.

The GTOs inform the central nervous system about muscle tension in the muscle by detecting changes in tension caused by the contraction of the muscle. The Golgi tendon organ consists of collagen bundles that are surrounded by a sheath of connective tissue.

There are some special muscle receptors that can sense the tension within a muscle, and Golgi tendon organs (GTOs) are one of them.

What are the functions of the Golgi tendon organs? The Golgi tendon organs have a number of functions. They play a significant role in the modulation of muscle tone, the prevention of excessive force during muscle contractions, and the fine-tuning of complex and coordinated movements.

In addition, the GTOs also function to prevent overstretching of the muscle and maintain muscle stiffness. These structures are therefore critical in protecting muscles from damage and ensuring their optimal performance.

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when one stationary object is replaced by another stationary object, the change between the two objects maybe perceived as the movement of a single object. this creates?

Answers

When one stationary object is replaced by another stationary object, the change between the two objects maybe perceived as the movement of a single object. This creates an optical illusion.

An optical illusion is defined as a visual phenomenon in which the information gathered by the eye is processed in a way that results in a false perception of reality or the visual impression of seeing something that is not present or incorrectly perceiving it. It is a misinterpretation of a visual stimulus caused by the brain's ability to misjudge sensory information.

It can happen when visual information is processed in the brain, and it can create an impression of movement that isn't there. This phenomenon occurs when an object is moving or when the eyes are moving around, but it can also happen when the object being looked at is stationary.

When one stationary object is replaced by another stationary object, the change between the two objects maybe perceived as the movement of a single object. This creates an optical illusion because the visual system is misled into thinking that the object is moving.

The brain continues to process visual information even when the object is stationary, creating the impression that the object is moving. This is why an optical illusion can be used to make a stationary object appear to move or to make a moving object appear to be stationary.

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a charge is passing through a static magnetic field. the velocity of the charge makes a 30o angle with the field. the force exerted by the field changes the kinetic energy of the charge.

Answers

The magnetic force exerted by the field on the charge is 0.5qvB.

F = qvBsin(θ)

where;

F = qvBsin(30)

F = 0.5qvB

Magnetic force is a fundamental force that arises due to the motion of electric charges. It is the force that acts between two magnetic poles or between a magnetic pole and a moving charged particle. Magnetic force is a vector quantity and is described in terms of its direction, magnitude, and point of application.

The force between two magnetic poles is governed by the inverse square law, which means that the force decreases as the distance between the poles increases. The direction of the magnetic force is perpendicular to the direction of motion of the charged particle and to the direction of the magnetic field in which it moves. The magnitude of the magnetic force is proportional to the charge of the particle, its velocity, and the strength of the magnetic field.

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what is refraction? what is refraction? the bending of waves due to a change in wave amplitude the bending of waves due to a change in wavelength the bending of waves due to a change in wave velocity the bending of waves due to a change in wave phase

Answers

Refraction is the bending of waves due to a change in their speed.

What is refraction? Refraction is a phenomenon in which waves bend due to a change in speed when they travel from one medium to another medium. It usually occurs when the waves pass from one medium to another medium, and the angle at which the waves hit the surface is not perpendicular.

It happens because waves travel at different speeds in different media. When waves pass through the medium, the refracted waves change direction, but their frequency and wavelength remain constant.

The most commonly observed examples of refraction are the bending of light rays in water, the splitting of white light into a rainbow, the mirages on hot days, and the apparent bending of objects partially submerged in water.

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A pendulum swings on a massless string back and forth between points 1 and 5. Neglecting forces due to the air, at which point(s) is the total power delivered by the forces acting on the pendulum equal to zero? (Mark all that apply.) NONE of the points 1-5. What is the angle between each force and the velocity at each point? Is the velocity zero or non-zero at each point?

Answers

The power delivered by the forces acting on the pendulum is equal to zero at points 2 and 4. The angle between each force and the velocity at each point is 90 degrees. The velocity is zero at points 1 and 5, and non-zero at points 2 and 4.

A pendulum is a mass attached to a string or rod that swings back and forth due to the force of gravity. When the pendulum swings, there are forces acting on it, including the force of gravity, tension in the string or rod, and air resistance. To determine the points at which the total power delivered by the forces acting on the pendulum is equal to zero, we need to consider the work done by each force.To do this, we use the equation for work:work = force x distance x cos(theta)where force is the magnitude of the force, distance is the distance traveled by the pendulum, and theta is the angle between the force and the direction of motion of the pendulum.

When the total work done by all forces acting on the pendulum is zero, the power delivered by those forces is also zero.Since the mass of the pendulum is negligible compared to the force of gravity, we can assume that the tension in the string is always perpendicular to the direction of motion of the pendulum. Therefore, the angle between the force of tension and the velocity of the pendulum is 90 degrees. The angle between the force of gravity and the velocity of the pendulum is also 90 degrees at points 1 and 5, where the pendulum comes to a stop and changes direction.The velocity of the pendulum is zero at points 1 and 5 because it comes to a stop and changes direction.

The velocity is non-zero at points 2 and 4 because the pendulum is moving at its maximum speed in these positions. Therefore, the power delivered by the forces acting on the pendulum is equal to zero at points 2 and 4, where the work done by the force of gravity is equal and opposite to the work done by the force of tension.

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