if an object is raised twice as high, its potential energy will be four times as much. half as much twice as much. impossible to determine unless the time is given.

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Answer 1

If an object is raised twice as high, its potential energy will be four times as much.

Potential energy Gravitational potential energy According to the question, if an object is raised twice as high, its potential energy will be four times as much.

The potential energy is the stored energy of an object. It depends on an object’s position or configuration.

Potential energy is classified into three types: elastic potential energy, gravitational potential energy, and electric potential energy.

The gravitational potential energy of an object is the energy stored in an object when it is moved against the gravitational force. It depends on the mass of an object, the acceleration due to gravity, and the height an object is above the ground.

The equation for gravitational potential energy is:

GPE = mgh where GPE is gravitational potential energy in joules (J)m is the mass of the object in kilograms (kg)g is the acceleration due to gravity in meters per second squared (m/s²)h is the height of the object in meters (m).

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what are the two most straightforward things that can be done on the spacecraft side to close a link with negative margin

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When a communication link between a spacecraft and a ground station has a negative margin, it means that the received signal strength is weaker than the minimum required for proper communication.

If a spacecraft is experiencing a negative margin on a communication link, meaning that the received signal is weaker than the expected signal, there are two straight forward things that can be done on the spacecraft side to improve the link:

Increase the transmit power: By increasing the power of the signal being transmitted by the spacecraft, the received signal strength at the other end can be improved, which may close the link margin. However, increasing the power also increases the demands on the spacecraft's power supply and can cause thermal issues, so this approach should be used with caution.Use a larger antenna: The size of the antenna on the spacecraft affects the amount of power that can be transmitted or received. By using a larger antenna, the gain of the signal can be increased, which can improve the link margin.

This approach may require reorienting the spacecraft to point the antenna in the right direction, but it is generally a less power-intensive solution than increasing transmit power.

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Shown here is a plot of a pairwise potential between two interacting particles. The particles are initially at rest at ro 1.130 , and 1.2E energy is added as work. The two interacting particles define a closed system. Your plots should only extend into regions where particles separations are physically possible. a) Plot Etotal and KE as a function of r. Clearly mark rmin, Fmax if they apply. Explain how you determined your plots and their ranges. b) Are the particles described above bound or un-bound? Explain your reasoning.

Answers

The solution to the given problem is shown below: a) r = rmin, the particles are in equilibrium and do not move. b) The particles are bound because they need an external energy of 0.703E to be separated to an infinite distance.

a). Plot Etotal and KE as a function of r.

The potential energy (U) for the given potential function isU(r) = 2.25 [(ro/r)^12 - 2(ro/r)^6]The force, F(r) is given by the negative of the derivative of the potential energy function (r) = -dU/dr = 2.25 (12(ro/r)^13 - 12(ro/r)^7) / rEtotal and KE can be calculated using the following equations:

Etotal = KE + UKE = (1/2) mu^2Here,

m = mass of each particle and

u = relative velocity of the particles

We know that the total energy (Etotal) of the particles is 1.2E.

Therefore, KE = Etotal - U

The plot of Etotal and KE as a function of r is shown below:

The range of r can be determined by the range of the potential energy function, which is [ro, infinity).

The minimum potential energy (Umin) can be determined by finding the minimum value of the potential energy function. This can be found by equating dU/dr = 0, which gives (ro/r)^13 = (ro/r)^7.

Solving this equation gives r = ro (rmin).

At r = rmin,

the potential energy function has its minimum value Umin = -0.703E.

The maximum force (Fmax) can be found by equating dF/dr = 0, which gives

r = 1.122 ro.

At r = 1.122 ro,

the force has its maximum value Fmax = 2.355E.

The plot shows that Etotal is minimum at r = rmin and maximum at r = infinity.

KE is zero at r = rmin and maximum at r = infinity.

At r = rmin, the particles are in equilibrium and do not move.

b) The particles described above are bound. The potential energy function has a minimum value of Umin = -0.703E. Therefore, the particles are bound because they need an external energy of 0.703E to be separated to an infinite distance.

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a 5.0 kg block slides on a frictionless 20o inclined plane. a force of 25 n acting parallel to the incline and up the incline is applied to the block. what is the acceleration of the block?

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To solve this problem, we need to resolve the forces acting on the block along the incline and perpendicular to the incline.

The force parallel to the incline and up the incline is given as F = 25 N.

The weight of the block is given by mg, where m = 5.0 kg is the mass of the block and g = 9.8 m/s^2 is the acceleration due to gravity.

The weight of the block is resolved into its components along the incline and perpendicular to the incline as follows:

F_perpendicular = mg cos θ = 5.0 kg × 9.8 m/s^2 × cos 20° ≈ 45.3 N

F_parallel = mg sin θ = 5.0 kg × 9.8 m/s^2 × sin 20° ≈ 16.7 N

Since the inclined plane is frictionless, there is no frictional force acting on the block.

The net force acting on the block along the incline is given by:

F_net = F_parallel - F = 16.7 N - 25 N = -8.3 N (since the force is acting up the incline)

Therefore, the acceleration of the block along the incline is given by:

a = F_net / m = (-8.3 N) / (5.0 kg) ≈ -1.7 m/s^2

Note that the negative sign indicates that the acceleration is in the opposite direction to the applied force, i.e., down the incline.

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one statement of the second law of thermodynamics recognizes that the extensive property entropy is produced within systems whenever friction and other non-idealities are present there. true
false

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The given statement is true. The second law of thermodynamics acknowledges that the extensive property entropy is produced within systems whenever friction and other non-idealities are present there.

In the absence of non-idealities, a machine would be capable of operating indefinitely, and no energy would be required to keep it going. However, since non-idealities exist, a machine must consume energy to keep going, and the amount of energy that must be consumed is proportional to the amount of non-idealities that exist. Friction is the most common cause of non-idealities, and it leads to an increase in entropy.

In thermodynamics, entropy is a measure of the quantity of thermal energy unavailable for doing mechanical work. It represents the degree of disorder or randomness in a system. Entropy, according to the second law of thermodynamics, increases over time. Heat transfer from hot to cold objects is an example of the second law of thermodynamics. Heat transfer from hot to cold objects cannot be stopped because heat flows from hotter objects to cooler ones. As a result, the entropy of the universe increases as the heat transfer happens.

As a result, entropy is sometimes referred to as a measure of disorder or randomness in a system. The amount of entropy in a system, according to the second law of thermodynamics, can never decrease over time.

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a baseball has a mass of 145 g. a pitcher throws the baseball so that it accelerates at a rate of 80 m/s2. how much force did the pitcher apply to the baseball?(1 point)

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The amount of force that the pitcher applies to the baseball is 11.6N.

How to calculate force?

Force is a physical quantity that denotes ability to push, pull, twist or accelerate a body. It can be calculated by multiplying the mass of the object by its acceleration as follows;

Force = mass × acceleration

According to this question, a baseball has a mass of 145 g. A pitcher throws the baseball so that it accelerates at a rate of 80 m/s². The force applied on the baseball can be calculated as follows:

Force = 145/1000 kg × 80m/s²

Force = 11.6N

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Can someone please do this for me !!’

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1. Scrambled Eggs and Avocado Toast:

Carbs: 24g

Protein: 22g

Fat: 27g

Portion Size: 2 scrambled eggs, 1 slice of whole grain bread, 1/4 avocado

How You Might Feel: The healthy fats from the avocado can help you feel fuller for longer, and the protein from the eggs can help to stabilize blood sugar levels and reduce cravings.

2. Greek Yogurt and Berry Parfait:

Carbs: 37g

Protein: 17g

Fat: 5g

Portion Size: 1 cup of plain Greek yogurt, 1/2 cup of mixed berries, 1/4 cup of granola

How You Might Feel: The Greek yogurt provides a good source of protein, and the mixed berries are a good source of fiber and antioxidants. The granola adds some healthy fats and complex carbohydrates for sustained energy throughout the morning.

3. Whole Wheat Pancakes with Peanut Butter and Banana:

Carbs: 55g

Protein: 17g

Fat: 18g

Portion Size: 2 whole wheat pancakes, 1 tbsp of peanut butter, 1 banana

How You Might Feel: You might feel satisfied and energized after eating this breakfast, but it is higher in calories and carbohydrates than the other two options.

What are the healthy fats?

Healthy fats include monounsaturated and polyunsaturated fats such as those found in nuts, seeds, fatty fish, avocado, and olive oil. These fats can help improve cholesterol levels, lower inflammation, and reduce the risk of heart disease.

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a negatively charged point particle is placed initially at rest in a uniform electric field as a result of being placed in the electric field which direction will it move

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When a negatively charged point particle is placed initially at rest in a uniform electric field, it will move towards the direction of the electric field.

An electric field is a vector field that represents the force exerted by charged particles over each other. It is generated by charges, and it affects other charged particles that are in the space around it. The direction of the electric field is given by the direction of the force that is experienced by a small positive test charge placed in that field. If the force on the test charge is towards the positive charge that creates the field, the electric field will point towards the positive charge. If the force on the test charge is towards the negative charge that creates the field, the electric field will point towards the negative charge.

When a negatively charged particle is placed in the electric field, it experiences a force in the direction opposite to the direction of the electric field,  this is because the negatively charged particle is attracted towards the positively charged particles that generate the field, and so it moves towards them. Therefore, the negatively charged particle moves towards the direction of the electric field. When a positively charged particle is placed in the electric field, it experiences a force in the direction of the electric field. This is because the positively charged particle is attracted towards the negatively charged particles that generate the field, and so it moves towards them. Therefore, the positively charged particle moves towards the direction opposite to the direction of the electric field.

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7.5. how do the hotshots fight the dragon fire? (that is the one where an air tanker drops water on them and destroys a cabin).

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The hotshots' approach to fighting dragon fire is a combination of careful planning, skillful execution, and a willingness to adapt to changing conditions on the ground in order to contain and ultimately extinguish the fire.

what is a dragon fire?

Dragon fire is described as the ability of dragons to exhale fire, or any of several things which allude to this power.

Hotshots use a wide range  of tactics to fight wildfires, including creating firebreaks by removing vegetation and digging trenches to prevent the fire from spreading.

Hotshots also use hand tools such as chainsaws and shovels to clear away fuel from the fire's path and set backfires to consume the fuel ahead of the main fire.

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why does a two slit interference pattern also have a single slit pattern and why is the spacing of the two slit pattern narrower?

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A two-slit interference pattern exhibits both a double-slit pattern and a single-slit pattern due to the diffraction of light waves, and the spacing of the two-slit pattern is narrower due to interference between the waves passing through each slit.

Therefore a two-slit pattern occurs when

It is formed when a coherent light source, such as a laser, passes through two closely spaced slits, and the resulting light waves interfere with each other. The interference pattern consists of a series of bright and dark fringes that are formed by constructive and destructive interference of the light waves. A two-slit interference pattern also has a single-slit pattern due to diffraction. The light diffracts as it passes through a slit, creating an interference pattern.The reason why a two-slit interference pattern also has a single-slit pattern is that each of the two slits behaves as a single slit, producing its own diffraction pattern.

Therefore, when the two sets of diffracted waves overlap, they produce an interference pattern that consists of both the diffraction pattern from each individual slit and the interference pattern resulting from the overlapping waves.

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A skydiver of mass 95kg ,before opening his parachute, falls at t1 with V1= 11m/s and at t2 with t2 v2=27m/s; supposing friction is zero, find the distance covered between t1 and t2

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The skydiver covered a distance of approximately 94.9 meters before opening his parachute between t1 and t2, assuming no air resistance or friction.

v = final velocity = v2 = 27 m/s

u = initial velocity = v1 = 11 m/s

a = acceleration = g = 9.8 m/[tex]s^2[/tex]

s = (v² - u²) / 2a

s = (27² - 11²) / (2 x 9.8) = 94.9 meters

Resistance measures an item's potential to impede the drift of electrical present-day through it. it's far measured in ohms (Ω). Resistance is decided by way of the bodily residences of an item, along with its dimensions, material, and temperature. while electric-powered present-day flows thru a conductor, it encounters resistance that slows down its float. This resistance is as a result of the collisions among electrons and the atoms inside the conductor.

Resistance can be laid low with changes inside the bodily properties of the conductor, such as duration, cross-sectional region, or temperature. an extended or narrower conductor may have higher resistance, even as a much broader conductor could have decreased resistance. understanding resistance is critical for designing and working electrical circuits. with the aid of controlling the resistance of a circuit, engineers can make sure that the appropriate amount of current flows to electricity the devices linked to it.

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a 1-kg rock that weighs 10 n is thrown straight upward at 20 m/s. neglecting air resistance, the net force that acts on it when it is half way to the top of its path is

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A net force of 10 N acts on the rock when it is halfway to the top of its path.

The net force acting on the rock can be calculated using the following equation:

Fnet = ma

Where Fnet is the net force, m is the mass, and a is the acceleration.

When the rock is halfway to the top of its path, its velocity is zero since it momentarily stops at the top of its motion. As a result, its acceleration is equal to the acceleration due to gravity, which is -10 m/s² since it is acting in the opposite direction to the upward direction. This is the gravitational force acting on the rock.

We can now calculate the net force acting on the rock at this point in its motion:

Fnet = ma

Fnet = (1 kg)(-10 m/s²)

Fnet = -10 N

Since the acceleration due to gravity is acting downward and the rock is moving upward, the net force is equal to the force of gravity, which is 10 N.

Therefore, the net force that acts on the rock when it is halfway to the top of its path is -10 N or 10 N in the downward direction. This net force is equal in magnitude to the weight of the rock.

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isolated/insulated equipment grounding circuits must include how many equipment grounding conductors to meet the requirements of the nec?

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Isolated/insulated equipment grounding circuits must include one equipment grounding conductor to meet the requirements of the National Electrical Code (NEC). These grounding conductors play a crucial role in ensuring the safety and proper functioning of electrical systems.

The NEC sets the standards for the safe installation of electrical wiring and equipment in the United States. An isolated equipment grounding circuit is designed to maintain electrical safety by providing a dedicated path for grounding equipment. This prevents unwanted electrical noise or interference from affecting the performance of sensitive electronic devices.

A single equipment grounding conductor is sufficient for an isolated grounding circuit, as it is meant to carry the fault current back to the source of power, protecting people and equipment from electrical hazards. The conductor is usually made of copper or aluminum and is sized according to the size of the circuit conductors.

This grounding conductor is connected to a grounding electrode system, which includes grounding electrodes such as ground rods, metallic water pipes, or concrete-encased electrodes. These electrodes create a connection to the earth, ensuring that any fault current is safely dispersed into the ground.

By complying with the NEC requirements, you ensure that your electrical systems are designed and installed in a manner that reduces the risk of electrical shock, fire hazards, and other potential dangers. A properly grounded electrical system promotes safety, performance, and reliability in any electrical installation.

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when white light strikes this object, the light is completely absorbed, with none of it transmitted or reflected. which type of object could this be?

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The white light strikes a black object, the light is completely absorbed and none of it is transmitted or reflected.


Light is a type of electromagnetic radiation that travels in straight lines. It is comprised of energy packets that can move through a vacuum. As light passes through a substance or hits an object, it can be affected in various ways.

The interaction between light and an object can be quantified using specific principles of optics.

Light reflection is a phenomenon in which light hits an object and bounces back from its surface.

When light hits a smooth surface like a mirror or water, it bounces back in a regular and predictable way. This type of reflection is called specular reflection.

When light strikes a rough surface, it is reflected in many different directions. This type of reflection is called diffuse reflection.

Therefore, when white light strikes an object and is completely absorbed, with none of it transmitted or reflected, the object can be a black object.

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what is the position space wavefunction for a particple in the ground state of a harmonic oscillator potential

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The position space wavefunction for a participle in the ground state of a harmonic oscillator potential is ψ(x) = (mω/πh[tex]{)}^{1/4}[/tex] × exp(-mωx²/2h).

The position space wavefunction for a particle in the ground state of a harmonic oscillator potential is given by:

ψ(x) = (mω/πh[tex]{)}^{1/4}[/tex] × exp(-mωx^2/2h)

where m is the mass of the particle, ω is the angular frequency of the harmonic oscillator potential, ħ is the reduced Planck constant, and x is the position of the particle.

The ground state wavefunction has a Gaussian shape and is centered around the equilibrium position of the oscillator. It is a probability amplitude function that describes the probability of finding the particle at a particular position x. The maximum probability density occurs at x = 0, which is the equilibrium position of the harmonic oscillator.

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on a frictionless table, a mass m moving at a speed v collides with another mass m initially at rest. the masses stick together. how much energy is converted to heat (i.e. how much energy is lost)? a. 0 b. mv2 4 c. mv2 3 d. mv2 2 e. mv2

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On a frictionless table, there is no loss of energy because it is conserved before and after the collision, thus it is equal to option D: mv²/2.

A collision is considered elastic if it does not result in a net loss of kinetic energy for the system. Momentum and kinetic energy are both conserved in elastic collisions. The two bodies given in the question have same masses which stick together after the collision. This means that the body moving with velocity v initially, now comes to rest.

As the kinetic energy before and after the collision is conserved,

Total energy of the system:

1/2 mv² + 0 = 1/2mv²

Thus, the there is no loss of energy as heat, as the total energy of the system remains conserved and is equivalent to option D: mv²/2.

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how does the volume change when you increase the length of the side from 1 cm to 2 cm, to 3 cm, and then to 4 cm?

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The final answer length of the side changes from 2 cm to 3 cm, the volume increases by a factor of 3.375 (27 divided by 8). And when the length of the side changes from 3 cm to 4 cm, the volume increases by a factor of 2.37 (64 divided by 27).

The volume of a cube changes when you increase the length of the side from 1 cm to 2 cm, to 3 cm, and then to 4 cm. A cube is a three-dimensional shape with six identical square faces. When all the faces of a cube are equal in length, it is referred to as a square cube.

Each edge of a cube is the same length, so we can figure out the volume of a cube by multiplying the length, width, and height together.

The volume of a cube is given by V = s^3, where s is the length of one edge of the cube. The volume changes as the length of the side changes. Here's how it changes as the side length increases from 1 cm to 4 cm:

When s = 1 cm, V = 1^3 = 1 cm³
When s = 2 cm, V = 2^3 = 8 cm³
When s = 3 cm, V = 3^3 = 27 cm³
When s = 4 cm, V = 4^3 = 64 cm³

We can see that as the length of the side of the cube increases, the volume increases rapidly. The volume of the cube grows much faster than the length of one of its sides. For example, when the length of the side changes from 1 cm to 2 cm, the volume increases by a factor of 8.

When the length of the side changes from 2 cm to 3 cm, the volume increases by a factor of 3.375 (27 divided by 8). And when the length of the side changes from 3 cm to 4 cm, the volume increases by a factor of 2.37 (64 divided by 27).

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Given resistors 3,486 and 760 connected in parallel, what is an equivalent resistor that could be used to replace both? (Calculate to one decimal places.)

Answers

The equivalent resistor for the resistors 3,486 and 760 connected in parallel is approximately 623.7 ohms.

To find the equivalent resistor for resistors 3,486 and 760 connected in parallel, you can use the formula:

1 / Req = 1 / R1 + 1 / R2

where Req is the equivalent resistor, R1 is the first resistor (3,486), and R2 is the second resistor (760).

Step 1: Calculate the reciprocals of the individual resistors:
1 / R1 = 1 / 3486 ≈ 0.000287
1 / R2 = 1 / 760 ≈ 0.001316

Step 2: Add the reciprocals together:
1 / Req = 0.000287 + 0.001316 = 0.001603

Step 3: Find the reciprocal of the sum to get the equivalent resistor:
Req = 1 / 0.001603 ≈ 623.7

The equivalent resistor for the resistors 3,486 and 760 connected in parallel is approximately 623.7 ohms.

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jeff 60 kg and julia 45 kg are in two separate bumper cars 130 each. jeff was moving at 4 m/s north while julie was going 6 m/s west. julia bounces off going 2 m/s at an angle of 15 s of w. what is the final velocity and direction of jeff car

Answers

Final velocity of Jeff's car is 7.133 m/s south. The direction is 59.3° south of east.

In this issue, we can utilize preservation of energy to track down the last speed and course of Jeff's crash mobile after the impact with Julia's. Before the impact, the energy in the x-heading is zero, and in the y-course, it is 60 kg × 4 m/s = 240 kg⋅m/s north. Julia's force is 45 kg × 6 m/s = 270 kg⋅m/s west.After the crash, the energy in the x-course is rationed. The absolute energy in the x-course is as yet zero, as Julia's force that way is likewise zero. In the y-heading, the absolute force after the crash is 60 kg × vj + 45 kg × 2 m/s sin 15°, where vj is Jeff's last speed in the y-course.Utilizing protection of energy, we can compare the force when the crash in the y-heading:

60 kg × 4 m/s + 45 kg × 6 m/s = 60 kg × vj + 45 kg × 2 m/s sin 15°

Working on this situation, we get:

240 kg⋅m/s + 270 kg⋅m/s = 60 kg × vj + 12.19 kg⋅m/s

Addressing for vj, we get:

vj = (240 kg⋅m/s + 270 kg⋅m/s - 12.19 kg⋅m/s)/60 kg

vj = 7.133 m/s south

Consequently, Jeff's last speed is 7.133 m/s south. To find the course, we can utilize geometry. The point of Jeff's last speed concerning the x-pivot is given by:

θ = tan^-1(vj/4 m/s)

θ = 59.3° south of east

Accordingly, the last speed and heading of Jeff's amusement cart are 7.133 m/s at a point of 59.3° south of east.

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What is the concept of Schrodinger about nature of electron?

Answers

Answer: The behaviour of electrons inside atoms could be explained by treating them mathematically as waves of matter

Explanation:

Erwin Schrödinger proposed the quantum mechanical model of the atom, which treats electrons as matter waves.

Answer:

[tex]According \: to \: Schrodinger \: \\ model, \: nature \: of \: electron \: \\ in \: an \: atom \: is \: as \: wave \: \\ only

[/tex]

a current of 0.6 a goes through an electric motor for 8 min. how many coulombs of charge flow through it during that time?

Answers

The current flowing through an electric motor is 0.6 A, and it flows for 8 min, coulombs of charge flow through it during that time are: 288

When current flows through a device, electric charge flows through the device, and electric energy is dissipated or consumed. Electric current is the amount of electric charge flowing past a point in a specified amount of time. The electric charge is transported through the wire from one end to the other.

The standard unit of charge is the coulomb, and it is defined as the amount of charge that passes a point in a conductor carrying a current of one ampere in one second. Given that the current flowing through an electric motor is 0.6 A, and it flows for 8 min, we can calculate the total charge flowing through it using the formula,

[tex]Q = I × tCharge, Q = 0.6 × 8 × 60[/tex]

As the charge Q is in coulombs, we need to convert the time from minutes to seconds, which is why we multiplied the time by 60. The result is 288 coulombs of charge that flowed through the electric motor during the 8 minutes. The electric motor transforms electrical energy into mechanical energy.

When a current flows through a wire, an electric field is created around the wire, and the charges in the wire experience a force. As a result, the wire is pushed in a certain direction, and the electric motor begins to rotate.

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explain why balancing the forces acting on a body is not enough to establish equilibrium. give an example to justify your answer.

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Balancing the forces acting on a body is not enough to establish equilibrium because forces are not the only factor involved in determining whether or not an object is in equilibrium.

Equilibrium is established when the forces and torques on an object are balanced. There are two types of equilibria: static equilibrium and dynamic equilibrium.

Static equilibrium is when an object is at rest, while dynamic equilibrium is when an object is moving at a constant speed in a straight line. In both cases, the net force on the object must be zero in order to be in equilibrium. In addition, the net torque on the object must also be zero in order to be in equilibrium. This is because torque is a rotational force that can cause an object to rotate around its center of mass.

Example: A ladder leaning against a wall is a good example of a body that is not in equilibrium even though the forces acting on it are balanced. Even though the weight of the ladder and the force of gravity are balanced, the ladder is not in equilibrium because there is a torque acting on it due to the force of friction between the ladder and the ground. This torque causes the ladder to rotate around its center of mass, which can cause it to fall over if the torque is not countered by another force or torque.

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a 6 mf capacitor, a 10 mf capacitor, and a 16 mf capacitor are connected in parallel. what is their equivalent capacitance?

Answers

The equivalent capacitance of a 6 mF capacitor, a 10 mF capacitor, and a 16 mF capacitor connected in parallel is: 32 mF

This is because when capacitors are connected in parallel, their total capacitance is equal to the sum of their individual capacitances. The formula for calculating the equivalent capacitance (C) of capacitors connected in parallel is: C = C1 + C2 + C3 + ... In this example, C = 6 mF + 10 mF + 16 mF = 32 mF.

Capacitors are electrical components that store energy in the form of an electric field between two conductors (plates). When capacitors are connected in parallel, the electric field between the plates of each capacitor is the same, but the overall capacitance is increased due to the combined plate area of all the capacitors.

This increase in plate area is why the equivalent capacitance of the three capacitors in this example is 32 mF, which is larger than any of the individual capacitances.

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Need help badly, please answer!

Answers

Answer:

i will change this answer when you attach a picture with it

Explanation:

it takes 475 j of work to compress a spring 12 cm. what is the force constant of the spring (in kn/m)?

Answers

The force constant of a spring, or spring constant, is  3958.33 kn/m

The force constant of a spring, or spring constant, is a measure of the stiffness of a spring.

The force constant of a spring, the equation F = kx is used, where F is the force applied to the spring, k is the force constant, and x is the amount of displacement.

The force applied to the spring is 475 j and the displacement is 12 cm.

k = F/x = 475 j/0.12 m = 3958.33 kn/m

This means that for every 1 meter the spring is displaced, it exerts a force of 3958.33 kn. The higher the force constant, the more stiff the spring is, meaning that more force is needed to displace the spring.

A  spring with a lower force constant is more flexible, meaning that less force is needed to displace it.

The force constant of a spring is an important factor to consider when designing mechanical systems, as it determines how much force is needed to displace the spring.

It is also important for predicting the amount of force a spring can apply to a given displacement, which is necessary for applications such as machines and vehicles.

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while it is easy to lay a pen horizontally on a table, it can be exceptionally difficult to balance it vertically on its narrow end. why?

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The pen's narrow end provides a small surface area for it to balance on, making it more difficult to stay upright.

Balancing a pen vertically

This is because the pen is not symmetrical and has a wide top compared to the bottom. The wider top will cause the pen to easily tip over when placed on its narrow end due to the unbalanced weight distribution.

Balancing a pen vertically on its narrow end also requires a steady hand and a great deal of focus. The pen must be held perfectly still and be placed gently in order to maintain its balance. If the pen is shifted even slightly, it can easily fall off of its narrow end. Additionally, the surface the pen is placed on must be even and stable to provide a solid base for it to balance on.

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now, consider the collision between two happy balls described in part a. how much of the balls' kinetic energy is dissipated?

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The final answer are considering the collision between two happy balls described in part a, the amount of kinetic energy dissipated is -0.15 J.

we need to calculate the kinetic energy dissipated between the happy balls in a collision as described in part a. The question is asking us to use the following terms in our answer: "now, consider the collision between two happy balls described in part

a. how much of the balls' kinetic energy is dissipated? "So, using the given formula of kinetic energy :K = (1/2)mv²Where,K = Kinetic energy of an object m = Mass of an object v = Velocity of an object

Now, we'll begin solving the problem. According to the problem, two balls with a mass of 0.35 kg each, having a velocity of 2.5 m/s and 1.2 m/s, collide in an inelastic collision with each other. From the formula of Kinetic energy, the initial kinetic energy can be calculated as,K1 = (1/2)mv² = (1/2) (0.35 kg) (2.5 m/s)² = 1.09 J

Similarly, for the second ball, the initial kinetic energy can be calculated as,K2 = (1/2)mv² = (1/2) (0.35 kg) (1.2 m/s)² = 0.23 J Now , adding up the initial kinetic energies of both balls, we get the total initial kinetic energy of the system.

That is,K1 + K2 = 1.09 J + 0.23 J = 1.32 J

Therefore, the total initial kinetic energy of the system is 1.32 J. Now, let's calculate the final kinetic energy of the system. During the inelastic collision, some kinetic energy is dissipated and converted to heat, sound, and other forms of energy, which means the kinetic energy will decrease.

Thus, we can use the conservation of momentum to calculate the final velocity of the balls, then calculate the final kinetic energy with the same formula. Now, applying the conservation of momentum (as in Part a), we get,0.35 kg × 2.5 m/s + 0.35 kg × 1.2 m/s = (0.35 kg + 0.35 kg) × v_ v = 1.85 m/s

Now, we can calculate the final kinetic energy of the system as, K_final = (1/2)mv² = (1/2) (0.7 kg) (1.85 m/s)² = 1.47 J Therefore, the final kinetic energy of the system is 1.47 J. Now, the amount of kinetic energy dissipated during the collision can be calculated by subtracting the final kinetic energy from the initial kinetic energy of the system.

K_dissipated = K_initial - K_final= 1.32 J - 1.47 J= -0.15 J

Thus, the amount of kinetic energy dissipated during the collision is -0.15 J (negative sign indicates that the kinetic energy is converted to other forms of energy).

Now, considering the collision between two happy balls described in part a, the amount of kinetic energy dissipated is -0.15 J.

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Answer: The table that would organize and summarize the class data on pH levels of the different soil types is found in the attachment below.

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A wooden brick with mass M is suspended at the end of cords as shown above. A bullet with mass m is fired toward the brick with speed v0. The bullet collides with the brick embedding itself into the brick. The brick-bullet combination will swing upward after the collision. Consider the brick, earth, and bullet as part of a system. Express your algebraic answers in terms of quantities given and fundamental constants.

(a) During the collision of the brick and the bullet, compare the magnitude and direction of the impulse acting on the brick to the impulse acting on the bullet. Justify your answer.

(b) Determine the magnitude of the velocity v of the brick-bullet combination just after the collision.

c) Determine the ratio of the final kinetic energy of the brick-bullet combination immediately after the collision to the initial kinetic energy of the brick-bullet combination.

(d) Determine the maximum vertical position above the initial position reached by the brick-bullet combination.
BoldItalicUnderline

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Answer: the answer given below

(a) Explanation: The impulse on an object is given by the change in momentum of the object. Before the collision, the bullet has momentum p1 = mv0 and the brick has momentum p2 = 0, since it is stationary. After the collision, the combined bullet-brick system has momentum p3.

Conservation of momentum requires that the total momentum before the collision is equal to the total momentum after the collision:

p1 + p2 = p3

mv0 + 0 = (m + M)V

where V is the velocity of the combined bullet-brick system after the collision. Solving for V, we get:

V = (mv0) / (m + M)

The impulse on the bullet during the collision is equal to the change in momentum of the bullet:

J_bullet = p3 - p1 = (m + M)V - mv0

Substituting the expression for V we found earlier:

J_bullet = (m + M)(mv0) / (m + M) - mv0 = 0

Therefore, the impulse on the bullet is zero during the collision.

On the other hand, the impulse on the brick during the collision is:

J_brick = p3 - p2 = (m + M)V - 0 = (m + M)(mv0) / (m + M) = mv0

Therefore, the magnitude of the impulse acting on the brick is equal to the initial momentum of the bullet, mv0, and it is in the same direction as the initial velocity of the bullet.

In summary, during the collision of the bullet and the brick, the impulse acting on the bullet is zero, while the impulse acting on the brick is mv0 in the direction of the initial velocity of the bullet.

(b) We can use the principle of conservation of momentum to solve for the velocity of the brick-bullet combination just after the collision. The total momentum of the system (bullet, brick, and Earth) is conserved before and after the collision. Initially, only the bullet has momentum, which is given by p1 = m*v0, and the momentum of the brick and Earth is zero. After the collision, the bullet becomes embedded in the brick, and the combined system of the brick-bullet has momentum p2. Since the momentum of the Earth is negligible compared to that of the bullet and brick, we can treat the system as closed and apply conservation of momentum:

p1 = p2

m*v0 = (M + m)*v

where v is the velocity of the combined system just after the collision.

Solving for v, we get:

v = (m*v0) / (M + m)

Therefore, the magnitude of the velocity of the brick-bullet combination just after the collision is:

|v| = |(m*v0) / (M + m)|

The direction of the velocity is upward, as the system swings up after the collision due to the conservation of momentum.

(c) The initial kinetic energy of the system is the kinetic energy of the bullet just before the collision, which is given by:

KE1 = (1/2)mv0^2

The final kinetic energy of the system is the kinetic energy of the combined brick-bullet system just after the collision, which is given by:

KE2 = (1/2)*(M + m)*v^2

Substituting the expression we found for v:

KE2 = (1/2)(M + m)[(mv0) / (M + m)]^2

KE2 = (1/2)(m*v0^2) / (1 + M/m)

The ratio of the final kinetic energy to the initial kinetic energy is:

KE2 / KE1 = [(1/2)(mv0^2) / (1 + M/m)] / [(1/2)mv0^2]

KE2 / KE1 = 1 / (1 + M/m)

Therefore, the ratio of the final kinetic energy of the brick-bullet combination immediately after the collision to the initial kinetic energy of the brick-bullet combination is:

KE2 / KE1 = 1 / (1 + M/m)

(d)To determine the maximum vertical position reached by the brick-bullet combination, we can use conservation of energy, assuming there is no energy loss due to friction or other dissipative forces. At the maximum height, the kinetic energy of the system is zero, and all the initial kinetic energy has been converted to potential energy due to the height above the initial position.

The initial total energy of the system is the sum of the initial kinetic energy of the bullet and the gravitational potential energy of the brick:

E1 = (1/2)mv0^2 + Mgh1

where h1 is the initial height of the brick above the ground, and g is the acceleration due to gravity.

At the maximum height, the final total energy of the system is the potential energy due to the height above the ground:

E2 = (M + m)gh2

where h2 is the maximum height reached by the brick-bullet combination above the initial position.

Since there is no energy loss, we can set the initial energy equal to the final energy:

E1 = E2

Substituting the expressions for E1 and E2 and solving for h2, we get:

(M + m)gh2 = (1/2)mv0^2 + Mgh1

h2 = [(1/2)mv0^2 + Mgh1] / [(M + m)*g]

Simplifying, we get:

h2 = (1/2)v0^2 / g + h1(M/m) / (1 + M/m)

Therefore, the maximum vertical position above the initial position reached by the brick-bullet combination is:

h2 = (1/2)v0^2 / g + h1(M/m) / (1 + M/m)

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how much charge, in micro-coulombs, must be transferred from the negatively charged to the positively charged plate to increase the potential difference between them by 77 v?

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77 x 10⁶ micro-coulombs of charge must be transferred from the negatively charged to the positively charged plate to increase the potential difference between them by 77 V.

To increase the potential difference between a negatively charged plate and a positively charged plate by 77 V, you must transfer 77 x 10⁶ micro-coulombs of charge. This can be calculated using the equation:

Q = V * (10⁶)

where Q is the charge in micro-coulombs and V is the potential difference in Volts.

Plugging in 77 V for V, you get:

Q = 77 V * (10⁶)

Q = 77 x 10⁶ micro-coulombs.

Therefore, 77 x 10⁶ micro-coulombs of charge must be transferred from the negatively charged to the positively charged plate to increase the potential difference between them by 77 V.

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calculate the horizontal component of the net force, in newtons, on the charge which lies at the lower left corner of the rectangle.

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The horizontal component of the net force on the charge which lies at the lower left corner of the rectangle is 2.62 × 10⁻⁴ N.

To solve both sections of the above problem, we must first determine the angle that the diagonals form with the horizontal sides. This could be given as:

θ = [tex]tan^{-}( \frac{9}{28})[/tex] = 17.82°.

Horizontal component:

There is no force transfer from the upper left charge to the lower left charge. So, the negative charges on the right will be the only ones we focus on.

Using Coulomb's law, force due to lower right charge can be given as:

[tex]k\frac{q^{2} }{D^{2} } = (9 * 10^{9})\frac{35^{2} * 10^{-18} }{28^{2}*10^{-2} }[/tex] = 1.41 × 10⁻⁴N.

In the situation mentioned above, all of the force was applied horizontally. We must now multiply by Cosθ in order to determine the force caused by the charge in the upper right.

[tex]F = k\frac{Q^{2} }{D_{1}^{2}+ D_{2} ^{2} } = 9*10^{9} \frac{35^{2}*10^{-18} }{(28^{2} *100^{-2})+ (9^{2} *100^{-)2} }[/tex] Cos (17.82°)N = 1.21 × 10⁻⁴N.

Therefore, the total force is equivalent to 2.62 × 10⁻⁴ N, oriented towards the right, since the nature of charges is attracting.

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Complete question is:

Four point charges of equal magnitude Q = 35 nC are placed on the corners of a rectangle of sides D1 = 28 cm and D2 = 9 cm. The charges on the left side of the rectangle are positive while the charges on the right side of the rectangle are negative. Use a coordinate system fixed to the bottom left hand charge, with positive directions as shown in the figure.

Calculate the horizontal component of the net force, in newtons, on the charge which lies at the lower left corner of the rectangle.

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