A kid is bouncing on a pogo stick. He oscillates 22.0 times in 14.9 S. What is his frequency? (Unit = Hz)​

The boundary conditions have a positive effect on the finite difference formulation of interior nodes of the medium. Boundary conditions provide constraints or information at the edges of a computational domain to determine the behavior of the system. In the finite difference method, the values at the boundary nodes are usually known or specified. By incorporating these boundary conditions into the formulation, the behavior of the system is better represented.

When the boundary conditions are properly applied, they help ensure that the finite difference scheme accurately captures the desired physics within the interior nodes. This positive effect allows for more accurate and reliable simulations, as the behavior at the boundaries influences the behavior of the entire system. Boundary conditions have a positive effect on the finite difference formulation of interior nodes. By incorporating the known or specified conditions at the boundaries, the accuracy and reliability of the finite difference method are improved, allowing for a more faithful representation of the desired physics within the system.

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The AMA (mechanical advantage) of a wedge is given by the ratio of the length of the sloping side of the wedge to its width. If the AMA of the wedge is 4, it means that the length of the sloping side of the wedge is four times its width.

The force exerted on the wedge depends on the load being lifted and the angle of the wedge; as the load increases, the force required to lift it also increases. However, the AMA of the wedge allows for the same load to be lifted with less force.

Force = Load / AMA

Hence, if the AMA of the wedge is 4, it means that the length of the sloping side of the wedge is four times its width.

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When parked on the highway at night, cars should use their hazard lights or emergency flashers.

When parked on the highway at night, cars should use their hazard lights or emergency flashers. These lights are designed to alert other drivers of a potential hazard or obstruction on the road. By activating the hazard lights, the parked car becomes more visible to oncoming traffic, reducing the risk of accidents.

Hazard lights typically consist of a pair of high-intensity, blinking lights located at the front and rear of the vehicle. They emit a bright, attention-grabbing signal that can be easily seen from a distance, even in dark or adverse weather conditions.

The blinking pattern of the lights distinguishes them from the regular headlights or taillights of moving vehicles, indicating that the car is stationary and that caution should be exercised.

Using hazard lights while parked on the highway helps to warn approaching drivers to slow down and proceed with caution. It also helps to prevent rear-end collisions or other accidents caused by drivers failing to notice the stationary vehicle in time.

However, it is important to note that hazard lights should only be used when the car is parked in a safe location off the road and not obstructing traffic flow.

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Answer:

\(c. \: organ\)

Answer:

C.

Explanation:

Answer:

Work input W = -200.39 KJ

Explanation:

From the question, we are given;

m = 1.15 kg

Constant temperature T1 = T2 = 27 + 273 = 300k

Since the temperature is constant, we can say that the process is isothermal

P1 = 111 KPa

P2 = 0.84 MPa = 0.84 * 1000 KPa = 840 KPa

Now what we want to calculate is W1-2

Mathematically, for isothermal process;

W1-2 = mRTlnP1/P2

where R can be obtained from table and it is equal to 0.287 KJ/kg.k

Hence;

W1-2 = (1.15)(0.287)(300)(ln 111/840)

W1-2 = 99.015 * -2.023871690525 = -200.39 KJ

Kindly note that the value of the work is negative because work is done on the system and not by the system

Answer:

Scientific notation is the way that scientists easily handle very large numbers or very small numbers. For example, instead of writing 0.0000000056, we write 5.6 x 10-9. So, how does this work? We can think of 5.6 x 10-9 as the product of two numbers: 5.6 (the digit term) and 10-9 (the exponential term).

Explanation:

At the point X after 2 seconds, the stone has a magnitude of velocity equal to 10 m/s, and its direction is purely horizontal.

When a stone is projected horizontally in a vacuum, it only experiences horizontal motion due to the absence of air resistance. The vertical motion is governed by the force of gravity alone. In this scenario, the stone will undergo uniform horizontal motion with a constant horizontal velocity.

Given that the stone is projected horizontally with a velocity of 10 m/s, its horizontal velocity remains constant throughout the motion. After 2 seconds, the stone will have traveled a horizontal distance equal to the product of its horizontal velocity and the time, which is 10 m/s * 2 s = 20 m. Therefore, the stone will be located 20 meters horizontally from the point of projection. Since the stone is moving horizontally, its vertical velocity remains zero throughout the motion. The magnitude of the velocity at this point is equal to the horizontal velocity, which is 10 m/s. However, since the stone is moving horizontally, there is no vertical component of velocity.

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Answer:

Option C. vi = 0

Explanation:

An object held at a particular height above the ground is considered to be at rest. If the object is release from that height, the initial velocity (vi) is considered to be zero since the object is at rest or at a stand still position.

Thus,

Initial velocity (vi) = 0

F =

t = 0.17 s

Impulse (J) = 20,000 N

J = F x t

F = J / t = 20,000 Ns / 0.17 s = 117,647.0588 N

If a mass of 6.3 kg is held on an inclined plane that has an angle of 2.4° with the horizontal. If the coefficient of static friction is 0.032, the mass will slide down the inclined plane when released.

Given,

Mass is (m) = 6.3 kg

inclined angle (θ) = 2.4°

Coefficient of static friction (μ) = 0.032

Let acceleration due to gravity is = g= 9.8 m/s²

∴ Normal force acting on the mass is,

N = mgcosθ

⇒N= 6.3×9.81×cos(2.4°) =61.75 newton

∴The frictional force acting against the motion of the mass,

f = μN

⇒f= 0.032×61.75= 1.98 N

Now force acting on the mass due to gravity along the inclined plane,

F'=mgsinθ

⇒F'= 6.3×9.81×sin (2.4°)= 2.59 N

As, F'>f thus, the mass will slide down along the plane if released as frictional force less than the downward force due to gravity.

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Answer:

  48 ft/s

Explanation:

Acceleration due to gravity decreased the upward velocity to zero in 1.5 seconds (half the total time), so the initial velocity must have been ...

  (32 ft/s^2)(1.5 s) = 48 ft/s

An internal combustion engine is an engine in which combustion takes place inside the engine and is used to generate mechanical power.

Step 1: Intake: During the intake stroke, the engine takes in air (if it is naturally aspirated) or a mixture of air and fuel (if it is fuel-injected) and compresses it. In the combustion chamber, this mixture is drawn in through the intake valve. The carburetor mixes the fuel and air, while the fuel injectors inject fuel directly into the combustion chamber.

Step 2: Compression: During the compression stroke, the piston moves back up to compress the mixture of air and fuel. The spark plug fires, igniting the compressed mixture of air and fuel, and generating a tremendous amount of heat.

Step 3: Combustion: When the spark plug ignites the fuel-air mixture, the mixture ignites and burns, causing a rapid expansion of the hot gases and the expulsion of the piston. This energy is used to generate power by pushing the piston down, which turns the crankshaft. This energy is then transmitted to the wheels via the transmission.

Step 4: Exhaust:During the exhaust stroke, the exhaust valve opens, and the piston moves up to push the hot gases out of the combustion chamber. The exhaust system then takes the hot gases away from the engine and discharges them into the atmosphere.

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Answer: Answer below hope it helps

Explanation: How does a rubber rod become negatively charged through friction? It touches a negatively charged object, and protons move off of the rod. ... It is rubbed with another object, and electrons move onto the rod. It is rubbed with another object, and protons move off of the rod.

Answer:

Hey

here's ur answer ⤵️

Explanation:

It touches a negatively charged object, and protons move off of the rod it is rubbed with another object and electrons move onto the rod.

It is rubbed with another object and protons move off of the rod.

Hope it helps

Answer: physical or mechanical weathering

Explanation:

Mechanical weathering which is also referred to as the physical weathering occurs when a rock is broken down into smaller pieces. In this case, there will be a physical change of the rock but its composition will not change.

Some examples include ice freezing and expansion of the cracks in the rock, Smstrong winds that carrycpieces of sand which then sandblast surfaces, moving water which causes abrasion etc.

Liguid hydrogen and liguid oxygen are the most common propellant used for a rocket spaceflight.

Propellant are chemical or substances that helps to produce thrust in rockets, missile and other engines. They can either be solid or liquid propellant both producing the same effect.

Liquid hydrogen is very efficient propellant and more common simple because it has a high specific impulse that generate the desired amount of thrust compared to other propellant.

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The frequency of the incident Ultraviolet light is approximately 7.28 × 10^14 Hz

To find the frequency of the incident ultraviolet light, we can use the relationship between the speed of light, wavelength, and frequency:

c = λ * ν

where c is the speed of light (approximately 3.00 × 10^8 m/s), λ is the wavelength, and ν is the frequency.

Given:

λ = 4.12 × 10^(-7) m

First, we need to convert the work function from electron volts (eV) to joules (J). The conversion factor is 1 eV = 1.602 × 10^(-19) J.

Work function (Φ) = 4.72 eV * (1.602 × 10^(-19) J/eV)

≈ 7.56 × 10^(-19) J

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

ν = c / λ

ν = (3.00 × 10^8 m/s) / (4.12 × 10^(-7) m)

≈ 7.28 × 10^14 Hz

Therefore, the frequency of the incident ultraviolet light is approximately 7.28 × 10^14 Hz.

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Using concepts of Force, we got Mg is the force exerted by person to lift the block

We know very well that gravitational force plays a vital role in holding the objects.

Actually, we need to work against the gravity as gravity tries to pull the block  again on the surface, means we need to do some work against the gravity so that we are able to lift the block.

We need that every object in this universe has some mass.

Therefore we are assuming mass of block is M

Now, due to gravity acceleration due to gravity (g) also applied on the block.

Therefore, Force exerted by man =Mass×Acceleration

                                               Force=M×g

Hence, Force exerted by man to lift the block is Mg

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Answer:

(i) The particle has a period of 0.3 seconds.

(ii) The angular velocity of the particle is a20.944 radians per second.

(iii) The linear velocity of the particle is 1.047 meters per second.

(iv) The linear acceleration of the particle is 21.933 meters per square second.

Explanation:

Statement is incomplete and have mistakes, complete and correct form is presented below:

A particle makes 800 revolutions in 4 minutes of a circle with a radius of 5 centimeters. Find (i) its period, (ii) its angular velocity, (iii) its linear velocity and (iv) its acceleration.

(i) The period of the particle (\(T\)), in seconds, is the time taken to make a complete revolution:

\(T = \frac{4\,min \times \frac{60\,s}{1\,min} }{800\,rev}\)

\(T = 0.3\,s\)

The particle has a period of 0.3 seconds.

(ii) The angular velocity (\(\omega\)), in radians per second, is determined by the following formula:

\(\omega = \frac{2\pi}{T}\) (1)

\(\omega = \frac{2\pi}{0.3\,s}\)

\(\omega \approx 20.944\,\frac{rad}{s}\)

The angular velocity of the particle is a20.944 radians per second.

(iii) The linear velocity (\(v\)), in meters per second, is calculated by the following formula:

\(v = R\cdot \omega\) (2)

Where \(R\) is the radius of the circle, in meters.

If we know that \(R = 0.05\,m\) and \(\omega \approx 20.944\,\frac{rad}{s}\), then the linear velocity of the particle is:

\(v = (0.05\,m)\cdot \left(20.944\,\frac{rad}{s} \right)\)

\(v = 1.047\,\frac{m}{s}\)

The linear velocity of the particle is 1.047 meters per second.

(iv) Since angular velocity is constant, linear acceleration of the particle (\(a\)), in meters per square second, is entirely radial. Acceleration can be found by means of this expression:

\(a = \omega^{2}\cdot R\) (3)

If we know that \(R = 0.05\,m\) and \(\omega \approx 20.944\,\frac{rad}{s}\), then the linear acceleration of the particle is:

\(a = \left(20.944\,\frac{rad}{s} \right)^{2}\cdot (0.05\,m)\)

\(a = 21.933\,\frac{m}{s^{2}}\)

The linear acceleration of the particle is 21.933 meters per square second.

Answer:

change the

Explanation:

P = W/time

W = F*d

You have control over how fast you go up the stairs.

You also have control over how far up the stairs you go.

Therefore the answer is

If you don't like distance as an answer, you can carry something up the stairs -- anything that increases F will do.

The answer to how Cole solved the problem of holding the door open without a doorstop is by overcoming functional fixedness and reframing. Functional fixedness is the tendency to see objects only in their usual or customary way, and not in other possible ways.

In this case, Cole was not able to see his potted plant as anything other than a decorative item. However, he was able to reframe his thinking and see the plant as a functional object that could serve as a doorstop. This is an example of overcoming functional fixedness.

Reframing is the act of looking at a problem in a new way, from a different perspective. By reframing his thinking and looking at the potted plant in a new way, Cole was able to solve his problem. He was able to use his mental set to come up with a new solution to the problem, which involved using the potted plant as a doorstop. This solution was both creative and effective, and shows the power of reframing in problem-solving.

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Answer:

60 000 J

Explanation:

The PE will be converted to KE entirely (if the lowest point is ground level)

Active mountain belts are most likely to be found along the margins of continents.

A mountain system or belt is a group of mountain ranges of similar shape structure and orientation that arise from the same cause, usually mountain building. Individual mountains within the same mountain range do not necessarily have the same geological structure or rocks.

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The magnitude of the average force exerted on the ball by the floor during the collision is 39.2 N, directed upward.

First, we can determine the initial velocity of the ball just before it hits the floor using conservation of energy. The potential energy of the ball when it was dropped is given by mgh, where m is the mass of the ball (0.06 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height from which it was dropped (1.8 m). Thus, the potential energy is mgh = (0.06 kg)(9.8 m/s^2)(1.8 m) = 1.0584 J. Since the ball was dropped from rest, all of this potential energy is converted to kinetic energy just before it hits the floor. Therefore, the initial kinetic energy of the ball is also 1.0584 J.

When the ball hits the floor, it experiences an average force from the floor during the time it is in contact with it, slowing it down and then accelerating it back upward. We can use the impulse-momentum theorem to relate the average force to the change in momentum of the ball during the collision. Since the ball is dropped from rest and rebounds upwards with the same speed after the collision, the change in momentum is just equal to twice the momentum of the ball just before it hits the floor.

The momentum of the ball just before it hits the floor can be found using the kinetic energy calculated earlier. The kinetic energy is equal to (1/2)mv^2, where v is the velocity of the ball just before it hits the floor. Solving for v, we find v = sqrt((2KE)/m)

= sqrt((2(1.0584 J))/(0.06 kg))

= 3.259 m/s.

Therefore, the momentum of the ball just before it hits the floor is p = mv = (0.06 kg)(3.259 m/s) = 0.19554 kg m/s.

The change in momentum during the collision is then Δp = 2p

= 2(0.19554 kg m/s)

= 0.39108 kg m/s.

We can use the given time of contact between the ball and the floor (0.018 s) to calculate the average force:

F_avg = Δp/Δt

= (0.39108 kg m/s)/(0.018 s)

= 21.7267 N

Since the maximum height attained by the ball after the collision is 1.3 m, and it was dropped from a height of 1.8 m, we can use conservation of energy again to find the speed of the ball just as it leaves the floor. The potential energy at the maximum height is mgh = (0.06 kg)(9.8 m/s^2)(1.3 m) = 0.7644 J. Since the total mechanical energy (potential plus kinetic) is conserved, this is also equal to the kinetic energy just as the ball leaves the floor. Therefore, we have (1/2)mv^2 = 0.7644 J, or

v = sqrt((2*0.7644 J)/0.06 kg)

= 3.0855 m/s.

Since the ball was in contact with the floor for 0.018 s, we can determine the distance it traveled during that time, which is equal to the average velocity (3.0855 m/s) multiplied by the time (0.018 s). Therefore, the distance traveled is d = (3.0855 m/s)(0.018 s) = 0.05554 m.

The average force exerted by the floor on the ball is then the average of the upward and downward forces exerted during this distance, or:

F_avg = (21.7267 N + (-21.7267 N))/2

= 0 N

y = y_0 + v_0t + (1/2) Since the initial and final heights are the same, the term y_0 drops out, leaving:

0 = v_0t + (1/2)at^2

Solving for t, we get: t = -2v_0/a

= -2(3.259 m/s)/(-9.8 m/s^2)

= 0.665 s

Therefore, the duration of the final portion of the collision is 0.665 - 0.018 = 0.647 s, and the distance traveled during this time is

d = (3.0855 m/s)(0.647 s)

= 2.0006 m.

The average force exerted by the floor on the ball during this final portion of the collision is then: F_avg = Δp/Δt

= mv_f/Δt

= (0.06 kg)(3.259 m/s)/0.647 s

= 30.3882 N

This force is directed upward, so the final answer is that the magnitude of the average force exerted on the ball by the floor during the collision is 39.2 N, directed upward.

In summary, we can find the average force exerted on a ball by a floor during a collision by using conservation of energy to find the initial velocity of the ball just before it hits the floor, and then using the impulse-momentum theorem to relate the average force to the change in momentum of the ball during the collision. For this particular problem, we also needed to use kinematic equations to find the final speed and distance traveled by the ball just as it leaves the floor, and estimate the average force exerted during a final portion of the collision when the ball was still in contact with the floor. The final answer is that the magnitude of the average force exerted on the ball by the floor during the collision is 39.2 N, directed upward.

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Answer:

Fracturing can emit seismic waves through the ground.

Explanation:

I believe the answer is seismic, I've studied this before.

So the best way to show this is with a diagram

As shown here, the sun's light is blocked by the moon. This prevents the Earth from getting the sun's light

So the correct answer is A

Answer:

The answer is A: Light from the sun is blocked from earths surface by the moon

Explanation:

hope this helps

Answer:

The x-component of \(F_{3}\) is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\(\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O\) (1)

Where:

\(\vec F_{1}\), \(\vec F_{2}\), \(\vec F_{3}\) - External forces exerted on the ring, measured in newtons.

\(\vec O\) - Vector zero, measured in newtons.

If we know that \(\vec F_{1} = (70.711,70.711)\,[N]\), \(\vec F_{2} = (-126.859, 46.173)\,[N]\), \(F_{3} = (F_{3,x},F_{3,y})\) and \(\vec O = (0,0)\,[N]\), then we construct the following system of linear equations:

\(\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N\) (2)

\(\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N\) (3)

The solution of this system is:

\(F_{3,x} = 56.148\,N\), \(F_{3,y} = -116.884\,N\)

The x-component of \(F_{3}\) is 56.148 newtons.

Answer: There are seven main stores of energy: magnetic

internal (thermal)

chemical

kinetic

electrostatic

elastic potential

gravitational potential

Explanation: So you can have any of these

When the mass is doubled, the time period increases to 2.82 s. When mass is halved, it decreases to 1.41 s. When amplitude is doubled, no change happens. When spring constant is doubled, T will be 1.41 s.

The time period of oscillation is the time taken to complete one oscillation and it depends on the mass and spring constant. Let m be the mass and K be the spring constant, then time period T is

   T = 2π\(\sqrt{\frac{m}{K} }\)

Here 2π √m/K = 2 s

a) When mass is doubled, m changes to 2m

     T' = 2π √2m/K = √2T = √2 × 2 = 2.82 sec

b) When the mass is halved, m will become m/2

 So, T = 2π\(\sqrt{\frac{m}{2}* \frac{1}{K} }\) = \(\frac{1}{\sqrt{2} }\) × 2 = 1.41 s

c) Amplitude does not have any effect on the time period. So it will remain the same. So T will be 2 s.

d) When spring constant is doubled, K will become 2K

   So, T' = 2π ×√(m/2K) = \(\frac{1}{\sqrt{2} }\) × 2 = 1.41.

So the factors that affect time period are mass and spring constant.

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Answer:

It is in contact with two vertical rails which are joined at the top. The rails are without friction and resistance. There is a horizontal uniform magnetic field of magnitude B perpendicular to the plane of the ring and the rails.

Explanation:

Not sure if this is correct but pls give thanks anyway!

Answer:

Hey, bro here is the explanation....

Explanation:

Hope it helps...