A block of mass = 4.00 kg is supported by a spring scale with unit of measure of Newtons. This spring scale is attached to top of an elevator. The elevator accelerates upward with a = 3.00 m/sec2. What is the reading on the spring scale to 3 significant figures?

Answers

Answer 1

Answer:

118 N

Explanation:

Given mass of the block, m = 4.00kg.

The acceleration of the elevator, a = 3.0 m/s^2.

As elevotar attaced with spring scale and accelerating upward

(block and elevator), so total force

[tex]F_N-mg=ma[/tex]

Here, mg is the weight of the block downward direction.

or

[tex]F_N=ma+mg=m(g+a)[/tex]

substitute the given value, we get

[tex]F_N=4kg(9.8m/s^2+3m/s^2)[/tex]

     = 117.6 N = 118 N.

Thus, the reading on the spring scale to 3 significant figures is 118 N.


Related Questions

When light goes from one material into another material having a higher index of refraction, it

Answers

Answer

When the light goes from one medium to another medium with higher refractive index the velocity of the ray decreases, wavelength of the ray also decreases.

But the frequency of the ray when it enters the medium of higher refractive index remain same.

So, we can conclude that speed of ray and wavelength decrease but frequency remain unchanged.

Final answer:

When light moves from a material to another material with a higher index of refraction, it changes its path due to refraction.

Explanation:

When light goes from one material into another material having a higher index of refraction, it changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the lower refractive index.

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A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0 cm. If the frequency is low, the penny rides up and down without difficulty. If the frequency is steadily increased, there comes a point at which the penny leaves the surface.

1) At what point in the cycle does the penny first lose contact with the piston?

A. midpoint moving up

B. midpoint losing down

C. highest point

D. lowest point

Answers

Answer: C

Explanation:

Highest Point

A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to 0.8 cm at the nozzle exit. If it takes 50 s to fill the bucket with water, determine (a) the volume and mass flow rates of water through the hose, and (b) the average velocity of water at the nozzle exit.

Answers

Answer:

Explanation:

Given

Volume of bucket [tex]V=10\ gallon[/tex]

Time taken to fill the bucket [tex]t=50\ s[/tex]

so volume flow rate is [tex]\dot{V}=\frac{10}{50}=0.2\ gal/s[/tex]

1 gal is equivalent to [tex]0.133\ ft^3[/tex]

[tex]\dot{V}=0.0267\ ft^3/s[/tex]

mass flow rate [tex]\dot{m}=\rho \times \dot{V}[/tex]

[tex]\dot{m}=62.4\times 0.0267[/tex]

[tex]\dot{m}=1.668\ lbs[/tex]

(b)Average velocity through nozzle exit

[tex]\dot{V}=Av_{avg}[/tex]

[tex]v_{avg}=\dfrac{0.0267}{\frac{\pi}{4}\times (0.0262)^2}[/tex]

[tex]v_{avg}=49.51\ ft/s[/tex]

Final answer:

The volume flow rate is 757.082 cm^3/sec and the mass flow rate is 757.082 g/sec. The average velocity of the water at the nozzle exit is approximately 476.677 cm/sec.

Explanation:

To solve this problem, we first need to figure out the volume and mass flow rates of the water. Given that 1 gallon = 3.78541 liters, and 1 liter = 1,000 cm3, a 10-gallon bucket contains 10 * 3.78541 * 1000 = 37,854.1 cm3. If it takes 50 seconds to fill this bucket, we can calculate the volume flow rate as volume/time = 37,854.1 cm3/50 s = 757.082 cm3/s.

The mass flow rate can be determined by multiplying this volume flow rate by the density of water (1 g/cm3), giving a result of 757.082 g/s.

To find the average velocity of the water at the nozzle exit, we use the equation of continuity which states that the volume flow rate must be constant at all points in the pipe. Therefore, the velocity at the nozzle can be found by dividing the flow rate by the cross-sectional area of the nozzle (pi * (d/2)2). This gives an average velocity of about 476.677 cm/s.

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A wire of arbitrary shape, which is confined to the x-y plane, carries a current i from point a to point b in the plane. show that if a uniform magnetic field b→ perpendicular to the x-y plane is present, the force that the wire experiences is the same as that which would be felt by a wire running straight from a to



b.

Answers

Answer:

See explanation

Explanation:

Solution:-

- A wire of arbitrary shape,which is confined to the x-y plane,carries a current I from point A to point B in the x-y plane.

- See diagram (attached) for clarity.

- Let’s assume that the horizontal distance between A and B is "s" and the vertical distance between A and B is "d". Then for the straight line path vector ( L ):

                    L = s i^ + d j^

- The force on the straight wire with current I is then:

                    F = I * ( L x B )

Where,  L: The path vector between points A and B

             B: The magnetic field strength vector

For the curved wire vector "ds = dx i^ + dy j^" and the force on the wire is:

                   F = ∫ [ I (ds x B) = I ∫ (dx i^ + dy j^) x B

When current "I" and magnetic field "B" are uniform then we can pull both of them out of the integral. Separate the integral and calculate each differential separately:

                  F = I ∫ (dx i^) x B + I ∫ (dy j^) x B

                     = I (s i^ x B) + I ( d j^ x B ) = I ( L x B )

- The force of curved and straight line have the same force:

                 F = I ( L x B ) acting on them.

                   

                     

In my solar system, we have a planet that is the innermost to our star that is exactly like the innermost planet in your solar system. The planet has a very large impact crater with a basin that covers a large region of the planet's surface, but few smaller craters have formed on top of it. What can one conclude from this?

Answers

Answer:

1. The planet doesn't have a thick enough atmosphere.

2. There have been multiple impacts on the planet.

Explanation:

As the planet is very close to the star, there is high possibility that it will not have an atmosphere. Just like Mercury doesn't have one. Presence of a very large crater with basin indicates that in the past a huge body had hit the planet and thus creating the crater with basin. Also, it must be very old.

Second observation that is given is the presence of smaller craters in the basin. This indicates impact craters created by smaller objects. If the planet had an atmosphere, these smaller objects would not be able to penetrate and reach the surface. Thus presence of these smaller crater indicate towards the planet not having any atmosphere.

Two technicians are discussing oil filters. Technician A says that the oil will remain perfectly clean if just the oil filter is changed regularly. Technician B says that oil filters can filter particles smaller that the human eye can see. Which technician is correct?
A)A only
B)B only
C)Both A and B
D)Neither A nor B

Answers

Answer:

C. Both technician A and technician B are correct

Explanation

Engine performs at its best with clean oil. That's why oil filters work to sift out particles (some so small that humans can't see ) to keep contaminants out of the oil. Dirty or clogged oil filters allow contaminants to sail straight to the engine where they can cause damage as well as affect fuel economy.

Over time If the oil filter is changed regularly it will filter perfectly clean oil,

Final answer:

Technician B is correct; oil filters can filter out small particles, while Technician A is incorrect; changing the oil filter alone doesn't keep the oil perfectly clean.

Explanation:

The question pertains to the role and effectiveness of oil filters in vehicle maintenance. Technician B is correct in saying that oil filters can filter particles smaller than the human eye can see. Modern oil filters are designed to remove very small particulates from engine oil, thus protecting the engine from wear and damage. However, Technician A is incorrect; changing the oil filter alone will not keep the oil perfectly clean over time. The oil itself can degrade and become contaminated with substances that an oil filter cannot remove, such as acids and water, which is why regular oil changes are also necessary. Therefore, the correct answer is B) Technician B only.

the two forces acting on a falling object are gravity and?....1.force, 2. friction, 3. air resistance, and 4. net force?​

Answers

The two forces acting on a falling object are gravity and air resistance (option 3). Gravity causes the object to accelerate toward Earth, while air resistance acts in the opposite direction, slowing the object down.

The two forces acting on a falling object are gravity and air resistance (option 3). When an object is dropped, it accelerates toward the center of the Earth due to the force of gravity. In a vacuum, where there is no air, the only force acting on a falling object is its weight, which is the force due to gravity acting on an object of mass m. However, in the real world, objects are not in perfect free-fall because they experience air resistance, which opposes the motion of the object as it falls through the air.

. If a substance has an excess number of electrons on its surface, what type of charge does it have?
A positive charge.

A negative charge.

A neutral charge.

A nuclear charge.

Answers

Answer:

A negative charge

Explanation:

If there is an excess of electrons in a substance, this will have a negative net charge. This is because electrons have negative charge

q = -1.67*10^{-19}

each electron will contribute with this charge to the total negative net charge of the substance.

hence, the answer is "A negative charge"

A substance with an excess number of electrons on its surface has a negative charge due to the negatively charged nature of electrons.

If a substance has an excess number of electrons on its surface, it will have a negative charge. This is because electrons have a negative charge, and having more electrons than protons results in a net negative charge for the substance. By contrast, if a substance had more protons than electrons, it would have a positive charge. Substances are typically electrically neutral when the number of electrons equals the number of protons, and this balance means there is no net electric charge.

An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Conversely, when an atom loses one or more electrons, it becomes a positively charged atom, also known as a cation. For instance, when a neutral sodium atom loses one electron, it becomes a sodium cation with a 1+ charge. Conversely, an oxygen atom that gains two electrons becomes an anion with a 2- charge.

A worker pushes a wheelbarrow with a horizontal force of 50 N on level ground over a distance of 5.0 m. If a friction force of 43 N acts on the wheelbarrow in a direction opposite that of the worker, what work is done on the wheelbarrow by the worker?a) 250 J
b) 215 J
c) 35 J
d) 10 J
e) None of these answers is correct.

Answers

Answer:

Work done will be equal to 35 J

So option (c) will be correct answer

Explanation:

We have given force F = 50 N

Frictional force f = 43 N

So net force will be equal to [tex]F_{net}=50-43=7N[/tex]

Distance covered on the ground d = 5 m

We have to find the work done

Work done is equal to [tex]W=F_{net}\times d=7\times 5=35J[/tex]

So option (C) will be the correct answer.

The correct option is A 250 J.

Given, Horizontal  force is 50 N.

Frictional force is 43 N.

The distance between worker and pushing of wheelbarrow is 5 m.

We know that, Work is a physics term describing the amount of energy transferred when it is moved over a distance by an external force.

So Work = Force . Displacement

 work = F D cos [tex]\Theta[/tex]......(eq 1)

Here [tex]\Theta[/tex] is 0 .

So, [tex]work = 50 \times 5 \times 1[/tex] ( cos 0 = 1 )

Work = 250 J

Hence 250 J of work is done on the wheelbarrow by the worker.

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Light having a speed in vacuum of3.0�108m/s enters a liquid of refractive index 2.0. In this liquid, its speed will be
A)0.75�108m/s
B)6.0�108m/s
C)1.5�108m/s
D)3.0�108m/s
E)None of the above choices are correct.

Answers

Explanation:

Given that,

Speed of light in vacuum, [tex]v=3\times 10^8\ m/s[/tex]

Refractive index of the liquid, n = 2

We need to find the speed of light in the liquid. Th refractive index of material is given by the ratio of speed of light in vacuum to the speed in the medium. i.e.

[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2}\\\\v=1.5\times 10^8\ m/s[/tex]

So, the correct option is (c). Hence, this is the required solution.

A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform in all directions, how much acoustic power is emitted by the whistle? The threshold of human hearing is 1.0 × 10-12 W/m2.

Answers

Answer:

Emitted power will be equal to [tex]7.85\times 10^{-5}watt[/tex]

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing [tex]I=10^{-12}W/m^2[/tex]

We have to find the emitted power

Emitted power is equal to [tex]P=I\times A[/tex]

[tex]P=I\times 4\pi R^2[/tex]

[tex]P=10^{-12}\times 4\times 3.14\times 2500^2=7.85\times 10^{-5}watt[/tex]

So emitted power will be equal to [tex]7.85\times 10^{-5}watt[/tex]

The layer of the sun that radiates most of the light that reaches earth is the

Answers

Answer:

photosphere

Explanation:

Photosphere, visible surface of the Sun, from which is emitted most of the Sun’s light that reaches Earth directly. Since the Sun is so far away, the edge of the photosphere appears sharp to the naked eye, but in reality the Sun has no surface, since it is too hot for matter to exist in anything but a plasma state—that is, as a gas composed of ionized atoms.

Source :

Photosphere astronomy

Written By:. Harold Zirin

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time it took for his voice to reach the earth via radio waves. (b) Someday a person will walk on Mars, which is from the earth at the point of closest approach. Determine the minimum time that will be required for a message from Mars to reach the earth via radio waves.

Answers

Answer:

a) It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

Gamma rays X-rays Ultraviolet rays Visible region InfraredMicrowave Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of [tex]3x10^{8}m/s[/tex].  

a) Find the time it took for his voice to reach the Earth via radio waves.

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

[tex]c = \frac{d}{t}[/tex]  (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

[tex]t = \frac{d}{c}[/tex]  (2)

The distance from the Earth to the Moon is [tex]3.85x10^{8} m[/tex], therefore.

[tex]t = \frac{3.85x10^{8} m}{3x10^{8}m/s}[/tex]

[tex]t = 1.28s[/tex]

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.

The distance from the Earth to the Mars at its closest approach is [tex]5.76x10^{10}m[/tex], therefore.

[tex]t = \frac{5.76x10^{10}m}{3x10^{8}m/s}[/tex]

[tex]t = 192s[/tex]

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

Calculate the electrostatic force that a small sphere A, possessing a net charge of 2.0 x 10^-6 Coulombs exerts on another small sphere, B with a net charge of -3.0 x 10^-6, when the distance between their centers is 10.0 meters k= 8.99 x10^9

Answers

Answer:

F = 5.33*10^-4N

Explanation:

to find the electrostatic force you use the Coulomb's law, given by the formula:

[tex]F=k\frac{q_Aq_B}{r^2}[/tex]

k: Coulomb's constant = 8.89*10^9 Nm^2/C^2

q_a: charge of A = 2.0*10^{-6}C

q_B: charge of B = -3.0*10^{-6}C

r: distance between the spheres = 10.0m

By replacing all these values you obtain:

[tex]F=(8.89*10^9Nm^2/C^2)\frac{(2.0*10^{-6}C)(-3.0*10^{-6}C)}{(10.0m)^2}=5.33*10^{-4}N[/tex]

hence, the forcebetween the spheres is about 5.33*10^-4N

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 126-mile trip in a typical midsize car produces about 2.99 x 109 J of energy. How fast would a 45.8-kg flywheel with a radius of 0.512 m have to rotate to store this much energy? Give your answer in rev/min.

Answers

ω = ?

mass = 45.8kg

r = 0.512m

E = 2.99*10⁹J

Kinetic Energy of rotation = I * ω²

K.E = I * ω²

I = ½ m*r²

I = ½ * 45.8 * (0.512)²

I = 6.0kgm²

K.E = ½ * I * ω²

ω = √(2K.E / I )

ω = √[( 2* 2.99*10⁹) / 6]

ω = 3.157*10⁴ rad/s

the intensity at the central maximum of a double-slit interference pattern is 4I_1. The intensity at the first minimum is zero. at what fraction of the distance from the central maximum to the first minimum is the intensity I_1? Thank you.

Answers

Final answer:

The fraction of the distance from the central maximum to the first minimum where the intensity is I1 is 1/2.

Explanation:

The intensity at the central maximum of a double-slit interference pattern is 4I1 and the intensity at the first minimum is zero. To find the fraction of the distance from the central maximum to the first minimum, we can use the formula for intensity in double-slit interference:

I = 4I1 cos2(πd/λ)

At the first minimum, the intensity is zero. Therefore, we have:

0 = 4I1 cos2(πd/λ)

Simplifying this equation, we find that cos2(πd/λ) = 0.

If cos2(πd/λ) = 0, then cos(πd/λ) = 0.

And if cos(πd/λ) = 0, then πd/λ = π/2.

Solving for d/λ, we get d/λ = 1/2. Therefore, the fraction of the distance from the central maximum to the first minimum where the intensity is I1 is 1/2.

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The diagram below shows eight different positions of the moon around Earth.
Which two positions show the location of the moon in its crescent phases as seen from Earth?

Group of answer choices

2 and 4

2 and 8

4 and 6

6 and 8

Answers

Answer:

6 and 8

Explanation:

The different positions of the moon, as seen from Earth, reflect how much light is being reflected off the moon from the sun.

Position 1 indicates the 3rd Quarter.

Position 2 indicates the Waning Gibbous.

Position 3 indicates the Full Moon.

Position 4 indicates the Waxing Gibbous.

Position 5 indicates the 1st Quarter.

Position 6 indicates the Waxing Crescent.

Position 7 indicates the New Moon.

Position 8 indicates the Waning Crescent.

You would read the diagram counterclockwise, with positions 7 to 3 as the moon increases light and positions 3 to 7 as the moon decreases light.

Therefore, we see that our 2 choices where we have a crescent is positions 6 and 8.

Which is true about the potential energy stored in an inductor?

Answers

Answer:

It depends on the current

Explanation:

The formula for the energy stored in an inductor is W =1/2 *L*I2

We can state that an electric current flowing in inductor , there is energy stored in the magnetic field. let consider the pure inductor L, and the power which is supplied to initiate the current in inductor.

consider the equation P = i*v

write the equation in terms of inductor.

P = L*i di/dt

P*dt = L*i di

rewrite the Energy equation W = \int_{0}^{t}P*dt

W = \int_{0}^{t} L*i di

by solving we will get the W = 1/2 *L*I2

From this information and formula we can state that he energy stored in an inductor depends on the current.

Answer:

The complete option for the question

a.it depends on the self induced emf

b it depends on the inverse of the self induced emf

C.it depends on the current

d.it depends on the inverse of current

Answer is.it depends on the current

Explanation:

The energy stored in an inductor is W=1/2LI^2

From this equation it can be seen that energy in inductor L depends on current L

The distance between slits in a double-slit experiment is decreased by a factor of 2. If the distance between fringes is small compared to the distance from the slits to the screen, the distance between adjacent fringes near the center of the interference pattern _______.

Answers

Final answer:

The distance between adjacent fringes near the center of the interference pattern in a double-slit experiment decreases by the same factor when the distance between the slits is decreased by a factor of 2.

Explanation:

The distance between adjacent fringes near the center of the interference pattern in a double-slit experiment is determined by the wavelength of the light and the distance between the slits. When the distance between the slits is decreased by a factor of 2, the distance between adjacent fringes near the center of the interference pattern also decreases by the same factor. This is because the interference pattern is directly related to the slit separation.

quzilet The Magnet Recognition Program for health care organizations is based on fourteen forces of magnetism related to five magnet model components. Which force of magnetism is assessed to review the structural empowerment of the organization?

Answers

Answer:

4. Personnel policies and programs

Explanation:

The Health care organizations that apply for Magnet status must demonstrate new ways of doing things and innovations in professional practice.

Personnel policies and programs is one of the forces of magnetism that impacts the structural empowerment of the organization.

The Personnel policies of an organization should provide an innovative environment in which the staff are developed and empowered. Empirical quality outcomes are reviewed by assessing the quality of care. New knowledge, innovations, and improvements are reviewed by assessing the quality improvement of the health care organization. Interdisciplinary relationships are assessed to review exemplary professional practice.

Final answer:

The Magnet Recognition Program evaluates 'Structural Empowerment' to review the structural empowerment of a healthcare organization. It accounts for professional engagement, community involvement, personnel policies and programs, professional development, and diversity.

Explanation:

The Magnet Recognition Program, a healthcare organization accreditation program initiated by the American Nurses Credentialing Center (ANCC), assesses the performance of hospitals based on 14 forces of magnetism that form five model components. The 14th force, known as Structural Empowerment, is evaluated to review the ability of the organization to provide an infrastructure that supports professional development and interprofessional collaboration. It includes professional engagement, community involvement, personnel policies and programs, professional development, and diversity.

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Tall towers support power lines h = 59 m above the ground and 19 m apart that run from a hydroelectric plant to a large city, carrying 60 Hz alternating current with amplitude 4 104 A. That is, the current in both of the power lines is 1 = (4 × 104 A) sin(2n-60 Hz . t). out in Base
(a) calculate the amplitude largest magnitude and direction of the magnetic field produce by the power lies at ne base of the tower, when a current of x 104 A in the left power line is headed out of the page, and a current of 4 x 104 A in the right power line is headed into the page. magnitude direction up
(b) This magnetic field is not large compared to the Earth's magnetic field, but it varies in time and so might have different biological effects than the Earth's steady field. For a person lying on the ground at the base of the tower, approximately what is the maximum emf produced around the perimeter of the body which is about 2 m long by half a meter wide? mw

Answers

Answer:

a) 4.25 x 10∧-5 T

b) 16.03 mv

Explanation:

the solution is shown in the pictures attached

Final answer:

The maximum magnetic field that could be induced at the base of the power line towers would be around 1.67 µT, directed upwards. For a person lying down near the base of these power lines, the approximate maximum emf produced around their perimeter would be about 5.6 mV. The magnetic field strength is quite weak and its likelihood to cause biological impact is low.

Explanation:

First, for (a), the field is the result of both sources, with one of the currents in opposite direction. So, we can just take the double of it, giving us that the magnitude of magnetic field is B = (2μ₀I)/(2πr). Plugging the known values in, we get B = 2 x (4π x 10⁻⁷ Tm/A) x (4x10⁴A) / (2πx19m) approximately equal to 1.67 µT directed upward.

For (b), we will consider the person as a rectangular coil. In this case, we can use Faraday's Law of Electromagnetic Induction which states that emf = -N ∆Φ/∆t, where ∆Φ is the change in magnetic flux and N is the amount of loops or turns. Here, N=1 as the person is considered as a single loop.

The flux changes from B*A to -B*A, so ∆Φ = 2BA. Resulting, emf = (2BA Δt) with Δt = 1/120 second (as it's 60 Hz, so frequency = 2 per every 1/60 seconds). Plugging all the known values, we get emf approximately equal to 0.0056 V or 5.6 mV. Even though it changes over time, this magnetic field strength is still very weak and thus, is less likely to cause biological effects.

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A swinging pendulum has a total energy of [tex]E_i[/tex]. The amplitude of the pendulum's oscillations is then increased by a factor of 4. By what factor does the total energy stored in the moving pendulum change? Ignore damping.

Answers

Final answer:

The total energy of the pendulum is directly proportional to the square of the amplitude. Hence, if the amplitude is increased by a factor of 4, the total energy in the system will increase by a factor of 16.

Explanation:

The subject of your question is related to the energy of a pendulum in simple harmonic motion. The total energy stored in an oscillating pendulum depends on the square of its amplitude. The total energy E of an oscillator is the sum of its kinetic energy K = mu² / 2 and the elastic potential energy of the force U(x)= k_x²/2 (formula for potential energy in a harmonic oscillator). These two forms of energy: kinetic energy and potential energy oscillate back and forth but their sum remains constant.

Since the total energy ETotal = (1/2)kA² is proportional to the square of the amplitude, if the amplitude of the pendulum's oscillations is increased by a factor of 4, this means you're multiplying the square of the quantity by 4. As a result, the energy in the pendulum increases by a factor of 4² which is 16.

So, as the amplitude of the pendulum increases by a factor of 4, the total energy increases by a factor of 16.

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Final answer:

When the amplitude of a swinging pendulum is increased by a factor of 4, the total energy stored in the moving pendulum increases by a factor of 16.

Explanation:

The total energy of a swinging pendulum is given by the sum of its potential energy and kinetic energy, which is proportional to the square of the amplitude.

In this case, when the amplitude is increased by a factor of 4, the total energy increases by a factor of 16.

This is because the total energy is directly proportional to the square of the amplitude, so increasing the amplitude by a factor of 4 results in an increase of the total energy by the square of that factor.

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A radiographer stands six feet from the x-ray source when performing a portable chest exam and receives an exposure of 2 mGy. If the radiographer performs a repeat exam using the same technical factors standing at a distance of three feet from the source, how much exposure will be received?

Answers

Answer:

  I₂ = 8 mG

Explanation:

The intensity of a beam is

          I = P / A

Where P is the emitted power which is 3) 3

           

Let's use index 1 for the initial position of r₁ = 6 ft and 2 for the second position r₂ = 3 ft

          I₁ A₁  = I₂  A₂

           I₂ = I₁ A₁ / A₂

The area of ​​the beam if we assume that it is distributed either in the form of a sphere is

           A₁ = 4π r²

We substitute

            I₂ = I₁ (r₁ / r₂)²

           I₂ = 2 (6/3)²

           I₂ = 2 4

           I₂ = 8 mG

Answer:

The amount of exposure that will be received at 3 ft is 8 mGy

Explanation:

Here, we note that the  amount of radiation exposure of the radiographer is given by the inverse square law. That is the amount of radiation exposure is directly proportional to the inverse square of the distance that is

[tex]\frac{Old \, \, Intensity}{New \, \, \, Intensity} = \frac{(New\, distance)^2}{(Old\, distance)^2} \therefore \frac{2}{New \, \, \, Intensity} = \frac{3^2}{6^2}[/tex]

Or New intensity = [tex]2\times \frac{36}{9}[/tex]  = 8mGy

Therefore, the amount of exposure that will be received at 3 ft = 8 mGy.

An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.7 s. A passenger in the elevator is holding a 5.1 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates

Answers

Answer: Tension = 53.6N

Explanation:

Given that

Height h = 1 m

Time t = 1.7 s.

Mass m = 5.1 kg 

From the equation of the motion we can get the acceleration of the elevator:

h = X0+ V0t + at2/2;

Th elevator starts from rest with a constant upward acceleration. Initial velocity Vo = 0, also Xo = 0; thus

a = 2h/t2 = 2 × 1/1.7^2

a = 0.69 m/s2.

Then we can find the tension in the cord by using the formula

T = mg + ma

= 5.1 (9.8 + 0.69)

= 5.1 × 10.5

= 53.6N

If a spring has a spring constant of 1.00 × 10^3 N/m, what is the restoring force when the mass has been displaced 20.0 cm?

Answers

The mass that should be displaced 20.0 cm should be 20000 N/m.

Calculation of the mass:

At the time When a spring should be stretched or compressed its length so it changed by an amount x from its equilibrium length due to which the restoring force is exerted.

here,

spring constant is [tex]k = 1.00 * 10^3 N/m[/tex]

mass is x = 20.0 cm

Now

As per Hooke's law

We know that

F = - kx

[tex]= - 1.00 *10 ^3 * 20.0[/tex]

F = 20000 N/m

Hence, The mass that should be displaced 20.0 cm should be 20000 N/m.

Learn more about mass here: https://brainly.com/question/18358825

To find the restoring force, use Hooke's Law: F = -kx. With a spring constant of 1.00 × 10³ N/m and a displacement of 20.0 cm (0.200 m), the restoring force is 200 N.

To answer this, we use Hooke's Law, which states:

F = -kx

Where:

F indicates the restoring force

k denotes the spring constant

x indicates the displacement

Given the spring constant k = 1.00 × 10³ N/m and displacement x = 20.0 cm (which we convert to meters, so x = 0.200 m), we substitute these values into the equation:

F = - (1.00 × 10³ N/m) × (0.200 m)

This simplifies to:

F = -200 N

The negative sign indicates that the force is directed opposite to the direction of displacement, which is typical for restoring forces. Therefore, the restoring force is 200 N.

A pilot, whose mass is 84.0 kg, makes a loop-the-loop in a fast jet. Assume that the jet maintains a constant speed of 345 m/s and that the radius of the loop-the-loop is 3.033 km. What is the apparent weight that the pilot feels (i.e., the force with which the pilot presses against the seat) at the bottom of the loop-the-loop?

Answers

Final answer:

The apparent weight that the pilot feels at the bottom of the loop-the-loop is calculated by summing the gravitational force and the centripetal force, resulting in an apparent weight of 4012.9 N.

Explanation:

The student's question is asking to calculate the apparent weight that a pilot feels at the bottom of a loop-the-loop. To find the apparent weight, we must calculate the normal force on the pilot, which is the combination of the gravitational force and the centripetal force required to keep the pilot in a circular path at the bottom of the loop.

Firstly, we calculate the gravitational force (weight) acting on the pilot: W = m × g, where m is the mass of the pilot, and g is the acceleration due to gravity. For the pilot with a mass of 84.0 kg, W = 84.0 kg × 9.8 m/s² = 823.2 N.

Next, we determine the centripetal force required for circular motion at the bottom of the loop: Fc = (mv²) / R, where m is the mass, v is the constant speed, and R is the radius of the loop. With v = 345 m/s and R = 3.033 km (or 3033 m), Fc = (84.0 kg × (345 m/s)²) / 3033 m = 3189.7 N.

The apparent weight of the pilot at the bottom of the loop is the sum of the gravitational force and the centripetal force: Apparent weight = W + Fc = 823.2 N + 3189.7 N = 4012.9 N.

Therefore, the pilot feels an apparent weight of 4012.9 N at the bottom of the loop-the-loop.

What are the methods that are used for heat transfer

Answers

Answer:

Heat can travel from one place to another in three ways: Conduction, Convection and Radiation. Both conduction and convection require matter to transfer heat. If there is a temperature difference between two systems heat will always find a way to transfer from the higher to lower system.

Explanation:

Hope this helps

A loudspeaker diaphragm is producing a sound for 4.6 s by moving back and forth in simple harmonic motion. The angular frequency of the motion is 2.57 x 104 rad/s. How many times does the diaphragm move back and forth

Answers

Number of times the  diaphragm move back and forth is  5.59×10^4

Explanation:

Given data,

ω=4.6 s

we have the formula

f=ω/2π

The number of times the diaphragm moves back and forth in 4.6 s is

Number of times= ft

Number of times= ft

                           =(ω/2π) t

                          =(7.54×10^4 rad/sec)(4.6 s)/2π

Number of time=5.59×10^4

Number of times the  diaphragm move back and forth is  5.59×10^4

             

Assume a beam of light hits the boundary separating medium 1, with index of refraction n1 and medium 2, with index of refraction n2. If total internal reflection occurs at the boundary that separates medium 1 and medium 2, then we know which of the following? a) n1= n2 b) n1< n2 c) n1> n2 d) n1 ≥ n2

Answers

Answer:

option C

Explanation:

Given,

Refractive index of medium 1 = n₁

Refractive index of medium 2 = n₂

For total internal reflection to take place light should move from denser medium to the rarer medium.

Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2

 n₁ > n₂

The correct answer is option C

When the person ran that blue light over the beads they_________,
The blue light bar that the person was using must therefore be emitting_____________.

A. Changed Colors; UV light
B. Exploded; Heat
C. Stayed White; Energy

Answers

Answer:

A Changed colors: UV light

Explanation:

Was a uv light, reacted just like the sun did.

Answer:

A changed colors

Explanation:

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