How much work, in kJ, can a spring whose spring constant is 6 kN/cm produce after it has been compressed 3 cm from its unloaded length?

Answers

Answer 1

Final answer:

The spring can produce 0.270 kJ of work after it has been compressed 3 cm from its unloaded length, with a spring constant of 6 kN/cm.

Explanation:

The work done by a spring is given by the expression ½ kx², where k is the spring constant and x is the displacement from the spring's equilibrium position. In this case, we are given a spring constant of 6 kN/cm and a compression of 3 cm. First, convert the spring constant to N/m by multiplying by 10³ (since 1 kN = 10³ N and 1 cm = 10⁻² m), which gives us 600,000 N/m. The work can then be calculated as:

W = ½ kx² = ½ (600,000 N/m) (0.03 m)² = 270 J

Convert joules to kilojoules, 1 kJ = 1,000 J, gives:

W = 0.270 kJ

Therefore, the spring can produce 0.270 kJ of work after being compressed 3 cm from its unloaded length.


Related Questions

A major-league pitcher can throw a baseball in excess of 44.7 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 16.1 m away from the point of release?

Answers

The baseball drops by 0.64 meter.

Explanation:

Consider the horizontal motion of ball

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 44.7 m/s

        Acceleration, a = 0 m/s²  

        Displacement, s = 16.1 m      

     Substituting

                      s = ut + 0.5 at²

                      16.1 = 44.7 x t + 0.5 x 0 x t²

                      t = 0.36 s

      Time taken to travel 16.1 m is 0.36  seconds

Now we need to find how much ball travel vertically during this 0.36 seconds.

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 0.36 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 0.36 + 0.5 x 9.81 x 0.36²

                      s = 0.64 m

     The baseball drops by 0.64 meter.

You are not harmed by contact with a charged metal ball, even though its voltage may be very high. Is the reason similar to why you are not harmed by the hotter- than-1000°C sparks from a fireworks sparkler?

Answers

Final answer:

You are not harmed by a charged metal ball or by fireworks sparks because the actual exposure (amount of charge and duration for the ball, and size and duration for the sparks) is insufficient to cause injury.

Explanation:

The reason you are not harmed by contact with a charged metal ball, even if its voltage may be very high, is similar to why you are not harmed by the hotter-than-1000°C sparks from a fireworks sparkler. In both cases, it is the amount of charge and the duration of the contact that determine the danger of the electric shock, not just the voltage alone. For the metal ball, even though the voltage is very high, the resulting current is usually very low because the charge moves to the outer surface and is quickly neutralized by surrounding materials. Consequently, the exposure to the charge is so brief that it doesn't cause harm. Similarly, with the sparks from a sparkler, they may be extremely hot, but they are small and lose their heat quickly upon contact with the skin, so they don't transfer enough energy to cause a burn.

How high will a 1.30-kg rock go from the point of release if thrown straight up by someone who does 80.0 J of work on it? Neglect air resistance.

Answers

Height it will go is 6.27 m

Explanation:

The work done on the stone is kinetic energy.

Kinetic energy = 80 J

We have

        Kinetic energy = 0.5 x Mass x Velocity²

        Mass = 1.30 kg

       80 = 0.5 x 1.30 x Velocity²

       Velocity = 11.09 m/s

We have equation of motion v² = u² + 2as

Initial velocity, u = 11.09 m/s  

Acceleration, a = -9.81 m/s²  

Final velocity, v = 0 m/s  

Substituting  

v² = u² + 2as

0² = 11.09² + 2 x -9.81 x s

s = 6.27 m  

Height it will go is 6.27 m

To find the height a 1.30-kg rock will reach when thrown vertically upwards after 80.0 J of work is done on it, convert the work into potential energy and use the formula PE = mgh. The rock will reach a height of about 6.14 meters.

Work-energy principle: The work done on an object when thrown vertically upwards is stored as potential energy at the highest point of the rock's trajectory. In this case, with 80.0 J of work done on a 1.30-kg rock, the rock will reach a height determined by the conversion from work to potential energy.

Calculating the height: By converting the work done into potential energy and using the formula PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration due to gravity (9.81 m/s^2), and h is the height, you can calculate the height the rock will reach.

Answer: The rock will reach a height of approximately 6.14 meters from the point of release.

An AM radio station broadcasts at a frequency of 600 kHz. If these waves have a speed of 3 x 108 m/s, then what is their wavelength in meters?

Answers

Answer: 500metres

Explanation:

Wavelength is defined as the distance between two successive crest and troughs. The relationship between the wavelength (¶), frequency (f) and velocity of the wave(v) is given by v= f¶

Therefore wavelength ¶ = v/f

Given the velocity = 3×10^8m/s

frequency = 600kHz = 600,000Hz

Wavelength =3×10^8/600,000

Wavelength = 500meters

Their wavelength in meters is 500meters

In solar thermal electric generation, the sun’s energy is concentrated by ___ to heat the fluid in a pipe that is used to generate electricity.

Answers

Answer:

Mirrors

Explanation:

In Solar thermal power generation system, mirrors are used to collect and focus sunlight  onto a fluid pipe to produce high temperature required to generate electricity.

There are two main components: mirrors capture and focus light onto a receiver. The heat transfer fluid is circulated in the receiver. Steam is produced which is converted to mechanical energy in a turbine. It powers a generator and electricity is produced.

In solar thermal electric generation, concave mirrors are used to concentrate sunlight onto a pipe in a system known as a parabolic trough. The heat is transferred via heat exchanger to produce steam, which generates electricity through a steam turbine-generator setup.

In solar thermal electric generation, the sun's energy is concentrated by concave mirrors or reflectors to heat the fluid in a pipe that is used to generate electricity. These mirrors are often part of a system known as a parabolic trough or concentrating collector, which has the approximate shape of a section of a cylinder, specifically one-quarter of a full cylinder. The mirrors focus sunlight onto a blackened pipe containing a special fluid.

Once this fluid is heated, it is then pumped through a heat exchanger where its thermal energy is transferred to water to produce steam. This steam drives a turbine connected to a generator, producing electricity through a conventional steam cycle. This entire process is a part of larger solar thermal power plants, which utilize solar collectors with two components: mirrors to collect and focus sunlight, and a receiver to collect the heat transferred to the fluid.

An electric dipole is formed from two charges, , spaced 1.0 cm apart. The dipole is at the origin, oriented along the y-axis. The electric field strength at the point (x,y) = (0,10) in cm is 360 N/C. a. Calculate the dipole moment and charge q. b. Calculate the electric field strength at the point (x,y) = (10, 0) in cm.

Answers

Answer:

q = 3.703*10^-9

dipole moment = 3.703*10^-11

E @ (10,0) = 331.977 N/C

Explanation:

Given:

Coordinates

+ q = (0,-0.5) cm

- q = (0,0.5) cm

E = 360 N/C @ (0,10)

Solution:

[tex]Electric Field Strength @ (0,10) = \frac{k*q}{R^2}\\[/tex]

Note: +q (away from charge) and -q (towards the charge)

[tex]E = k*q*(\frac{1}{R^2_{1}} - \frac{1}{R^2_{2} })\\ 360 = (9*10^9)*q*(\frac{1}{0.095^2} - \frac{1}{0.105^2 })\\\\q =1.99 *10^-9[/tex]

Dipole moment = q.(space between two charges)

= (3.703*10^-9 * 0.01)

= 3.703*10^-11

Electric Field Strength @ (10,0)

[tex]E = 2*k*q*(\frac{1}{R^2})*cos(Q)\\ \\Q = arctan(\frac{10}{0.5}) = 87.138 degrees\\R = \sqrt{10^2 + 0.5^2} = 10.01249 cm\\E = 2* (9*10^9)*(3.703*10^-9)*\frac{cos (87.138)}{0.1001249^2}\\E = 331.977 N/C[/tex]

Answer:

a. dipole moment = 2.003 × 10⁻¹³ Cm, charge = 2.003 × 10⁻¹¹ C b. 8.96 N/C

Explanation:

a. The electric field due to a dipole is given by

E = p/2πεy³ where p = dipole moment and y = distance of dipole to point of electric field = 10 cm = 0.10 m. E = electric field strength at (0,10) = 360 N/C

So, E = p/2πεy³

p = 2πεy³E = 2π × 8.854 × 10⁻¹² × (0.10 m)³ × 360 N/C = 2.003 × 10⁻¹³ Cm

Also, p = qd where q = charge and d = distance of charges apart = 1.0 cm = 0.01 m

q = p/d = 2.003 × 10⁻¹³ Cm/0.01 m = 2.003 × 10⁻¹¹ C

b. The electric field at point (10,0)

E = qx/[πε√(d² + 4x²)]³ which is the electric field at an axis perpendicular to the dipole where x = 10 cm = 0.10 m

E = 2.003 × 10⁻¹¹ C × 0.10/π × 8.854 × 10⁻¹²√(0.01² + 4×(0.1)²)³  = 8.96 N/C.

A nerve signal is transmitted through a neuron when an excess of Na+ ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately 10.0 μm in diameter, and measurements show that about 5.60×1011 Na+ions per meter (each of charge +e) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of nearly point charges moving along the axon. Let us look at a 0.100 mm length of the axon and model it as a point charge.

1) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 cm below the skin? Note that distance to the skin is much greater than the length of the charged axon so that the axon charge is effectiely point-like.,
2) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100 mm length of the axon? Answer in C

3) the field is directed away from the axon or the field is directed into the axon

Answers

Final answer:

The electric field produced at the surface of the body due to the sudden influx of charge in the axon can be calculated using Coulomb's law. If the charge that enters each meter of the axon is distributed uniformly along it, the total charge that enters a 0.100 mm length of the axon can be calculated. The electric field produced by the influx of Na+ ions is directed away from the axon.

Explanation:

1) The electric field produced at the surface of the body due to the sudden influx of charge in the axon can be calculated using Coulomb's law. The formula for electric field is E = k * (Q / r^2), where k is the electrostatic constant, Q is the charge, and r is the distance. By substituting the given values, we can find the magnitude and direction of the electric field.

2) If the charge that enters each meter of the axon is distributed uniformly along it, we can calculate the total charge that enters a 0.100 mm length of the axon by multiplying the charge per meter by the length in meters.

3) Since the excess of Na+ ions enter the axon, the electric field produced by them will be directed away from the axon.

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A 66-kg diver jumps off a 9.7-m tower. (a) Find the diver's velocity when he hits the water. (b) The diver comes to a stop 2.0 m below the surface. Find the net force exerted on the diver while in the water.

Answers

Answer:

(a)  13.795 m/s.

(b) -3140.28 N.

Explanation:

(a) Using newton's  equation of motion,

v² = u² + 2gs.......................... Equation 1

Where v = final velocity, u = initial velocity, s = height of the tower, g = acceleration due to gravity.

Given: s = 9.7 m, u = 0 m/s ( jump from a height), g = 9.81 m/s².

Substitute into equation 1

v² = 0² + 2×9.81×9.7

v² = 190.314

v = √(190.314)

v = 13.795 m/s.

Hence the velocity of the driver when he hits the water = 13.795 m/s.

(b)

F = ma.................... Equation 2

Where F = force exerted on the diver, m = mass of the diver, a = acceleration of the diver below the water surface.

Also using

v² = u² + 2as ............ Equation 3

Note: At the point when the diver enters the water, u = 13.795 m/s, and at the point when the diver comes to a complete stop, v = 0 m/s

Given: s = 2.0 m, u = 13.795 m/s, v = 0 m/s

Substitute into equation 3

0² = 13.795²+2(2a)

0 = 190.30203 + 4a

-4a = 190.30203

a = 190.30203/-4

a = -47.58 m/s²

Also given: m = 66 kg,

Substitute into equation 3

F = (-47.58)(66)

F = -3140.28

Note: The Force is negative because it act against the motion of the diver.

Hence the net force exerted on the diver while in the water = -3140.28 N.

A client has developed dystrophic calcification as a result of macroscopic deposition of calcium salts. The tissue that would be most affected would be?

Answers

Answer:

Tissues that are damaged or injured.

Explanation:

Dystrophic calcification involves the deposition of calcium in soft tissues despite no disturbance in the calcium metabolism, and this is often seen at damaged tissues.

Examples of areas in the body where dystrophic calcification can occur include atherosclerotic plaques and damaged heart valves.

Suppose you go outside and look at three stars. Star A is blue, Star B is white, and Star C is red. Which star is the hottest and which star is the coldest?

Answers

Answer:

star A is hottest and star c is coldest

Explanation:

from WEIN'S LAW which relates stellar color to stellar temperature, the visible light output of very hot stars is dominated by blue color while the stars that are cool emit most of their visible light as red .

Final answer:

Star A, the blue star, is the hottest due to its blue color, which is a consequence of its high temperature. Meanwhile, Star C, the red star, is the coldest because red light has a longer wavelength than blue or white light.

Explanation:

The color of a star can give us an idea of its temperature thanks to a concept called black-body radiation. This concept tells us that the hotter a star is, the shorter the wavelength of light it will emit, which in turn defines its color. Blue stars are the hottest, as blue light has a shorter wavelength than red or white light. As for white stars, they are intermediate in temperature, since white light is a combination of many different colors (wavelengths). Finally, red stars are the coldest of the three, because red light has a longer wavelength than blue or white light.

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1.

The collision shown here is ____________.

elastic

inelastic

2.

Any work being done is done against the force of __________.

friction

mass

3.

With an inelastic collision, you have to use the conservation of ____.

matter

momentum

Answers

Answer:

2. Friction

3. Momentum

Explanation:

For elastic collision, the total kinetic energy before and after collision are same while for inelastic collision, kinetic energy is converted to other forms apart from kinetic energy for example we get sound energy or rotational energy. Since in this case there's no diagram, we can't make a conclusion. However, in real life, all collisions are inelastic

2

Work is always done against frictional force

3

The law of conservation of momentum is used in an inelastic collision

Answer:

1) Inelastic

2) Friction

3)Momentum

Explanation:

USATestprep

Mark made a business trip of 214 miles. He averaged 53mph for the first part of the trip and 54mph for the second part. If the trip took 4 hours, how long did he travel at each rate?

Answers

Answer:

He travel for 2 h at the first part and 2 h for the second part.

Explanation:

given information:

business trip's distance, [tex]s_{tot}[/tex] = 214 miles

v₁ = 53 mph

v₂ = 54 mph

t₁ = t

t₂ = 4 - t

now lets find the distance of each speed

s₁ = v₁ t₁ = 53 t₁

s₂ = v₂ t₂ = 54 t₂

[tex]s_{tot}[/tex] = s₁ + s₂

214 = 53 t₁ + 54 t₂

214 = 53t + 54 (4-t)

214 = 53t + 216 - 54 t

t = 216 - 214

 = 2 h

t₁ = t = 2 h

t₂ = 4 - t = 4 -2 = 2h

How does the turbine make electricity

Answers

Answer:

A wind turbine turns wind energy into electricity using the aerodynamic force from the rotor blades.

Explanation:

A dragster and driver together have mass891.3 kg.The dragster, starting from rest,attains a speed of 25.2 m/s in 0.62 s.Find the average acceleration of the drag-ster during this time interval.

Answers

Answer:

40.645 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S.I unit of acceleration is m/s². Mathematically,

a = (v-u)/t................................. Equation 1

Where a = acceleration, v = velocity, u = initial velocity, t = time.

Given: v = 25.2 m/s, u = 0 m/s, t = 0.62 s.

Substituting into equation 1

a = (25.2-0)/0.62

a = 25.2/0.62

a = 40.645 m/s²

Hence the average acceleration of the drag-star = 40.645 m/s²

Johannes Kepler used decades of Tycho Brahe's observational data to formulate an accurate description of planetary motion. Kepler spent almost 30 years of his life trying to develop a simple description of planetary motion based on a heliocentric model that fit Tycho's data. What conclusion did Kepler eventually come to that revolutionized the heliocentric model of the solar system?

Answers

Explanation:

Johannes Kepler, working with the data carefully collected by Tycho Brahe without the help of a telescope, especially those related to the retrograde motion of Mars, realized that the motion of the planets could not be explained by his model of perfect polyhedra. Coming to the conclusion that all the planets move in elliptical orbits, with the Sun in one of the foci.

In modern radiotherapy, superconducting wire coils are used to create magnetic fields that focus particle beams in order to deliver the maximum dose of ionizing radiation to a tumor site while minimizing the destruction of healthy tissue. For what type of radiation would this technique NOT be useful?

Answers

Answer:Gamma Photons

Explanation: Gamma photons are photons that are parts of the electromagnetic radiation,they are of a very short wave length of less than about 0.01nanometer.,which means they travel with a very high speed and a very high energy of about 100,00volts. Gamma photons can KILL HEALTHY AND LIVING CELLS, that is why it is mainly used to KILL CANCEROUS CELLS/TUMORS but in modern radiology the use of superconducting wire coils are not employed or used to create magnetic fields for Gamma photons.

An airplane has been loaded in such a manner that the CG is located aft of the aft CG limit. One undesirable flight characteristic a pilot might experience with this airplane would be ______.a. a longer takeoff run.
b. difficulty in recovering from a stalled condition.
c. stalling at higher-than-normal airspeed.

Answers

Answer:

Option B is correct ( difficulty in recovering from a stalled condition.)

Explanation:

Option B is correct ( difficulty in recovering from a stalled condition.)

If you load airplane in such manners CG has changed then it will have huge impact on the longitudinal stability. Due to this airplane would face  difficulty in recovering from a stalled condition.

Stall is the phenomena in which air resistance is increased on the wings which causes decrease in lift. So it is important to remain with in CG limit put by manufacturer to avoid stall condition.

One undesirable flight characteristic a pilot might experience with this plane is  : ( B ) difficulty in recovering from a stalled condition.

CG Limit effect

When loading an aircraft the CG limit should not be exceeded as stated by the manufacturer, this is  because when the aircraft is loaded above the limit the longitudinal stability of the aircraft will be affected because more air resistance will be exerted on the wings of the plane which leads to a condition known as stall

Hence we can conclude that one undesirable flight characteristic a pilot might experience with this plane is  difficulty in recovering from a stalled condition.

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Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?

Answers

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s[/tex]

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

[tex]t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s[/tex]

For second drop time is given by

[tex]t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s[/tex]

Distance from second drop

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m[/tex]

Distance from second drop is 0.83999 m

Distance from third drop

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m[/tex]

Distance from third drop is 0.20999 m

A force F at an angle θ above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is µ. The work done by the frictional force is:a. – µFdcosθb. –Fdcosθc. –µmgdd. –µmgdcosθ

Answers

Answer:

option B

Explanation:

given,

Force = F

angle of inclination, = θ

distance = d

coefficient of friction = μ

work done =frictional Force x displacement

W = f. s

friction force on the incline will be equal to horizontal of force component. so,

f = - F cos θ

Work done will be equal to

W = - F d cos θ

Hence, the correct answer is option B

The engine of a 1540-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to a speed of 100 km/h at full power on a level road. Is your answer realistic?

Answers

Answer:

7.92 s

Explanation:

Power: This can be defined as the rate at which energy is used. The S.I unit of

Watt (W).

Mathematically power can be represented as,

P = E/t

Pt = E............................. Equation 1

Where P = power of the engine, t = time, E = Energy.

But,

E = 1/2m(Δv)²................. Equation 2

Where m = mass of the automobile, Δv = change in velocity of the car = final velocity - initial velocity.

Substitute the value of E in equation 2 into equation 1

Pt = 1/2m(Δv)²................................. Equation 3

making t the subject of the equation,

t = 1/2m(Δv)²/P ........................ Equation 4

Given: m = 1540 kg, P = 75 kW = 75000 W, Δv = 100-0 = 100 km/h (initial velocity of the car = 0 km/h)

100(1000/3600) m/s = 27.78 m/s

Substitute into equation 4

t = 1/2(1540)(27.78)²/75000

t = 594230.87/75000

t = 7.92 second.

Thus The time required to accelerate the car = 7.92 seconds.

It is not realistic period the time period is too short.

Final answer:

The car takes approximately 7.8 seconds to accelerate from rest to a speed of 100 km/h, given the power rating and mass that was provided in the question.

Explanation:

The process to solve this physics problem involves steps in calculating the work done on the car and then the time required for this work at the given power. To begin with, we can translate the speed of 100 km/h to m/s, which would be 27.8 m/s. Afterward, using the kinetic energy formula (K.E = 1/2 * m * v²), the work done on the car to achieve this speed can be calculated, which is 585840 Joules.

The next step is to compute the time. We know that Power = Work Done / Time, where power has been provided (75 kW = 75000 Watts) and we just computed the work done, so we can rearrange this equation to solve for time. This gives a resulting time of approximately 7.8 seconds. This may seem a short amount of time, but it's plausible for a high-powered sports car to reach 100 km/h in around this time frame.

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Which threats are characterized by possibly long periods of preparation (years is not uncommon), tremendous financial backing, a large and organized group of attackers, and attempts to subvert insiders or to plant individuals inside a potential target in advance of a planned attack?

Answers

Answer:

Highly structured or planned threat

Explanation:

This type of threat is an highly structured threat that is been planned using every available tools and machineries. The attack of this type consist of three phases, the preparation phase; where all the information and resources needed is been planned and source for. The execution phase; the attackers plane how to carry out the action, the techniques and technology needed for the success of the operation is carried out and tested. The post execution phase; the attackers planned there escape from any trouble that may arise and how not to leave any spot that will make them to be identified.

This type of threat requires to put the security outfit or the entire agency on red alert to ensure effective engagement before the attackers execute their actions.

When two or more different capacitors are connected in parallel across a potential source (battery), which of the following statements must be true? (There could be more than one correct choice.)

A) The potential difference across each capacitor is the same.

B) The capacitor with the largest capacitance has the most charge.

C) Each capacitor carries the same amount of charge.

D) The capacitor with the largest capacitance has the largest potential difference across it.

Answers

Answer:

Option A and B

Explanation:

In the parallel connection, the capacitors that are connected in parallel have the same potential difference across each capacitor as the battery. Therefore, statement A is correct.

Charge (Q), capacitance (C) and voltage across capacitor (V) is related with each other as follows:

Q = C ×V

As capacitance is not same for all the capacitors, therefore each capacitor carries different charge. Therefore, statement C is incorrect.

As potential difference across each capacitor is same, therefore, capacitor with the largest capacitance carry the most charge. Therefore, statement B is correct.

Therefore, statements A and B are correct.

Answer:

Option A and B is correct

Explanation:

When two or more capacitors are connected in parallel, the equivalent capacitance of the circuit is the sum of capacitance of all the capacitors.

C = C₁ + C₂+ C₃+....

The potential difference across each capacitor is equal. Thus, option A is correct.

The total charge is the sum of the charge on stored in each capacitor.

Q = Q₁ + Q₂+ Q₃+....

Q = CV

Since the voltage is same across each capacitor, the capacitor which has largest capacitance has the most charge. Thus, option B is correct.

The wave model of light is needed to explain many of the phenomena discussed in this chapter. Which of the following can be understood WITHOUT appealing to the wave model?a. sharp-edged shadows b. double-slit interference c. thin-film interference d. single-slit diffraction

Answers

Answer:

the correct answer is A Sherp-edged shadows

Explanation:

The wave model describes the path of light as the propagation of a cosenoidal wave, in this case when two beams of light interact, they do so through the supervision of the cosines of each one.

In the lightning model, the light wave characteristics are not taken into account, this model is used for the so-called geometric optics, which explains many phenomena, image formation, reflection phenomena and shadow formation defined in the corners , since in this model the light is nine rectilinear from one point to another.

With this explanation we see that the correct answer is to

Final answer:

Sharp-edged shadows can be understood without the wave model of light, as they can be explained by the ray model. Double-slit interference, thin-film interference, and single-slit diffraction require the wave model to account for light's wave properties, such as diffraction and interference.

Explanation:

The phenomena that can be understood without appealing to the wave model of light is sharp-edged shadows. This is because sharp-edged shadows can be explained using the ray model of light, which treats light as straight lines that travel from a source and create shadows when they are obstructed by an object. The other phenomena listed, including double-slit interference, thin-film interference, and single-slit diffraction, require an understanding of the wave properties of light, such as diffraction and interference, to be properly understood.

According to Huygens's principle, diffraction is the bending of waves around the edges of an object or through an opening. Interference occurs when waves superimpose and either constructively or destructively interfere with each other, giving rise to the various patterns seen in these phenomena. The ray model of light is insufficient to explain these effects because it does not account for the wave-like behaviors of light, such as bending and superposition of waves.

The windshield of a car has a total length of arm and blade of 9 ​inches, and rotates back and forth through an angle of 93degrees. What is the area of the portion of the windshield cleaned by the 7​-in wiper​ blade?

Answers

Answer:Area cleaned by the 7-in wiper blade=22

72sq in

Explanation: In a windshield the wiper moves in a semi circle. The angle in a semi circle is 180°

Let A= Total area

A= 2×9×3.1461×(93/180)

A=29.22squareinch

Let a be area cleaned by the 7-inch wiper

a=29.22 - 2(9-7)×3.1461×(93/180)

a=29.22 - 6.502

a= 22.72sqin

Answer:

The area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

Explanation:

Given

Length of blade = 9 inches

Angle of rotation = 93°

We're to calculate the area of the portion of the windshield cleaned by the 7​-in wiper​ blade?

We'll solve this by using area of a sector.

Area of a sector = ½r²θ

where θ is in radians.

So, angle of rotation (93°) must first be converted to radians

Converting 93º to radians, we get 31π/60

The area of the region swept out by the wiper blade = (area of the sector where r = 9 and

θ = 31π/60) - (area of the sector where r = (9-7) and θ = 31π/60).

We're making use of 9-7 because that region is outside the boundary of the 7in blade

So Area = ½*9²*31π/60 - ½*2²*31π/60

Area = ½*31π/60(9²-7²)

Area = 31π/120 * (81 - 49)

Area = 31π/120 * 32

Area = 992π/120

Area = 62.49151386765697 in²

Area ≈ 62.49 in²

Hence, the area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

A whistle of frequency 592 Hz moves in a circle of radius 64.7 cm at an angular speed of 13.8 rad/s. What are (a) the lowest and (b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle?

Answers

Answer:

(a) lowest frequency=577 Hz

(b) highest frequency=608 Hz

Explanation:

Given data

f(whistle frequency)=592 Hz

ω(angular speed)=13.8 rad/s

r(radius)=64.7 cm=0.647 m

To find

(a) Lowest frequency

(b) highest frequency

Solution

From Doppler effect

f=f×{(v±vd)/(v±vs)}

Where

v is speed of sound

Vd is speed detector relative to the medium(vd=0)

Vs is the speed of the source

Since

v=rω

For (a) lowest frequency

[tex]f^{i}=f(\frac{v}{v+rw} )\\f^{i}=(592Hz)(\frac{343m/s}{343m/s+(0.647m)(13.8rad/s)} )\\f^{i}=577Hz[/tex]

For (b) highest frequency

[tex]f^{i}=f(\frac{v}{v-rw} )\\f^{i}=(592Hz)(\frac{343m/s}{343m/s-(0.647m)(13.8rad/s)} )\\f^{i}=608Hz[/tex]

A motorcycle, which has an initial linear speed of 6.3 m/s, decelerates to a speed of 1.9 m/s in 4.6 s. Each wheel has a radius of 0.68 m and is rotating in a counterclockwise (positive) direction.a) the constant angular acceleration (in rad/s2) andb) the angular displacement (in rad) of each wheel?

Answers

Answer:

Explanation:

Given

initial velocity of motorcycle [tex]u=6.3\ m/s[/tex]

final velocity [tex]v=1.9\ m/s[/tex]

time taken [tex]t=4.6\ s[/tex]

radius of wheel [tex]r=0.68\ m[/tex]

using v=u+at

where a=acceleration

[tex]1.9=6.3+a\times 4.6[/tex]

[tex]a=-0.956\ m/s^2[/tex]

angular acceleration [tex]\alpha =\frac{a}{r}[/tex]

[tex]\alpha =\frac{-0.956}{0.68}=-1.406\ rad/s^2[/tex]

[tex]Displacement=average\ velocity\times time[/tex]

[tex]Displacement=\frac{u+v}{2}\times t[/tex]

[tex]Displacement=\frac{1.9+6.3}{2}\times 4.6=18.86\ m[/tex]

Angular displacement [tex]\theta =\frac{Linear\ displacement}{radius}[/tex]

Angular displacement [tex]\theta =\frac{18.86}{0.68}=27.73\ rad[/tex]

Answer:

Explanation:

initial velocity, u = 6.3 m/s

final speed, v = 1.9 m/s

time, t = 4.6 s

radius, r = 0.68 m

(a)

initial angular velocity, ωo = u / r = 6.3 / 0.68 = 9.26 rad/s

final angular velocity, ω = v / r = 1.9 / 0.68 = 2.79 rad/s

Let α be the angular acceleration

Use first equation of motion

ω = ωo + αt

2.79 = 9.26 + α x 4.6

α = - 1.41 rad/s²

(b) Let Ф be the angular displacement

Use third equation of motion

ω² = ωo² + 2 αФ

2.79² = 9.26² - 2 x 1.41 x Ф

Ф = 27.6 rad

Suppose that volumes of four stars in the Milky Way are 2.7 x 1018 km3, 6.9 x 1012 km3, 2.2 x 1012 km3, and 4.9 x 1021 km3. What is the order of the stars from least to greatest volume?

Answers

Answer:

The order is Star 3 < Star 2 < Star 1 < Star 4

Explanation:

Lets name the stars with their written order.

Star 1: 2.7 x 10^18 km3

Star 2: 6.9 x 10^12 km3

Star 3: 2.2 x 10^12 km3

Star 4: 4.9 x 10^21 km3

Star with lowest power of 10 has the least volume. Therefore, star 2 and star 3 would be the stars with least volume. Star 2's coefficient is better than Star 3. Thus, the order will be Star 3 < Star 2 < Star 1 < Star 4.

Snowflakes begin to form when water in the atmosphere freezes it causes the water molecules to bond into a hexagonal shape. Is this sentence correct?

Answers

Answer:

Yes! This sentence is correct.

Explanation:

So the complete question along with options is:

Q. Snowflakes begin to form when water in the atmosphere freezes it causes the water molecules to bond into a hexagonal shape.

a) no change

b) freezes, causing

c) freezes, it causes

d) freezes, this causes.

The coorect option is c.

What is the sharpness of the light and the amount of voltage when you put the magnet through the lower number of coils first vs the higher number of coils

Answers

Answer:hjvuihi

Explanation:lk

Two tiny beads are 25 cm apart with no other charged objects or fields present. Bead A has a net charge of magnitude 10 nC and bead B has a net charge of magnitude 1 nC. Which one of the following statements is true about the magnitudes of the electric forces on these beads?A. The force on A is 100 times the force on B.B. The force on B is 10 times the force on A.C. The force on B is 100 times the force on A.D. The force on A is 10 times the force on B.E. The force on A is exactly equal to the force on B.

Answers

Answer:

E. The force on A is exactly equal to the force on B.

Explanation:

The force between two charges is given by

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

where

[tex]q_1[/tex] = Charge on particle 1

[tex]q_2[/tex] = Charge on particle 2

r = Distance between the charges

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

[tex]F=\dfrac{8.99\times 10^9\times 10\times 10^{-9}\times 1\times 10^{-9}}{(25\times 10^{-2})^2}\\\Rightarrow F=0.0000014384\ N[/tex]

This force will be exerted on both the charges equally.

So, The force on A is exactly equal to the force on B

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