a spring is compressed by 1 cm when a force of 5.0N is applied. what is the potential energy stored in the spring when it is compressed by 10.cm

Answers

Answer 1

The potential energy stored in the spring is 0.025 J.

What is potential energy stored in a spring?

This is the energy stored in a spring when its is stretched or compressed through a certain length.

To calculate the potential energy stored in a spring, we use the formula below.

Formula:

E = Fe/2............... Equation 1

Where:

E = Potential energy of the springF = Force applied to the springe = Extension or Compression

From the question,

Given:

F = 5 Ne = 1 cm = 0.01 m

Substitute these values into equation 1

E = 5×0.01/2E = 0.025 J

Hence, the potential energy stored in the spring is 0.025 J.

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Related Questions

50 POINTS! What is simple slope/ what is the simple slope equation?

Answers

Answer:

u have to divide the 2numbers and multiply by 2

Explanation:

A spaceship of mass 2.40x10⁶kg is to be accelerated to a speed of 0.700 c. (b) How much fuel would it take to provide this much energy if all the rest energy of the fuel could be transformed to kinetic energy of the spaceship?

Answers

The fuel required to gain minimum energy is (3.02 x 10²³ / ΔH) kg.

We need to know about relativistic energy to solve this problem. The rest energy of the object can be determined by

Eo = m₀ . c²

where Eo is rest energy, m₀ is rest mass and c is the speed of light (3 x 10⁸ m/s).

The total energy of object can be described as

E = Eo / √(1 - v²/c²)

where E is total energy, v is the object speed.

From the question above, we know that :

m₀ = 2.4 x 10⁶ kg

c = 3 x 10⁸ m/s

v = 0.7c

Find the rest energy

Eo = m₀ . c²

Eo = 2.4 x 10⁶ . (3 x 10⁸)²

Eo = 2.16 x 10²³ joule

Determine the total energy

E = Eo / √(1 - v²/c²)

E = 2.16 x 10²³ / √(1 - (0.7c)²/c²)

E = 2.16 x 10²³ / 0.71

E = 3.02 x 10²³ joule

Assume that the Heat of Combustion for Space fuel = ΔH J/kg

m = E / ΔH

m = 3.02 x 10²³ /  ΔH kg

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Q|C Spherical waves of wavelength 45.0 cm propagate outward from a point source. (c) Explain how the phase of the wave at a distance of 240 cm compares with the phase at 60.0cm at the same moment.

Answers

We are given distance of the center of the source for two waves:

[tex]D_{1} =240 cm[/tex]

[tex]D_{2} =60 cm[/tex]

C) Calculating how their phases compares:

ΔФ240[tex]=\frac{D_{2} }{D_{1} } .[/tex]ΔФ60

[tex]\frac{240}{60} .[/tex]ΔФ60 = 4.ΔФ60

How can you determine a wave's phase?

The period, or length of each cycle, is determined by dividing the frequency by 1, so 1/100 corresponds to a period of 0.01 seconds. The phase shift equation is given by ps = 360 * td / p, where td is the time interval between waves, p is the wave period, and ps is the phase shift in degrees.

Phase: The "Phase" of a waveform refers to the location of the moving particle and is expressed in "Radians or degrees."

Phase difference, also known as "Phase angle," is the amount of time that one wave precedes or follows another.

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in 7 seconds, a projectile goes from 0 to 181m/s. What is the acceleration l? (round to the nearest whole number)

Answers

Answer:

84.83 ft/s^2 or 85 ft/s^2

Explanation:

A = Δ v Δ t

Acceleration is the rate of change of an objects speed; in other words, it's how fast velocity changes. According to Newton's second law, acceleration is directly proportional to the summation of all forces that act on an object and inversely proportional to its mass. It's all common sense - if several different forces are pushing an object, you need to work out what they add up to (they may be working in different directions), and then divide the resulting net force by your object's mass.

This acceleration definition says that acceleration and force are, in fact, the same thing. When the force changes, acceleration changes too, but the magnitude of its change depends on the mass of an object. This is not true in a situation when the mass also changes, e.g., in rocket thrust, where burnt propellants exit from the rocket's nozzle.

In the 17th century, Sir Isaac Newton, one of the most influential scientists of all time, published his famous book Principia. In it, he formulated the law of universal gravitation which states that any two objects with mass will attract each other with a force exponentially dependent on distance between these objects (specifically, it is inversely proportional to the distance squared). The heavier the objects are, the greater is gravitational force. It explains, for example, why planets orbit around the very dense Sun.

In Principia, Newton also includes three laws of motion which are central to understanding the physics of our world. The acceleration calculator is based on three various acceleration equations, where the third is derived from Newton's work:

a = (v_f - v_i) / Δt,

a = 2 * (Δd - v_i * Δt) / Δt²,

a = F / m,

where:

a is the acceleration,

v_i and v_f are respectively the initial and final velocities,

Δt is the acceleration time,

Δd is the distance traveled during acceleration,

F is the net force acting on an object that accelerates,

m is the mass of this object.

If you already know how to calculate acceleration let's focus on the units of acceleration. You can derive them from the equations we listed above. All you need to know is that speed is expressed in feet per second (imperial/US system) or in meters per second (SI system) and time in seconds. Therefore, if you divide speed by time (as we do in the first acceleration formula), you'll get acceleration unit ft/s² or m/s² depending on which system you use.

Alternatively, you can use the third equation. In this case, you need to divide force (poundals in US and newtons in SI) by mass (pounds in US and kilograms in SI) obtaining pdl/lb or N/kg. They both represent the same thing, as poundal is pdl = lb * ft/s² and the newton is N = kg * m/s². When you substitute it and reduce the units, you'll get (lb * ft/s²) / lb = ft/s² or (kg * m/s²) / kg = m/s².

There is also a third option that is, in fact, widely used. You can express acceleration by standard acceleration, due to gravity near the surface of the Earth which is defined as g = 31.17405 ft/s² = 9.80665 m/s². For example, if you say that an elevator is moving upwards with the acceleration of 0.2g, it means that it accelerates with about 6.2 ft/s² or 2 m/s² (i.e., 0.2*g).

after albemarle's heat wave, the rest of june was cooler. in fact, at the end of the month, meteorologists reported that the average high for all 30 days in june was just $80^\circ$. what was the average high (in degrees fahrenheit) from june 11 through june 30? (enter your answer as a number without units.)

Answers

The average high from June 11 through June 30 will be 76°F.

a)

The average high temperature for the previous 10 days must first be determined:

(84 + 88 + 91 + 92 + 96 + 94 + 89 + 78 + 87 + 81)/ 10

= 880/10

= 88°

According to the query, 82° is the "average" high temperature on a June day in Albemarle.

therefore,

88° - 82° = 6° is greater than normal

(b)The total of all temperatures for 30 days is 30*80 if the average temperature for 30 days is 80°.

= 2400

As calculated above, the sum of temperatures for the first 10 days is 880. Consequently, the total temperature for the following 20 days is

2400 - 880 = 1520

The mean for last 20 days is

1520/20

= 76°

The average high from June 11 through June 30 will be 76°F

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An x-ray beam with wavelength 0. 210 nm is directed at a crystal. As the angle of incidence increases, you observe the first strong interference maximum at an angle 70. 0 ∘. What is the spacing d between the planes of the crystal?

Answers

By using Bragg's diffraction, the spacing d is 0.112 nm.

We need to know about Bragg's diffraction to solve this problem. Bragg's diffraction is described how monochromatic light travels through a crystal medium. The monochromatic light should follow

2d.sin(θ) = λ

where d is the spacing between the planes of the crystal, θ is the incident light angle and λ is the wavelength.

From the question above, we know that

θ = 70⁰

λ = 0.210 nm = 0.21 x 10¯⁹ m

By substituting the given parameters, we get

2d.sin(θ) = λ

2d.sin(70⁰) = 0.21 x 10¯⁹

2d . 0.94 = 0.21 x 10¯⁹

d = 1.12 x 10¯¹⁰ m

d = 0.112 nm

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For flowing water, what is the magnitude of the velocity gradient needed to produce a shear stress of 1. 0 n/m 2 ?

Answers

The magnitude of the velocity gradient of water is 892.86 s^-1 or 892.86/s

Given:

The shear stress in the fluid is 1 N/m^2

Solution:

The formula for velocity gradient is given below

(dµ / dy)  =  τ / µ

Here τ  is sheer stress, µ is the dynamic viscosity and dµ / dy represents the velocity.

In the above equation, put the values  N/m^2 for τ and 1.12 x 10^-3 Ns/m^2  for μ.

(dµ / dy)  =  1 / 1.12 x 10^-3 Ns/m^2

(dµ / dy)  = 892.86 s^-1

Thus, the magnitude of the velocity gradient of water is 892.86 s^-1 or 892.86/s

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also known as a pallet truck, a(n) is a device used to lift and move heavy or stacked pallets.

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A pallet jack also known as a pallet truck, is a device used to lift and move heavy or stacked pallets.

A pallet jack, additionally called a pallet truck, pallet pump, pump truck, hand truck, scooter, canine, or jigger is a device used to lift and pass pallets. Pallet jacks are the most fundamental shape of a forklift and are meant to transport pallets within a warehouse.

A pallet jack is the most fundamental form of forklift and is intended to transport pallets in a warehouse or trailer. Pallet jacks are some of the maximum important gear located in warehouses and are used for transporting small masses for quick distances

The pallet jack has a deal with a lever. Slide the prongs into function beneath a pallet and crank the handle up and all the way down to carry the prongs off of the floor. Push or pull the pallet on your favored location and keep the lever right down to decrease the prongs returned to the ground.

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1. Two point charges, q1 and q2, of 4.00 μC each, are placed at x1=-16.0 cm and x2 = 16.0 cm away from the origin on the x-axis. A charge q3 of 1.00 μC is placed 12.0 cm away from the origin on the y-axis.

a. Find the distance from q3 to q1 and from q3 to q2.

b. Find the magnitude and the direction of the force F13 exerted by q1 on q3.

c. Find the magnitude and the direction of the force F23 exerted by q2 on q3.

d. Find the magnitude and the direction of the force F12 exerted by q1 on q2.

Answers

The distance between charge 1 and 3 is equal to 20 cm and the force between charge 1 and 3 is equal to force between charge 1 and 3 that is 0.9 Newton and the force between charge 2 and 3 is equal to 1.41 Newton.

What are charges?

Electric charges are basic property of matter carried by some elementary particles that governs how the particles are affected by an electric or magnetic field. Electric charge, which can be positive or negative, occurs in discrete natural units and is neither created nor destroyed.

What is coulomb's law?

Coulomb's law states that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

Given:

Magnitude on charge 1 = 4 μC

Magnitude on charge 2 = 4 μC

Magnitude on charge 3 = 1 μC

Distance between charge 1 and 2 = 32 cm

Distance of charge 3 from center of charge 1 and 2 = 12 cm

Distance between charge 1 and 3 can be find using Pythagoras theorem.

Distance between charge 1 and 3 = √(256 + 144)

Distance between charge 1 and 3 = 20cm

Similarly Distance between charge 2 and 3 = 20cm

Force between charge 1 and 3 can be found using coulomb law.

Force between charge 1 and 3 = (9 × 4 × 1)/40

Force between charge 1 and 3 = 36/40

Force between charge 1 and 3 = 0.9 Newton

Similarly using symmetry, force between charge 2 and 3 = 0.9 Newton

Force between charge 1 and 2 =(9 × 4 × 4)/102

Force between charge 1 and 2 = 1.41 newton

Therefore the distance between charge 1 and 3 is equal to 20 cm and the force between charge 1 and 3 is equal to force between charge 1 and 3 that is 0.9 Newton and the force between charge 2 and 3 is equal to 1.41 Newton.

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What was the control group for the Michelson-Morley experiment?

Answers

The control group for the Michelson-Morley experiment was Interferometer.

The Michelson Morley Experiment was one of the unsuccessful tests that proved the luminiferous ether idea did not exist.

Michelson and Morley attempted to explain how the Earth orbited the sun and how the passage of substances such as ether over the Earth's surface might cause measurable "ether wind."

They attempted to show the idea that the speed of light depends on the size of the ether wind as well as the direction of the beam in relation to it when the light is emitted from an Earthly source. It was supposed that ether was immobile.

The experiment's goal was to measure the speed of light in different directions in order to determine the speed of the ether relative to Earth, so proving its existence.

The theorized material Luminiferous Ether serves as a medium for the transmission of electromagnetic waves such as light rays and X-rays. Ether was thought to be a transmission medium for light propagation.

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i go 540km at a speed of 100 km per hour how long does it take?

Answers

Answer: 5.4 hours.

Explanation:

So, this would simply be divison. So, we can take 540 km, and seperate it into 500 km, and 40 km.

That being said, we can divide 500 km by 100, so we can see how many hours it takes for 500 km, which is 5.

Now, we divide 40/100, which gives us 0.4.

Add the two together, and it takes 5.4 hours to reach 540 km at a speed of 100 km/hr

the gauge pressure at the bottom of a cylinder of liquid is 0.40 atm . the liquid is poured into another cylinder with twice the radius of the first cylinder. part a what is the gauge pressure at the bottom of the second cylinder?

Answers

The gauge pressure at the bottom of the second cylinder is 0.10 atm.

What do you mean by gauge pressure?

The pressure of a system above the atmospheric pressure is known as gauge pressure, commonly referred to as overpressure. The pressure from the weight of the atmosphere is included in gauge pressure readings because gauge pressure is zero-referenced against ambient air (or atmospheric) pressure. This indicates that gauge pressure varies in response to both weather and height above sea level. Gauge pressure measurement is adequate for the majority of industrial applications because every operation in a refinery or manufacturing facility operates at the same air pressure. Absolute pressure is a measurement of pressure that is based on a reference pressure of zero or absolutely no pressure. The only place where this happens naturally is in a perfect vacuum, which can only be found in space.

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Switch on your light source and make sure it's shining onto a wall.hold your largest cardboard square between the light source and the wall.wha do you observe?

Answers

Light source will pass through and form a shadow, when I hold the largest cardboard between the light source and the wall. Since, the cardboard is not so thick it will allow rays to pass through it.

Light rays always travel in a straight line. A shadow is formed when the light rays hit an opaque object. Sun is a natural source of light. The rays from the sun fall on objects on earth and forms shadows. Light rays do not interfere with each other if they come cross each other. Light rays travel from the direction of source to the direction of eye. They show property of reflection wherein after colliding a surface they trace their path back.

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a 1250 kg car traveling at a speed of 25.0 m/s rounds a 175 m radius curve. assuming the road is level, determine the coefficient of static friction between the car’s tires and the road.

Answers

Coefficient of static friction between the car’s tires and the road is 0.3

What is static friction ?

A force that holds an object at rest is called static friction. The definition of static friction is: The resistance people feel when they attempt to move a stationary object across a surface without actually causing any relative motion between their body and the surface they are moving the object across.

Give an illustration of static friction.

When something is sitting on a surface, static friction affects it. For instance, every time you place your foot on the trail when trekking in the woods, there is static friction between your shoes and the surface.

Assume that road is level

Friction force provides centripetal force to the car

The weight of the car is = 1250 kg

speed of the car is = 25 m/s

curve radius =r= 175m

we know the formula of centripetal force is given by

[tex]f_{s} =\frac{mv^{2} }{r}[/tex]         where m = mass of the object

                                   v=  velocity

                                    r=  curve radius

and we know  the static force is given by f= μN

                                                                     here, N= mg

so now,

μmg=[tex]\frac{mv^{2} }{r}\\[/tex]

now put the all give values then we get,

[tex]$$\mu_r=\frac{v^2}{r g}=\frac{(25 \mathrm{~m} / \mathrm{sec})^2}{(175 \mathrm{~m})\left(9.8 \mathrm{~m} / \mathrm{sec}^2\right)}$$[/tex]

μ=[tex]\frac{625}{1715}[/tex]

μ= 0.3 , This is the coefficient of friction between the car tire and the rode

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What does a radio wave do to the charges in the receiving antenna to provide a signal for your car radio?

Answers

Radio waves allows the charges in the receiving antenna to respond to the electric field portion of the carrier wave.

What is a wave?

This refers to a type of disturbance which transports energy from one place to another without the actual movement of the particles and of various types such as  microwaves, radio waves etc.

The radio wave allows the electrons in the rod to vibrate with amplitudes which emulates those of the carrier wave and in this case, it is the antenna thereby resulting in the signals in the car radio being provided as a result of these type of interactions.

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What is the potential difference across a hand-held fan that has a resistance of. 120 ohms and a current of 0,005A flowing through it?

Answers

Answer: The answer is 0.6 volts

Explanation:

As the formula of  Resistance says that

Resistance= Voltage÷ Current  

Where its going to be

Voltage= Resistance÷ Current

Resistance= 120 ohms

Current= 0.005A

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Which of the following pairs can be the components of a velocity of 30 m s¹ (east)? (a) 20 m s' (east), 10 m s' (east)
(b) 35 m s¹ (east), 5 m s¹ (west)
(c) 20 m s¹ (east). 15 m s¹ (west)​

Answers

A and B. The pairs that can be the components of a velocity of 30 m s¹ (east) are 20 m s' (east), 10 m s' (east) and 35 m s¹ (east), 5 m s¹ (west).

What is resultant velocity?

The resultant velocity of an object is the sum of its individual vector velocities.

For a resultant velocity of 30 m/s east, the component velocities that can equal the resultant velocity is calculated as follows;

35 m/s east and 5 m/s west

let the east direction be positive

let the west direction be negative

R = 35 m/s - 5 m/s

R = 30 m/s east

Another pair

20 m/s east and 10 m/s east

R = 20 m/s + 10 m/s

R = 30 m/s

Thus, the pairs that can be the components of a velocity of 30 m s¹ (east) are (a) 20 m s' (east), 10 m s' (east) and (b) 35 m s¹ (east), 5 m s¹ (west).

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a 2-kg mass attached to the end of a spring with a spring constant of 200 n/m moves in simple harmonic motion. find the period of the oscillations.

Answers

By angular speed, the period of oscillation is 0.63 second.

We need to know about the angular speed of harmonic oscillation. the angular speed can be determined as

ω² = k / m

where ω is angular speed, k is spring constant,  and m is the mass.

The angular speed can be calculated by

ω = 2π/T

Hence,

ω² = k / m

(2π/T)² = k / m

where T is period.

From the question above, we know that:

m = 2 kg

k = 200 N/m

By substituting to the equation, we get

(2π/T)² = k / m

(2π/T)² = 200 / 2

(2π/T)² = 100

square root

2π/T = √100

T = 2π / √100

T = 0.63 second

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In the figure shown, if angle i is increased, angle r will _____.

Answers

Answer: I am assuming it is decreases

Explanation: a straight line is 180 degrees. 2 angles both add up to 180 degrees on a straight line. When one increases, the other has to decrease to keep it 180 degrees on a straight line

22. сс Peter exerts a horizontal force of 500 N on a box of mass 2.0 kg which also experiences a friction force of 200 N. If it takes 4.0 s to move the box 3.0 m, what is the efficiency in moving the box? Efficiency 2 Energy coming in energy going X 100 %. (c) ABUD 29% 40% 60% 71% FH301/P1/19 ont.​

Answers

The efficiency of moving the box is 60%.

Given in the question

Force exerted by peter = 500 N

Force due to friction on box = 200 N

Distance Traveled by box = 3 meters

Now work done by the force is given by

Work Done = (Magnitude of Force) × (Distance traveled)

Let us find work done by each force,

Work done by peter

Work Done = (Magnitude of Force) × (Distance traveled)

Put in the value, we get

Work Done =500 × 3

Work Done by peter = 1500 J

Similarly, Work done by friction

Work Done = (Magnitude of Force) × (Distance traveled)

Put in the value, we get

Work Done =200 × 3

Work Done by Friction = 600 J

As we know the work done by friction force is always negative,

Work Done by Friction = -600 J

So, net Work done = Work Done by peter + Work Done by Friction

net Work done = 1500 - 600

Net Work done = 900 J

Now Efficiency = (Net work done/work done by peter) × 100

Put in the values, we get

Efficiency = (900/1500) × 100

Efficiency = (3/5) × 100

Efficiency = 300/5

Efficiency = 60 %

So, the efficiency of moving the box is 60%.

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An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first sees the car, the locomotive is 350 m from the crossing and its speed is 12 m/s. If the engineer’s reaction time is 0.56 s, what should be the magnitude of the minimum deceleration to avoid an accident? Answer in units of m/s^2.

Answers

The magnitude of the minimum deceleration to avoid an accident is 2,186.5 m/s².

Minimum acceleration to avoid accident

The minimum acceleration to avoid accident is calculated as follows;

s = ut + ¹/₂at²

where;

u is the initial velocitya is the minimum accelerationt is time of motion

Substitute the given parameters and solve for the acceleration as follows;

350 = 12(0.56) +  ¹/₂(0.56²)a

350 = 6.72 + 0.157a

0.157a = 350 - 6.72

0.157a = 343.28

a = 343.28/0.157

a = 2,186.5 m/s²

Thus, the magnitude of the minimum deceleration to avoid an accident is 2,186.5 m/s².

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Construct Explanations Flora is a pilot. She knows that it is very dangerous to put an airplane into a dive that is too steep. Explain what the danger is, in terms of potential and Kinetic energy.​

Answers

One of the main causes of fatalities in the commercial aviation sector is loss of airplane control while in flight. There are numerous causes of upsets in airplanes, but none of them are statistically significant.

Eliminating one cause of upsets won't definitely result in fewer fatalities and accidents involving loss of control. Reducing the number of upset causes requires ongoing training. Additionally, many environmental factors that cause upsets can't always be avoided, which makes avoidance the best course of action. Therefore, in order to recover an upset airplane, pilots must possess the requisite knowledge and abilities.

All makers of big, swept-wing commercial jet aircraft use the same aerodynamic concepts.

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Two long, parallel conductors, separated by 10.0cm, carry currents in the same direction. The first wire carries a current I₁ = 5.00A, and the second carries I₂ = 8.00 A. (b) What is the force per unit length exerted by I₁ on I₂ ?

Answers

The force per unit length exerted by I₁ on I₂ is [tex]8\times10^-^5 \ N/m[/tex] .

Two long parallel conductors separated by a distance 'r' having current [tex]I_1[/tex] and [tex]I_2[/tex] have force per unit length which is given by  [tex]f=\frac{u_0I_1I_2}{2\pi r}[/tex] .....(1)  where [tex]u_0[/tex] is permittivity of free space whose value is [tex]4\pi \times10^-^7Tm/A[/tex]

It is given that two long, parallel conductors, separated by 10.0cm having current  I₁ = 5.00A and  I₂ = 8.00 A.

If the current flowing through the wire are in same direction then they the wire attract each other and if the current flowing through the wire are opposite in direction then they the wire repel each other

Putting r = 10.0cm [tex]=10\times10^{-2}m[/tex] ,  I₁ = 5.00A and  I₂ = 8.00 A in equation (1) , we get

[tex]f=\frac{4\pi \times10^-^7\times5.00\times8.00}{2\times\pi\times10.00\times10^{-2}}\\\\f=\frac{2\times40\times10^-^5}{10} \\\\f=8\times10^{-5}N/m[/tex]

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Suppose you have found a piece of material. You are not sure what the material is. You want to find out if it is a good conductor or a good insulator. Describe in words what you would do to determine if the material is an electrical conductor.​

Answers

Answer:

Touch the open ends of the two wires to each other to form a circuit and test the bulb. Touch the two open ends of the wire to each material you are testing, one at a time. If the bulb lights up, it is a good conductor. If it doesn't, it is a good insulator.How do you know if an object is a good conductor and poor conductor of electricity?

Image result for Suppose you have found a piece of material. You are not sure what the material is. You want to find out if it is a good conductor or a good insulator. Describe in words what you would do to determine if the material is an electrical conductor.​

Difference between Good Conductor and Bad Conductor of Electricity. Good conductors are those materials which allow electricity to pass through them easily. Bad conductors are those materials which do not allow electricity to pass through them easily.

A⁵⁷ Fe Fe nucleus at rest emits a 14.0keV photon. Use conservation of energy and momentum to find the kinetic energy of the recoiling nucleus in electron volts. Use M c² = 8.60 × 10⁻⁹J for the final state of the ⁵⁷Fe nucleus.

Answers

Kinetic energy is K = 1.82 × 10⁻³ eV.

⁵⁷Fe nucleus emits 14 keV photons at rest.

We have, Mc² = 8.6 × 10⁻⁹ J in the final state of the nucleus.

Let p(nucleus) be the moment of the nucleus and p(photon) be the momentum of the photon.

Then, by conservation of momentum since the total moment is zero.

p(nucleus) = p(photon) = E(γ) / c = 140 keV / c

The energy for the recoiling nucleus is given as:

E² = p²c² + (mc²)²

Now, Mc² = 8.60 × 10⁻⁹J =5.38 × 10¹⁰ eV = 5.38  × 10⁷ keV

Therefore,

(Mc² + K²) ² = (14 keV)² + (Mc²)²

(1 + K²/Mc²)² =(14/Mc²)² +1

The term (1 + K²/Mc²)² is less than 1  therefore, (14/Mc²)² + 1 is equal to 1 by clacualtor.

Now using binomial theorem,

(1 + K²/Mc²)² = √[1 + ((14/Mc²)²] = 1 + (1/2)(14/Mc²)²

K = 1.82 × 10⁻³ eV

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What is the mass of an object that requires 100N (kg-m/s2) of force in order to accelerate it at 10m/s2 (Please use G-R-E-S-A) pls pasagot

Answers

The calculated mass is 10 kg.

Mass (kg) times acceleration (m/s2) equals force (N). A constant mass item will therefore accelerate in direct proportion to the force exerted. When two objects of differing masses are subjected to the same force, the heavier object accelerates more slowly than the lighter object. Mass times acceleration, or F=m x a, equals force. This means that in order to move an object at the same speed as an object with smaller mass, a stronger force is required.

So, Mass times acceleration equals force

100=Mass×10

To calculate mass on its own, divide 100 by 10 on both sides.

Mass = 10 kg

Because this formula was developed using kg rather than another unit of mass like the slug, the mass is expressed in kg.

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20 POINTSSS PLS HELP
Imagine if the Earth suddenly sped up. What do you think would happen to its orbit?

Answers

Answer:

down below

Explanation:

I assume you're asking about it's orbit around the sun. I believe the Earth's orbit around the sun would be shorter. If the Earth sped up, one can assume that means the days are shorter and the rate at which the Earth orbits the sun would also speed up.

This sounds like an open ended question, so I don't think there can be a wrong answer.

Hope this helps! :)

A ball is projected 125 meters straight upward and then falls the same distance back to its starting point. Neglecting air resistance, its total time in the air is about.

Answers

The ball that is projected 125 meters straight upward has a total time in the air of: 10.1 s

The formulas for the vertical launch upward and the procedures we will use are:

y max = v₀²/(2*g)t max = v₀/ gt(of)=2*t max

Where:

v₀ = initial velocityg = gravityy max = maximum heightt max = time to reach maximum heightt(of) =  time of flight

Information about the problem:

g = 9.8 m/s²y max= 125 mv₀ = ?t max =?t(of) =?

Applying the maximum height formula and clearing the initial velocity we get:

y max = v₀²/(2*g)

v₀ = √(y max * (2*g))

v₀ = √( 125 m * (2 * 9.8 m/s²))

v₀ = √( 125 m * 19.6 m/s²)

v₀ = √2450 m²/s²

v₀ = 49.497 m/s

Applying the maximum time formula we get:

t max= v₀ / g

t max= 49.497 m/s / 9.8 m/s²

t max = 5.050 s

Applying  the time of flight formula, we get:

t(of) =2 * t max

t(of) =2 * 5.050 s

t(of) = 10.1 s

What is vertical launch upwards?

In physics vertical launch upwards is the motion described by an object that has been launched vertically upwards in which the height and the effect of the earth's gravitational force on the launched object are taken into account.

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The time taken now is 10 s

What is the time taken?

We know that in this case, the object was projected vertically upward and we know that in this direction the acceleration due to gravity is negative. Then the ball falls straight down to  the ground. We have to use the equation of kinematics under gravity to approach the problem.

We now have that;

h = ut + 1/2gt^2

Considering the downward motion were g is positive u = 0 m/s

h =  1/2gt^2

t = 2h/g

t = √2h/g

t = √ 2 * 125/10

t = 5 s

Given that the time taken to go up is the same as the time taken to come down;

Total time spent in air = 2 (5 s) = 10 s

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The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00μT.(b) What If? At one instant, the two conduce tors in a long household extension cord carry equal 2.00-\mathrm{A} currents in opposite directions. The two wires are 3.00 \mathrm{~mm} apart. Find the magnetic field 40.0 \mathrm{~cm} away from the middle of the straight cord, in the plane of the two wires.

Answers

The magnetic field 40.0 cm away from the middle of the straight cord, in the plane of the two wires is [tex]7.5\times 10^{-9}T[/tex]

The magnitude and direction of the magnetic field due to a straight  wire carrying current can be calculated using the previously mentioned Biot-Savart law. Let "I" be the current flowing in a straight line and "r" be the distance. Then the magnetic field produced by the wire at that particular point is given by  [tex]B=\frac{u_0I}{2\pi r}[/tex]  ...(1)Since the wire is assumed to be very long, the magnitude of the magnetic field depends on the distance of the point from the wire rather than the position along the wire.

Let [tex]B_1[/tex] be inside the plane and [tex]B_2[/tex] be outside the plane . It is required to calculate magnetic field at point A .

Direction of magnetic field at A is calculated using Right hand rule.

[tex]B_{net}=B_2-B_1\\\\B_{net}=\frac{u_0I}{2\pi (0.4-\frac{3\times10^{-3}}{2}) } -\frac{u_0I}{(0.4+\frac{3\times10^{-3}}{2})} \\\B_{net}=7.5\times 10^{-9}T[/tex]

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A straight wire carrying a 3.00-A current is placed in a uniform magnetic field of magnitude 0.280 T directed perpendicular to the wire. (b) Explain why you can't determine the direction of the magnetic force from the information given in the problem.

Answers

The reasons are discussed below -

We have a straight current carrying wire placed in a uniform magnetic field.

We have to explain why we can't determine the direction of the magnetic force from the information given in the problem.

What is the magnitude of force acting on a current (I) carrying wire of length (L) placed in a Magnetic field (B)?

The force on the current carrying wire will be -

F = IBL sinθ

According to the question -

It is impossible to determine the direction of magnetic force because of the following reasons -

Firstly, the length of the wire present in the region of magnetic field is given. If we assume it to be, then -

F = IB sin(θ) x ∞ = ∞ (which is not a valid result)

Secondly, Although it is mentioned that a straight current carrying wire is placed perpendicular to the uniform magnetic field, there is no mention about  the exact orientation of the magnetic field direction. For example, if we consider the wire along +x axis then perpendicular could be any of the four directions -

+k, -k, +j, -j

Hence, these are the reasons why we can't determine the direction of the magnetic force from the information given in the problem.

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