Two small, identical conducting spheres repel each other with a force of 0.030 N when they are 0.65 m apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of 0.075 N. What is the original charge on each sphere? (Enter the magnitudes in C.)

Answer:

Over tens of millions of years, the giant supercontinent Pangaea began to slowly drift apart, forming the continents as they are known today.

Explanation:

i took the active

Answer:

the continents were once connected as one landmass.

Explanation:

right on edge 2022

Answer:

Explanation:

Force equal to resistive force will be applied for movement . So force applied

F = 220 N .

displacement = 2800 m

work done against resistive force

= force x displacement

= 220 x 2800 J

= 6.16 x 10⁵ J .

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

[tex]C=2\pi r[/tex]    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

[tex]r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m[/tex]

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

[tex]L=Lo[1+\alpha \Delta T][/tex]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

[tex]L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m[/tex]

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

[tex]r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m[/tex]

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

Explanation:

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).

Answer:

a. Convert  120 × 10⁸ g to i mg = 1.2 × 10¹³ mg ii. to g = 1.2 × 10⁷ kg

b. Convert 200000 × 10³ m to i. micrometers = 0.2 × 10³ μm  ii. to cm = 2 × 10⁶ cm iii. to km = 2 × 10⁵ km

Explanation:

a. i. To convert the mass = 120 × 10⁸ g to mg, We know that 1000 mg = 10³ mg = 1 g, Since we are converting to mg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10³ mg = 120 × 10¹¹ mg = 1.2 × 10² × 10¹¹ mg = 1.2 × 10¹³ mg

ii. To convert the mass = 120 × 10⁸ g to kg, We know that 1000 g = 10³ g = 1 kg, 1 g = 10⁻³ kg. Since we are converting to kg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10⁻³ kg = 120 × 10⁵ kg = 1.2 × 10² × 10⁵ kg = 1.2 × 10⁷ kg

b. i.To convert the length = 200000 × 10³ m to micrometers, We know that 1/1000000 μm = 10⁻⁶ mg = 1 m, Since we are converting to micrometers, μm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000000 μm = 200000/1000000 × 10³ μm = 0.2 × 10³ μm

ii. To convert the length = 200000 × 10³ m to cm, We know that 100 cm = 10² cm = 1 m, 1 m = 10⁻² cm = 1/100 cm. Since we are converting to cm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/100 cm = 200000/100 × 10³ cm = 2000 × 10³ cm  = 2 × 10³ × 10³ cm = 2 × 10⁶ cm

iii. To convert the length = 200000 × 10³ m to km, We know that 1000 m = 10³ m = 1 km, 1 m = 10⁻³ km = 1/1000 km Since we are converting to km, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000 km = 200000/1000 × 10³ km = 200 × 10³ km = 2 × 10² × 10³ km = 2 × 10⁵ km

Answer:

Yes it is normal

Explanation:

When you exercise, your heart beat goes up, resulting in people saying that their heart feels like it is "beating out of their chests".

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

[tex]m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0[/tex]

with m = 1 slug and k = 9 lb/ft, the equation becomes

[tex]\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0[/tex]

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity  v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

The time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.

Given data:

The mass suspended from spring is, m = 1 slug.

The spring constant is, k = 9 lb/ft.

The magnitude of upward velocity is, v = 3 ft/s.

The magnitude of downward velocity is, v' = 3 ft/s.

The given problem can be resolved by framing a differential equation that governs the motion of spring. The differential equation governing the motion of spring is,

[tex]m \dfrac{d^{2}x}{dt^{2}}=-kx\\\\\\\dfrac{d^{2}x}{dt^{2}}+\dfrac{k}{m}x=0[/tex]

With m = 1 slug and k = 9 lb/ft, the equation becomes

[tex]\dfrac{d^{2}x}{dt^{2}}+\dfrac{9}{1}x=0\\\\\\\dfrac{d^{2}x}{dt^{2}}+9x=0[/tex]  

Now, the characteristic equation is,

D² + 9 = 0

D = ±√-9 = ±3i

And the general solution of the above equation is,

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂²

              = √[(-1)² + (-√3/3)²]

              = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3

and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    

where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6.

t = 5π/18 + 2nπ/3

here, n is a natural number.

Thus, we can conclude that the time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.

Learn more about the differential equation here:

https://brainly.com/question/14620493  

Answer:

[tex]C_{pb}=0.501\ kJ/kg.K[/tex]

Explanation:

Given that

[tex]m_1=0.35 kg[/tex]

[tex]T_1=-27.5^oC[/tex]

[tex]m_2=0.214 kg[/tex]

[tex]T_2=25^oC[/tex]

[tex]T=16.4^oC[/tex]

We know that

[tex]C_{pw}=4.187 kJ/kg.K[/tex]

By using energy conservation

Heat lost by water = Heat gain by block

[tex]m_2\times C_{pw}\times (T_2-T)=m_1\times C_{pb}\times (T-T_1)[/tex]

[tex]0.214\times 4.187\times (25-16.4)=0.35\times C_{pb}\times (16.4+27.5)[/tex]

[tex]C_{pb}=0.501\ kJ/kg.K[/tex]

Therefore the specific heat of the block will be 0.501 kJ/kg.K

Answer is given below

Explanation:

Answer:

Work done by the force acting on a body is defined as the product of force and displacement of the body in the direction of the force. It is a scalar quantity. Force acting on the body must produce a displacement for the work is to be done by the force. Thus, for the work to be done, the following conditions must be fulfilled:

A force must be applied and the applied force must produce a displacement in any direction except perpendicular to the direction of the force.

Mathematically,

[tex]work = force \times \: displacement[/tex]

( in the direction of force)

i.e W= F•D

The SI unit of force is Newton (N) and that of displacement of a metre (M). So the unit of work is Newton metre(Nm) which is joule (J).

Thus, one joule work is said to be done when one Newton force displaces a body through one metre in its own direction.

Hope this helps...

Good luck on your assignment..

Answer:

Most conductors are made from elements metallic elements

Metallic elements are those elements grouped as alkali metals, alkaline earth metals, basic metals, rare earth elements, actinides, and basic metals

Conductors of electricity carry electricity by means of the movable charged particles present in the conducting material

The movable charged particle are the electrons which are most mobile in metals because of their crystalline structure and available valence electron, which are the electrons in the outermost orbit of an atom and hence freest to move about within the material mass

As such the availability of free electrons determine conductivity of materials

Explanation:

Understanding the relationship between electrical conductivity and the structure of atoms and molecules will contribute to the design of efficiency and reliability of technologies deployed to remote locations such as the Moon, by ensuring the best possible output from the electric control inputs and the redesigning of existing electrical installations using efficient amount of manufacturing materials, thereby saving the environment and costs.

Free electo

Electric conductors possess movable electrically charged particles, referred to as "electrons" in metals. When an electric charge is applied to a metal at certain points, the electrons will move and allow electricity to pass through. Materials with high electron mobility are good conductors and materials with low electron mobility are not good conductors, instead referred to as "insulators."

Silver is the best conductor of electricity because it contains a higher number of movable atoms (free electrons). For a material to be a good conductor, the electricity passed through it must be able to move the electrons; the more free electrons in a metal, the greater its conductivity. However, silver is more expensive than other materials and is not normally used unless it is required for specialized equipment like satellites or circuit boards. Copper is less conductive than silver but is cheaper and commonly used as an effective conductor in household appliances. Most wires are copper-plated and electromagnet cores are normally wrapped with copper wire. Copper is also easy to solder and wrap into wires, so it is often used when a large amount of conductive material is required.

Answer:

0.03 Joules have been converted into other forms of energy as the direct result of the collision.

Explanation:

Let's start studying the conservation of momentum for the system:

[tex]P_i=P_f\\(0.25\,kg)\,{0.6\,m/s)+(0.5\,kg)\,(0\,m/s)=(0.25\,kg+0.5\,kg)\, v_f \\\\\\ 0.15\,kg\,m/s=0.75\,kg\,\,v_f\\v_f=0.15/0.75\,\,m/s\\v_f=0.2\,\,m/s[/tex]

Now that we know the speed of the newly created object, we can calculate how the final kinetic energy differs from the initial one:

[tex]K_i=\frac{1}{2} (0.25)\,(0.6)^2+\frac{1}{2} (0.5)\,(0)^2=0.045\,\,J\\ \\K_f=\frac{1}{2} (0.75)\,(0.2)^2=0.015\,\,J\\[/tex]

Then, when we subtract one from the other, we can estimate how much kinetic energy has been converted into other forms of energy in the collision:

0.045 J - 0.015 J = 0.03 J

Answer:

option d is answer because pd is measured in volt.

Answer:

The answer is Revising

Answer:

D

Explanation:

Revising

Answer:

d. Both kinetic and sliding friction

Explanation:

Kinetic friction, commonly known as sliding friction, happens when a body with its surfaces in contact is in relative motion with another. It's the frictional force slowing it down, and finally stopping a moving body. One can describe sliding friction as the resistance any two objects create while sliding against each other. It is often documented as the force required to hold a surface moving along another surface. It is determined by two variables- one is material of the object and another is its weight.

Answer:

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Explanation:

Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:

[tex]T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]

En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:

[tex]T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]

Donde:

[tex]T_{min}[/tex], [tex]T_{max}[/tex] - Tensiones mínima y máxima, medidas en newtons.

[tex]m[/tex] - Masa de la bola, medida en kilogramos.

[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.

[tex]L[/tex] - Distancia con respecto al eje de rotación, medida en metros.

[tex]v[/tex] - Rapidez tangencial, medido en metros por segundo.

Se elimina la aceleración centrípeta de ambas expresiones por igualación:

[tex]T_{min} + m\cdot g = T_{max} - m\cdot g[/tex]

Ahora, la diferencia entre las tensiones máxima y mínima es:

[tex]T_{max} - T_{min} = 2\cdot m \cdot g[/tex]

Si [tex]m = 1\,kg[/tex] y [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], entonces:

[tex]T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]T_{max}-T_{min} = 19.614\,N[/tex]

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Answer:

force = 45.12N

Explanation:

f = ma

f = 0.06 x 752 = 45.12N

Answer:

Explanation:

We shall apply the concept of impulse which is given as follows .

Impulse = force x time

Impulse = change in momentum

If u be the initial velocity of golf ball and v be the final velocity , m be the mass

change in momentum

= mu - ( - mv )

= mu+ mv

If F be the force applied and t be the duration of touch with the ball

Impulse = F x t

F x t = mu + mv

mv = Ft - mu

For given mu , greater the value of t , greater will be the value of v

so v is increased when t is increased .

Increased value of v will help in achieving greater distance attained by

golf ball

Answer:

I'm thinking Distillation

Explanation:

I'm not sure, but convection and conduction are insulators. It's between radiation and distillation if im not mistaking.

Answer:

The magnitude of the average force on the wall during the collision is 6 N.

Explanation:

Given;

mass of snowball, m = 120 g = 0.12 kg

velocity of the snowball, v = 7.5 m/s

duration of the collision between the snowball and the wall, t = 0.15 s

Magnitude of the average force can be calculated by applying Newton's second law of motion;

F = ma

where;

a is acceleration = v / t

a = 7.5 / 0.15

a = 50 m/s²

F = ma

F = 0.12 x 50

F = 6 N

Therefore, the magnitude of the average force on the wall during the collision is 6 N.

Answer: Because of the fine bore of the tube.

Explanation:

Temperature is the degree of hotness and coldness. And thermometer is the instrument use to measure temperature.

The two most common types of themometric fluids for thermometer are alcohol and mercury.

What makes a clinical thermometer suitable for measuring small changes in body temperature is because of the fine bore of the tube which makes it possible for small temperature changes to cause large changes in the length of mercury columns, making the thermometer very sensitive to temperature changes.

The most prominent feature of the thermometer is the kink or constriction of bore near the bulb.

Answer:

xxx

Explanation:

Answer:

See the attachment below.

Explanation:

Best Regards!

Answer:

The net force acting on the person is 5.34 N acting upwards

Explanation:

The information given are;

Mass of the person = 55.0 kg

Density of the person = 1010 kg/m³

Body volume of the person = Mass/Density = 55/1010 = 0.0545 m³

Weight of the person = Mass × Gravity = 55 × 9.81 = 539.55 N↓

Where the density of water = 1000 kg/m³, we have;

Upthrust = weight of the water displaced = Mass of the water displaced × Gravity

Mass of the water displaced = Density of water × Volume of the water

Mass of the water displaced = Density of water × Body volume of the person

                                                = 1000 × 0.0545 = 54.5 kg

The Upthrust = 54.5 × 9.81 = 534.21 N↑

Given that the net force, [tex]F_{Net}[/tex] acting is the sum of the mass of forces acting  on the person which are the weight acting downward and the upthrust acting upwards

We have;

F(Net) = 534.21↑ - 539.55↓ = 5.34 N↑

The net force acting on the person = 5.34 N upwards.

Answer:

-5.34

Explanation:

it says specifically to watch your sign

Answer:

B. particles in solid objects do not move

Explanation:

The kinetic theory model states that matter is made up of particles that are constantly moving but the movement of particles depends on whether the substance exists in the solid, liquid, or gaseous state.

So, according to the Kinetic Theory Model state, particles in solid objects do not move as all the particles in solids are closely packed.

Hence, the correct option is "B”.

Answer:

1 Nm

Explanation:

Given;

Force = F = (4, 3, 3)N

Position 1 = P = (3, 3, -1)m

Position 2 = Q = (2, -1, 4)m

The object moves along a straight line path from P to Q, therefore, the distance vector (d) is given by;

d = Q - P

d = (3, 3, -1) - (2, -1, 4)

d = (1, 4, -5)m

Now the work done (W) by the force (F) to move through the distance (d) is the dot product of the two vectors: F and d. i.e

W = F . d

For clarity, let's write vectors F and d in vector unit notation as follows;

F = 4 i + 3 j + 3 k

d = 1 i + 4 j - 5k

Therefore,

W = (4 i + 3 j + 3 k ) . (1 i + 4 j - 5k)

W = (4 + 12 - 15)

W = 1

Therefore, the workdone by the force is 1 Nm

Answer:

Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;

The closest (minimum) distance of Neptune from the Earth is 29 AU

The farthest (maximum) distance of Neptune fro the Earth is 31 AU

Explanation:

The following parameters are given;

The distance from the Earth to the Sun = 1 AU

The distance of Neptune from the Earth = 30 AU

We have;

When the Sun is between the Earth and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 + 1 = 31 AU

When the Earth is between the Sun  and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 - 1 = 29 AU

Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU

The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.

Answer:

29 AU

Explanation:

Answer:

≈933.3kg/m^3

Explanation:

Density=Mass/Volume

11200kg/12.0= 933.3333kg/m^3

Answer:

1. it helps to change the direction.

2. it helps us to walk on ground.

3. it helps the vechils to break while moving.

4. helps in changing one form of enegry to another form. eg when we rub our hands we feel heat energy.

5. it opposites the force.

6. it helps us to change shape of objects.eg we roll the dough to make it roti.

7. it changes the state of body from rest motion.eg when we push any obj from inclined plane it moves.

i all know is just 7..

Answer:

Distance = 30m

Displacement = 6m W

Explanation:

Given the following:

Movement 1 = 18m W

Movement 2 = 12m E

Diatance is a scalar quantity with only magnitude and no direction. That is, in Calculating the distance moved by the locomotive, the direction of travel or movement of the object is not considered. It only measures the total amount of movement made during the Time of motion.

Therefore, total distance traveled equals :

Movement 1 + movement 2

18m + 12m = 30m

B) Displacement also measures the movement made by an object. However, Displacement is a vector quantity and therefore, considers both magnitude and direction of travel of the object. Therefore, it measures the overall change in position of the object from its starting position.

Therefore, Displacement of the locomotive equals:

18m W - 12m E = 6m E

Answer:

The black lines represent the energy absorbed by the electrons.

Explanation:

Atoms emit lights when they are excited. These lights are of particular wavelengths that match with different colors. A series of colored lines appear along with spaces in the middle of the two colors. The middle of the colors is filled with dark spaces. Each spectral line of an element represents a specific characteristic of the element. These colored lines appearing in the form of series are termed to be the atomic spectrum of the element. Identification of the elements is done through the line of the spectrum they possess.

Answer:

(B) The black lines represent the energy absorbed by the electrons.

Explanation: