When looking at the chemical symbol, the charge of the ion is displayed as the
-superscript
-subscript
-coefficient
-product

Answers

Answer 1

Answer:

superscript

Explanation:

When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.

An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.

Answer 2

Answer:

superscript

Explanation:


Related Questions

An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)

Answers

Answer:

  K_{f} / K₀ =1.12

Explanation:

This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.

Initial moment. With arms outstretched

         L₀ = I₀ w₀

the wo value is 5.0 rad / s

final moment. After he shrugs his arms

         [tex]L_{f}[/tex] = I_{f}  w_{f}

indicate that the moment of inertia decreases by 11%

        I_{f} = I₀ - 0.11 I₀ = 0.89 I₀

        L_{f} = L₀

        I_{f} w_{f}  = I₀ w₀

        w_{f} = I₀ /I_{f}    w₀

let's calculate

        w_{f} = I₀ / 0.89 I₀   5.0

        w_{f} = 5.62 rad / s

Having these values ​​we can calculate the change in kinetic energy

         [tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)

         K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²

         K_{f} / K₀ =1.12

In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.

Answers

Answer:

The  angular speed is [tex]w = 5.89 \ rad/s[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 1.6 s[/tex]

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         [tex]\theta = n * 2 * \pi[/tex]

substituting values

        [tex]\theta = 1.5 * 2 * 3.142[/tex]

       [tex]\theta = 9.426 \ rad[/tex]

 The angular speed is mathematically represented as

         [tex]w = \frac{\theta }{t}[/tex]

substituting values

         [tex]w = \frac{9.426}{1.6}[/tex]

          [tex]w = 5.89 \ rad/s[/tex]

     

2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the

Answers

Answer:

for the first interference m = 1   y = 2,839 10-3 m

for the second interference m = 2   y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

                d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

           tan θ = y / L

with the angle it is small,

          tan θ = sin θ / cos θ = sin θ

   

we substitute

         sin θ = y / L

         d y / L = m λ

the distance between the central maximum and an interference line is

        y = m λ L / d

let's reduce the magnitudes to the SI system

     λ = 546 nm = 546 10⁻⁹ m

     d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

      y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

      y =  m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m

When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.

Answers

Answer:

335.79cm/s

Explanation:

When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;

v = fλ        ------------------(ii)

From the question;

f = 6.3Hz

λ = 53.3cm

Substitute these values into equation (i) as follows;

v = 6.3 x 53.3

v = 335.79cm/s

Therefore, the speed of the wave pulses along the wire is 335.79cm/s

Wind gusts create ripples on the ocean that have a wavelength of 3.03 cm and propagate at 3.37 m/s. What is their frequency (in Hz)?

Answers

Answer:

Their frequency is 111.22 Hz

Explanation:

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration and is expressed in units of length (m).

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

The propagation speed of a wave is the quantity that measures the speed at which the wave's disturbance propagates throughout its displacement. The speed at which the wave propagates depends on both the type of wave and the medium through which it propagates. Relate wavelength (λ) and frequency (f) inversely proportional using the following equation:

v = f * λ.

Then the frequency can be calculated as: f=v÷λ

In this case:

λ=3.03 cm=0.0303 m (1m=100 cm)v= 3.37 m/s

Replacing:

[tex]f=\frac{3.37 \frac{m}{s} }{0.0303 m}[/tex]

Solving:

f=111.22 Hz

Their frequency is 111.22 Hz

A skater on ice with arms extended and one leg out spins at 3 rev/s. After he draws his arms and the leg in, his moment of inertia is reduced to 1/2. What is his new angular speed

Answers

Answer:

The new angular speed is [tex]w = 6 \ rev/s[/tex]

Explanation:

From the  question we are told that

      The angular velocity of the spin is  [tex]w_o = 3 \ rev/s[/tex]

       The  original moment of inertia is  [tex]I_o[/tex]

        The new moment of inertia is  [tex]I =\frac{I_o}{2}[/tex]    

Generally angular momentum is mathematically represented as

      [tex]L = I * w[/tex]

Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so

         [tex]I * w = constant[/tex]

=>       [tex]I_o * w _o = I * w[/tex]

where w is the new angular speed  

  So  

          [tex]I_o * 3 = \frac{I_o}{2} * w[/tex]

=>        [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]

=>         [tex]w = 6 \ rev/s[/tex]

At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)

Answers

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

Four identical charges particles of charge 1Uc, 2Uc,
3Uc and 4Uc
are placed at x = lm, x=2m,
x=3m and
x=5m. The electric field intensity
at origin is?

Answers

Answer:

17.94 kN/C is the electric field intensity at the origin due to the charges.

Explanation:

From the question, we are told that

The distance of 1 μC from origin = 1 m

The distance of 2 μC from origin = 2 m

The distance of 3 μC from origin = 3 m

The distance of 4 μC from origin = 5 m

Therefore, for us to find the electric field intensity, we'll solve below:

The formula for Electric field intensity = ( k * q ) / ( r * r )

where , r is distance ,

k = 9 * 10^9 ,

and , q is charge .

now ,

electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]

=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C

=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C

=> electric field intensity at the origin = 17.94 kN/C

The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period

Answers

Answer:

Q = 23.49 C

Explanation:

We have,

Electric current drawn by the air conditioner is 16.2 A

Time, t = 1.45 s

It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,

[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]

So, the charge of 23.49 C is passing through the circuit during this period.

Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
Assuming that the skateboarder's acceleration is gsin 18.0 ∘, find his speed when he reaches the bottom of the ramp in 3.50 s .

Answers

Answer:

Explanation:

v= u + at

v is final velocity , u is initial velocity . a is acceleration and t is time

Initial velocity u = 0 . Putting the given values in the equation

v = 0 + g sin 18 x 3.5

= 10.6 m /s

For a skateboarder who starts from the rest, the speed when he reaches the bottom of the ramp will be 10.6 m/s.

What are Velocity and Acceleration?

The term "velocity" refers to a vector measurement of the rate and direction of motion. Velocity is the rate of movement in a single direction, to put it simply. Velocity can be used to determine how fast a rocket is heading into space and how fast a car is moving north on a congested motorway.

There are several types of velocity :

Instantaneous velocityAverage VelocityUniform VelocityNon-Uniform Velocity

The pace at which a person's velocity changes is known as acceleration. This implies that an object is accelerating if its velocity is rising or falling. An object that is accelerating won't have a steady change in location every second like an item moving at a constant speed does.

According to the question, the given values are :

Time, t = 3.50 sec

Initial Velocity, u = 0 m/s

Use equation of motion :

v = u+at

v = 0+ g sin 18 × 3.5

v = 10.6 m/s.

So, the final velocity will be 10.6 m/s.

To get more information about Velocity and Acceleration :

https://brainly.com/question/14683118

#SPJ2

An electron, moving west, enters a magnetic field of a certain strength. Because of this field the electron curves upward. What is the direction of the magnetic field?

Answers

Answer:

Towards the west.

Explanation:

The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.

Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.

According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward

can I get help please?​

Answers

Answer shown on photo

A "laser cannon" of a spacecraft has a beam of cross-sectional area A. The maximum electric field in the beam is 2E. The beam is aimed at an asteroid that is initially moving in the direction of the spacecraft. What is the acceleration of the asteroid relative to the spacecraft if the laser beam strikes the asteroid perpendicularly to its surface, and the surface is not reflecting

Answers

Answer:

Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m

Explanation:

The maximum electric field in the beam = 2E

cross-sectional area of beam = A

The intensity of an electromagnetic wave with electric field is

I = cε[tex]E_{0} ^{2}[/tex]/2

for [tex]E_{0}[/tex] = 2E

I = 2cε[tex]E^{2}[/tex]    ....equ 1

where

I is the intensity

c is the speed of light

ε is the permeability of free space

[tex]E_{0}[/tex]  is electric field

Radiation pressure of an electromagnetic wave on an absorbing surface is given as

P = I/c

substituting for I from above equ 1. we have

P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex]    ....equ 2

Also, pressure P = F/A

therefore,

F = PA    ....equ 3

where

F is the force

P is pressure

A is cross-sectional area

substitute equ 2 into equ 3, we have

F = 2ε[tex]E^{2}[/tex]A

force on a body = mass x acceleration.

that is

F = ma

therefore,

a = F/m

acceleration of the asteroid will then be

a = 2ε[tex]E^{2}[/tex]A/m

where m is the mass of the asteroid.

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly:___________.
A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c

Answers

Answer:

0.85c

Explanation:

Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²

Rest mass of proton [tex]M_{0P}[/tex]  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV

for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV

Recall that the rest energy, and the total energy are related by..

[tex]E[/tex] = γ[tex]E_{0}[/tex]

which can be written in this case as

[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1

where [tex]E[/tex] = total energy of the kaon, and

[tex]E_{0}[/tex] = rest energy of the kaon

γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]

where [tex]\beta = \frac{v}{c}[/tex]

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2

where [tex]E_{K}[/tex] is the total energy of the kaon, and

[tex]E_{0P}[/tex] is the rest energy of the proton.

From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89

1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1

squaring both sides, we get

3.57( 1 - [tex]\beta^{2}[/tex]) = 1

3.57 - 3.57[tex]\beta^{2}[/tex] = 1

2.57 = 3.57[tex]\beta^{2}[/tex]

[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72

[tex]\beta = \sqrt{0.72}[/tex] = 0.85

but, [tex]\beta = \frac{v}{c}[/tex]

v/c = 0.85

v = 0.85c

The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.

Answers

Answer and Explanation:

There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e.  [tex]B \alpha \frac{1}{r^2}[/tex]

The magnetic field is a volume of vectors

And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit

Therefore for a closed path,  we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.

find the value of k for which the given pair of vectors are not equal
2ki +3j​ and 8i + 4kj

Answers

Answer:

5

Explanation:

Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)?

Answers

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

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