define limitations in the operation conditions of a pn junction

Answers

Answer 1

Answer:

Such limitations are given below.

Explanation:

Each pn junction provides limited measurements of maximum forwarding current, highest possible inversion voltage as well as the maximum output level.If controlled within certain adsorption conditions, the pn junction could very well offer satisfying performance. In connector operation, the maximum inversion voltage seems to be of significant importance.


Related Questions

Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

Answers

Answer:

7.2N/C

Explanation:

Pls see attached file

A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upward speed of 1.00 m/s. The ball is in contact with the ground for 0.0140 s.

Required:
What is the average force exerted by the ground on the ball during this time? Also explain whether it's upwards or downwards.

Answers

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist and did not keep good records, the credit for the discovery of the Leyden jar usually goes to physicist Pieter Musschenbroek from Leyden, Holland. Musschenbroek accidentally discovered the Leyden jar when he tried to charge a jar of water and shocked himself by touching the wire on the inside of the jar while holding the jar on the outside. He said that the shock was no ordinary shock and his body shook violently as though he had been hit by lightning. The energy from the jar that passed through his body was probably around 1 J, and his jar probably had a capacitance of about 1 nF.A) Estimate the charge that passed through Musschenbroek's body.
B) What was the potential difference between the inside and outside of the Leyden jar before Musschenbroek discharged it?

Answers

Answer:

a) q = 4.47 10⁻⁵ C

b)     ΔV = 4.47 10⁴ V

Explanation:

A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor

           U = Q² / 2C

         Q = √ (2UC)

let's reduce the magnitudes to the SI system

   c = 1 nF = 1 10⁻⁹ F

let's calculate

         q = √ (2 1 10⁻⁹-9)

         q = 0.447 10⁻⁴ C

         q = 4.47 10⁻⁵ C

b) for the potential difference we use

             C = Q / ΔV

            ΔV = Q / C

            ΔV = 4.47 10⁻⁵ / 1 10⁻⁹

            ΔV = 4.47 10⁴ V

Two 2.0-cm-diameter insulating spheres have a 6.70 cm space between them. One sphere is charged to +70.0 nC, the other to -40.0 nC. What is the electric field strength at the midpoint between the two spheres?

Answers

Answer:

Explanation:

The distance of middle point from centres of spheres will be as follows

From each of 2 cm diameter sphere

R  = 1 + 6.7 / 2 = 4.35 cm = 4.35 x 10⁻² m

Expression for electric field = Q / 4πε R²

Electric field due to positive charge

E₁ = 70  x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴

= 33.3 x 10⁴ N/C

Electric field due to negative  charge

E₂ = 40  x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴

= 19.02 x 10⁴ N/C

E₁ and E₂ act in the same direction so

Total field = (33.3 + 19.02 ) x 10⁴

= 52.32 x 10⁴ N/C .

Water is boiled at 1 atm pressure in a 20-cm-internal diameter polished copper pan on an electric range. If it is observed that the water level in the pan drops by 8.00 cm in 15 minutes, determine the inner surface temperature of the pan.

Answers

Answer:

11.3298W

Explanation:

The rate of heat transfer is determined from the enthalpy of vaporization at the give pressure obtained and the mass flow rate. The mass flow rate is determined from the volume of the boiled water, the given time interval and the specific volume of the saturated liquid.

Given that

1atm as the atmospheric pressure

Internal diameter = 20cm = 0.2m

Time = 15mins = (15×60)secs

Latent heat of vaporization (hevap) = 2256.6

Q = mh(evap)

= m/∆t . hevap

= V/αliq∆t ×h(evap)

D^2π∆h/4αliq ∆t × hevap

= 0.2^2 ×π×0.8×2256.5/4×0.001043×15×60

=0.04×3.142×0.08×2256.6/2.00256

= 22.68876/2.00256

Q = 11.3298W

A ranger needs to capture a monkey hanging on a tree branch. The ranger aims his dart gun directly at the monkey and fires the tranquilizer dart. However, the monkey lets go of the branch at exactly the same time as the ranger fires the dart. Will the monkey get hit or will it avoid the dart?

Answers

Answer:

Yes the monkey will get hit and it will not avoid the dart.

Explanation:

Yes, the monkey will be hit anyway because the dart will follow a hyperbolic path and and will thus fall below the branches, so if the monkey jumps it will be hit.

No, the monkey will not avoid the dart because dart velocity doesn't matter. The speed of the bullet doesn’t even matter in this case because a faster bullet will hit the monkey at a higher height and while a slower bullet will simply hit the monkey closer to the ground.

Two charged particles of equal magnitude (+Q and +Q) are fixed at opposite corners of a square that lies in a plane. A test charge +q is placed at the third corner of the square. What is the direction of force on the test charge due to other two charges?

Answers

Answer:

The test charge will take the south-west direction indicated in option 6.

Explanation:

The image is shown below.

Since all the charges are positively charged, they will all repel each other. If we consider the force on +q due to +Q and +Q, then we can proceed as follows

The +Q particle at the top left corner of the cube will exert a vertical downward force on +q in the -ve y-axis.

The +Q particle at the bottom right corner of the cube will exert a force on +q towards the horizontal left on the -ve x-axis.

Both of these forces will act at angle of 90°, and therefore, the resultant force will act at an angle of 45° to horizontal and vertical forces.

The result is that the +q charge will move in a south-west direction of the cube.

A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it

Answers

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. [tex]Q[/tex], the amount of charge stored in this capacitor, will stay the same.

The formula [tex]\displaystyle Q = C\, V[/tex] relates the electric potential across a capacitor to:

[tex]Q[/tex], the charge stored in the capacitor, and[tex]C[/tex], the capacitance of this capacitor.

While [tex]Q[/tex] stays the same, moving the two plates apart could affect the potential [tex]V[/tex] by changing the capacitance [tex]C[/tex] of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

[tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex],

where

[tex]\epsilon[/tex] is the permittivity of the material between the two plates.[tex]A[/tex] is the area of each of the two plates.[tex]d[/tex] is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of [tex]\epsilon[/tex]. Neither will that change the area of the two plates.

However, as [tex]d[/tex] (the distance between the two plates) increases, the value of [tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex] will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula [tex]\displaystyle Q = C\, V[/tex] can be rewritten as:

[tex]V = \displaystyle \frac{Q}{C}[/tex].

The value of [tex]Q[/tex] (charge stored in this capacitor) stays the same. As the value of [tex]C[/tex] becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

a wave with a high amplitude______?

Answers

. . . is carrying more energy than a wave in the same medium with a lower amplitude.

Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of 38.0° with the normal to the surface, while in the glass it makes an angle of 26.0° with the normal. What is the index of refraction of the glass?

Answers

Answer:

n_glass = 1.404

Explanation:

In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]         (1)

n1: index of refraction of vacuum = 1.00

θ1: angle of the incident light respect to normal of the surface = 38.0°

n2: index of refraction of glass = ?

θ2: angle of the refracted light in the glass respect to normal = 26.0°

You solve the equation (1) for n2 and replace the values of all parameters:

[tex]n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404[/tex]

The index of refraction of the glass is 1.404

Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis through their centers. Two of the flywheels rotate in one direction at 8.94 rad/s, but the other one rotates in the opposite direction at 3.42 rad/s.

Required:
Calculate the magnitude of the net angular momentum of the system.

Answers

Answer:

the angular momentum is 1015.52 kg m²/s

Explanation:

given data

mass of each flywheel, m = 65 kg

radius of flywheel, r = 1.47 m

ω1 = 8.94 rad/s

ω2 = - 3.42 rad/s

to find out

magnitude of the net angular momentum

solution

we get here Moment of inertia that is express as

I = 0.5 m r²    .................1

put here value and we get

I = 0.5 × 65 × 1.47 × 1.47

I = 70.23 kg m²

and

now we get here Angular momentum that is express as

L = I × ω    ...........................2

and Net angular momentum will be

L = 2 × I x ω1 - I × ω2

put here value and we get

L = 2 × 70.23 × 8.94 - 70.23 × 3.42

L = 1015.52 kg m²/s

so

the angular momentum is 1015.52 kg m²/s

The magnitude of the net angular momentum of the system will be "1015.52 kg.m²/s".

Momentum

According to the question,

Flywheel's mass, m = 65 kg

Flywheel's radius, r = 1.47 m

ω₁ = 8.94 rad/s

ω₂ = 3.42 rad/s

We know,

The moment of inertia (I),

= 0.5 m r²

By substituting the values,

= 0.5 × 65 × 1.47 × 1.47

= 70.23 kg.m²

hence, The angular momentum be:

→ L = I × ω or,

     = 2 × I × ω₁ - l × ω₂

     = 2 × 70.23 × 8.94 - 70.23 × 3.42

     = 1015.52 kg.m²/s

Thus the above answer is correct.

Find out more information about momentum here:

https://brainly.com/question/25121535

a certain volume of dry air at NTP is allowed to expand five times of it original volume under adiabatic condition.calculate the final pressure.(air=1.4)​

Answers

Answer:

Final pressure 0.105atm

Explanation:

Let V1 represent the initial volume of dry air at NTP.

under adiabatic condition: no heat is lost or  gained by the system. This does not implies that the constant temperature throughout the system , but rather that no heat gained or loss by the system.

Adiabatic expansion:

[tex]\frac{T_1}{T_2} =(\frac{V_1}{V_2} )^{\gamma -1}[/tex]

273/T2=(5V1/V1)^(1.4−1)

273/T2=5^0.4

Final temperature  T2=143.41 K

Also

P1/P2=(V2/V1)^γ

1/P2=(5V1/V1)^1.4

Final pressure P2=0.105atm

An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aperture of diameter D . Which aperture diameter gives the best resolution? D=(1/2)????r D=????r D=2????r

Answers

Explanation:

As per Rayleigh criterion, the angular resolution is given as follows:

[tex]\theta=\frac{1.22 \lambda}{D}[/tex]

From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.

How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the

ground and onto a shelf in your closet?

O A. 2.45 J

OB. 4J

C. 39.2 J

D. 20 J

Answers

Answer:

Option C - 39.2 J

Explanation:

We are given that;

Mass; m = 2 kg.

Distance moved off the floor;d = 10 m.

Acceleration due to gravity;g = 9.8 m/s².

We want to find the work done.

Now, the Formula for work done is given by;

Work = Force × displacement.

In this case, it's force of gravity to lift up the boots, thus;

Formula for this force is;

Force = mass x acceleration due to gravity

Force = 2 × 9.8 = 19.2 N

∴ Work done = 19.6 × 2

Work done = 39.2 J.

Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.

Answer:39.2J

Explanation: I just answered this question and this was the correct answer. 4J is the wrong answer.

The ball tends to come back to the centerline of the flow when it is pushed by an external disturbance. Explain this phenomenon using the curvature of streamlines.

Answers

Answer is given below

Explanation:

given data

we will consider here

Ping-Pong ball weighs = 3.1 g

diameter =  4.2 cm

solution

Whenever the ball is pushed, the length of the airflow along the outer edge increases and it accelerates. According to Bernoulli's equation. As the speed increases, the pressure decreases, so the pressure at the outer end is reduced. As the pressure at the outer edge is low, the extra air jet pushes it back to the center line.

Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached to an eye anchor embedded in a wall. Rope segment BC creates an angle of ϕ = 51.0 ∘ with the floor and rope segment CD creates an angle θ with the horizontal. If both ropes BCA and CD can support a maximum tensile force Tmax = 120 lb , what is the maximum weight Wmax of the crate that the system can support? What is the

Answers

Answer:

Wmax = 63.65 ≈ 64 lb

Explanation:

Find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the zero of potential energy at sea level.

Answers

Answer:

P = 1470980 J

Explanation:

We have,

Mass of the hiker is 79 kg

It is required to find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level.

It is given by :

[tex]P=mgh\\\\P=79\times 9.8\times 1900\\\\P=1470980\ J[/tex]

So, the potential energy of 1470980 J is associated with a hiker.

If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will Group of answer choices

Answers

Answer:

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

Explanation:

The power dissipation by an electrical circuit is given by the following formula:

Power Dissipation = (Voltage)(Current)

P = VI

but, from Ohm's Law, we know that:

Voltage = (Current)(Resistance)

V = IR

Substituting this in formula of power:

P = (IR)(I)

P = I²R   ---------------- equation 1

Now, if we double the current , then the power dissipated by that circuit will be:

P' = I'²R

where,

I' = 2 I

Therefore,

P' = (2 I)²R

P' = 4 I²R

using equation 1

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

When one person was talking in a small room, the sound intensity level was 60 dB everywhere within the room. Then, there were 14 people talking in similar manner simultaneously in the room, what was the resulting sound intensity level?
A. 64 dB
B. 60 dB
C. 69 dB
D. 79 dB
E. 71 dB

Answers

Answer:

E= 71dB

Explanation:

See attached file for step by step calculation

A 0.410 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 35.0 pC charge on its surface. What is the potential (in V) near its surface

Answers

Answer:

The  potential is  [tex]V = 153.659 \ V[/tex]

Explanation:

From the question we are told that

     The diameter of the plastic sphere is  [tex]d = 0.410 \ cm = 0.0041 \ m[/tex]

      The magnitude of the charge is  [tex]q = 35.0 pC = 35.0 *10^{-12} \ C[/tex]

The radius of the plastic sphere is  mathematically evaluated as

          [tex]r = \frac{d}{2}[/tex]

=>     [tex]r = \frac{0.0041}{2}[/tex]

       [tex]r = 0.00205 \ m[/tex]

The  potential near the surface is mathematically represented as

         [tex]V = \frac{k * q}{r }[/tex]

Where k is the Coulombs constant with value [tex]9 *10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

substituting values  

       [tex]V = \frac{9*10^9 * 35 *10^{-12}}{0.00205}[/tex]

       [tex]V = 153.659 \ V[/tex]

       

If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density

Answers

The ball with greater mass has more density

1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.
a. Compute the acceleration of the crate?

Answers

Answer:

The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]

Explanation:

Recall the formula that relates force,mass and acceleration from newton's second law;

[tex]F=m\,a[/tex]

Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:

[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]

According to the model in which active galactic nuclei are powered by supermassive black holes, the high luminosity of an active galactic nucleus primarily consists of

Answers

Answer:

the high luminosity of an active galactic nucleus primarily consists of light emitted by hot gas in an accretion disk that swirls around the black hole

Find the pressure difference (in kPa) on an airplane wing if air flows over the upper surface with a speed of 125 m/s, and along the bottom surface with a speed of 109 m/s. [Express answer in TWO decimal places]

Answers

Answer:

P= 2414.9 Pa

Explanation:

given

density of air , p = 1.29 kg/m³

speed of air over the upper surface , v₁ = 125 m/s

speed of air over the lower surface , v₂ = 109 m/s

the pressure difference on an airplane wing , P = 0.5 × p × ( v₁² - v₂²)

P = 0.5 × 1.29 × ( 125² - 109²)

P= 0.645(3744)

P = 2414.9 Pa

the pressure difference on an airplane wing is 2414.9 Pa

A current carrying wire is oriented along the y axis It passes through a region 0.45 m long in which there is a magnetic field of 6.1 T in the z direction The wire experiences a force of 15.1 N in the x direction.1. What is the magnitude of the conventional current inthe wire?I = A2. What is the direction of the conventional current in thewire?-y+y

Answers

Answer:

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

Explanation:

- To find the direction of the conventional current in the wire you use the following formula:

[tex]\vec{F}=i\vec{l}\ X\ \vec{B}[/tex]       (1)

i: current in the wire = ?

F: magnitude of the magnetic force on the wire = 15.1N

B: magnitude of the magnetic field = 6.1T

l: length of the wire that is affected by the magnetic field = 0.45m

The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).

The direction of the current must be in the +y direction (+^j). In fact, you have:

^j X ^k = ^i

The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):

[tex]F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A[/tex]

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

What is surface tension??​

Answers

Answer:

Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects (e.g. water striders), usually denser than water, to float and slide on a water surface.

Explanation:

Answer:

It is the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area.

A charging bull elephant with a mass of 5500 kg comes directly toward you with a speed of 4.70 m/s . You toss a 0.160-kg rubber ball at the elephant with a speed of 7.50 m/s(a) When the ball bounces back toward you, what is its speed? (b) How do you account for the fact that the ball's kinetic energy has increased?

Answers

Answer:

v2 = - 16.899 m/s

velocity of ball increases so that the kinetic energy of the ball increases.

Explanation:

given data

mass of elephant, m1 = 5500 kg

mass of ball, m2 = 0.160 kg

initial velocity of elephant, u1 = - 4.70 m/s

initial velocity of ball, u2 = 7.50 m/s

solution

we consider here final velocity of ball = v2

so  collision formula is express as for v2

[tex]v_{2}=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )u_{1}+\left ( \frac{m_{2}-m_{1}}{m_{1}+m_{2}} \right )u_{2}[/tex]      .................1

put here value and we get

[tex]v_{2}=\left ( \frac{2\times 5500}{5500+0.160} \right )(-4.70)+\left ( \frac{0.16-5500}{5500+0.160} \right )(7.50)[/tex]  

solve it we get

v2 = - 16.899 m/s

here negative sign shows that the ball bounces back towards you

and

here we know the velocity of ball increases so that the kinetic energy of the ball increases.

and due to this effect, it will gain in energy is due to the energy from the elephant mass

Which jovian planet should have the most extreme seasonal changes? a. Saturn b. Neptune c. Jupiter d. Uranus

Answers

Answer:

D). Uranus.

Explanation:

Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.

As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, option D is the correct answer.

A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evaporate

Answers

gain heat energy and evaporate

For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.

When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.

For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.

Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.

Learn more: https://brainly.com/question/5019199

If a key is pressed on a piano, the frequency of the resulting sound will determine the ________, and the amplitude will determine the ________ of the perceived musical note.

Answers

Answer:

If a key is pressed on a piano, the frequency of the resulting sound will determine the ___PITCH_____, and the amplitude will determine the _____LOUDNESS___ of the perceived musical note.

Explanation:

The frequency of a vibrating string is primarily based on three factors:

The sounding length (longer is lower, shorter is higher)

The tension on the string (more tension is higher, less is lower)

The mass of the string, normally based on a uniform density per unit length (higher mass is lower, lower mass is higher)

To make a shorter string (such as in an upright piano) sound the same fundamental frequency as a longer string (such as in a 9' grand piano), either the thickness of the string must be increased (which increases the density and the mass) or the tension must be decreased, and usually it's a bit of both.

Thicker strings are often stiffer and that creates more inharmonic partials, and lower tension is associated with other problems, so the best way to make a string sound lower is the make it longer, but it is not practical to make a piano from strings that are all the same density and tension, because the lowest strings would have to be ridiculously long. Nine feet is already a great demand on space for a single musical instrument, and of course those pianos are extremely expensive and difficult to move.

And alsoBesides the pitch of a musical note, perhaps the most noticeable feature in how loud the note is. The loudness of a sound wave is determined from its amplitude. While loudness is only associated with sound waves, all types of waves have an amplitude. Waves on a calm ocean may be less than 1 foot high. Good surfing waves might be 10 feet or more in amplitude. During a storm the amplitude might increase to 40 or 50 feet.

Many things can influence the amplitude.

What is producing the sound?

How far are you from the source of the sound? The farther away the smaller the amplitude.

Intervening material. Sound does not travel through walls as well as air.

Depends on what is detecting the wave sound. Ear vs. microphone.

Answer:

The frequency will determine the pitch

the amplitude will determine the loudness

Explanation:

The frequency of a sound refers to the number of vibrations made by the sound wave produced in a unit of time. This usually affects how high or how low a note is perceived in music. High-frequency sounds have higher pitches, while low-frequency sounds have lower pitches.

The amplitude of a sound wave refers to the height between the wave crests and the equilibrium line in a sound wave. It shows how loud a sound will be. High amplitude sounds are loud while low amplitude sounds are quiet.

Other Questions
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