It takes a sample of molecules 10 s to diffuse 1.0 mm. how long will it take to diffuse 2.0 mm ?

Answers

Answer 1

The time taken to diffuse 2.00 mm will be 20 seconds.

The velocity of diffusion(v) is given by the following formula:

v = Distance/Time

We are given that it takes a sample of molecules 10 seconds to diffuse 1.00 mm.

Thus, its velocity is

v = 1.0 mm/10 s = 0.1 mm/s

In the second case, the distance is given, so we can calculate the time required for diffusion by again applying the same formula

0.1 = 2.00 mm/Time

On rearranging above equation, we get

Time = 20 seconds

Thus, the time taken to diffuse 2.00 mm will be 20 seconds.

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Related Questions

Explain When might a scientist use a
model?

Answers

Hi! Scientists use models for a lot of reasons. Primarily to study objects in detail that they wouldn't be able to normally. For example, to study something small like a cell, making a model could be easier. Also, things that are too dangerous to study in real life can be turned into models as well.

To put it shortly, scientists use models to observe things in better detail.

I can include a link to a website if that will help :)

https://study.com/academy/lesson/why-scientists-use-models-simulations.html

Good luck!

Identify the charges that are negative. mentum. . ​

Answers

Answer:

Charges B and C are negative

Explanation:

• We are certain of a law of magnetism that states "Field lines move from positive charge to negative charge"

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Answer:

B and C are negative.

Explanation:

This is because according to the law of charges in electric field line it says that the lines forming the electric field move from the positive terminal to the negative terminal,hence making B and C be negative as the lines are moving into them.

how many meters are in 45 minutes

Answers

Answer:

*I think you meant seconds

45 min = 2,700 sec

Explanation:

Hope it helps!!!

:}

A battery being tested has five cells testing normal and one cell showing very low. What action should be taken?
A. None, the battery is normal
B. Recharge the entire battery
C. Recharge only the cell that is very low
D. Replace the battery

Answers

It’s B Recharge the entire battery:)

The entire battery should be charged, as a battery being tested has five cells testing normal and one cell showing very low, so option B is correct.

What is Battery?

A battery is a component that turns chemical energy into electrical energy and stores it. This process is called electrochemistry, and the structure that supports a battery is referred to as an electrochemical cell. One or more electrochemical cells can be used to create a battery, as in Volta's original pile. Two electrodes and an electrolyte are the main components of an electrochemical cell.

Some popular batteries are only intended for one usage (known as primary or disposable batteries). The electrons only travel in one direction from the anode to the cathode. Either the buildup of reaction products on the electrodes prevents the reaction from continuing, or their electrodes get depleted as they release their positive or negative ions into the electrolyte, and it is finished. The battery is eventually disposed of (or, ideally, recycled; however, that is a whole different Nova topic).

Therefore, the entire battery should be charged, as a battery being tested has five cells testing normal and one cell showing very low.

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Two cats jump off a roof that is 23m off the ground. Cat A jumps directly up with a velocity of 7m/s, and cat B jumps directly down with a velocity of 7m/s. How far up does cat A go? How fast does cat A go just before it hits the ground? How fast does cat B go just before it hits the ground? How much longer is cat A in the air than cat B?

Answers

The height reached by cat A is 2.5 m.

The speed of cat A before it hits the ground is 36.4 m/s.

The speed of cat B before it hits the ground is 22.39 m/s.

The time spent in air by cat A than cat B is 1.43 seconds.

Time of motion of cat A

The time of motion of cat A is calculated as follows;

h = vt - ¹/₂gt

-23 = 7t - 4.9t²

4.9t² - 7t - 23 = 0

solve the quadratic equation using formula method;

a = 4.9, b = -7, c = - 23

t = 3.0 seconds

Time of motion of cat B

The time of motion of cat B is calculated as follows;

h = vt - ¹/₂gt

23 = 7t + 4.9t²

4.9t² + 7t - 23 = 0

solve the quadratic equation using formula method;

a = 4.9, b = 7, c = -23

t = 1.57 seconds

Height reached by cat A

h = u²/2g

h = (7²)/(2 x 9.8)

h = 2.5 m

Speed of cat A before it hits the ground

v = u + gt

v = 7  + 3(9.8)

v = 36.4 m/s

Speed of cat B before it hits the ground

v = u + gt

v = 7  + 1.57(9.8)

v = 22.39 m/s

Time spent in air by cat A than cat B

Time difference = 3 s - 1.57 s = 1.43 seconds

Thus, the height reached by cat A is 2.5 m.

The speed of cat A before it hits the ground is 36.4 m/s.

The speed of cat B before it hits the ground is 22.39 m/s.

The time spent in air by cat A than cat B is 1.43 seconds.

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Why does it make sense for the valence electrons to stay the same within a group?.

Answers

It makes sense for the valence electrons to stay the same within a group because these elements have same chemical properties.

We have a group from Periodic table.

We have to investigate the reason behind the fact that why elements with same number of valence electrons share the same group.

What is Periodic Table?

Periodic Table is a table of the chemical elements arranged in order of atomic number, usually in rows, so that elements with similar atomic structure (similar chemical properties) appear in vertical columns.

According to the question -

It is true that the number of valence electrons in the outermost shell of the elements of same group are equal. The number of valence electrons in the outermost orbit of the elements of the same group is same as the group number to which they belong. The elements in the same group have same number of valence electrons in their outermost shell. Due to this the elements have almost same Physical properties, Chemical properties and Reactivities of the elements.

Hence, it makes sense for the valence electrons to stay the same within a group because the elements in the same group have same properties.

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A car driver measures a tire pressure of 220 kpa. What is the absolute pressure in the tire?

Answers

321 kPa is the calculated pressure.

The measurement of pressure in relation to the pressure of a complete vacuum, or absolute zero pressure, is known as absolute pressure. In the same way as temperature must be expressed by its absolute unit, the Kelvin, the ideal gas law also requires the measurement of absolute pressure. Absolute pressure sensors have a permanently sealed absolute vacuum chamber that is exposed to the side of the sensor that is not in contact with the pressure media. Since the diaphragm employs the sealed vacuum as its reference and zero point, ambient pressure has no effect on its deformation.

Gauge pressure plus atmospheric pressure equals absolute pressure (220 + 101 = 321 kPa).

The pressure as compared to atmospheric pressure is known as gauge pressure.

Gauge pressure and ambient pressure are added to get absolute pressure.

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A cook puts 9.00g of water in a 2.00 -L pressure cooker that is then warmed to 500°C . What is the pressure inside the container?

Answers

If a cook puts 9.00g of water in a 2.00 -L pressure cooker that is then warmed to 500°C, the pressure inside the container is 15.8 atm.

What is the pressure inside the container?

From the ideal gas law;

PV = nRT

Where P is pressure, V is volume, n is amount of substance, T is temperature and R is the ideal gas constant (0.08206Latm/molK).

Given the data in the question;

Mass of water m = 9.00gVolume V = 2.00LTemperature = 500°C = ( 500 + 273.15)K = 773.15KPressure P = ?

First we determine the amount of water n.

We know that the molar mass of water is 18.01528 g/mol.

Hence

n = mass / molar mass

n = 9.00g / 18.01528 g/mol

n = 0.4995mol

Now, plug the values into the equation above and solve for P

PV = nRT

P = nRT / V

P = ( 0.4995mol × 0.08206Latm/molK × 773.15K ) / 2.00L

P = ( 31.6906Latm ) / 2.00L

P = 15.8 atm

If a cook puts 9.00g of water in a 2.00 -L pressure cooker that is then warmed to 500°C, the pressure inside the container is 15.8 atm.

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5. If a bullet leaves the muzzle of a rifle with a speed of 600 ms and the barrel of the rifle is 0.800m long, at what rate is the bullet accelerated while in the barrel?

Answers

The acceleration of the bullet leaving the barrel is 2.25 x 10⁵ m/s².

Given,

Velocity of the bullet leaving the barrel, v = 600 m/s

Distance travelled by the bullet = Size of the rifle = s = 0.800 m

According the Third Equation of Motion which relates velocity with the distance and acceleration says that,

v² = u² + 2as ;

where 'v' and 'u' are final and initial velocity of the bullet respectively

'a' is the acceleration of the bullet

's' is the distance travelled by the bullet

here u = 0 m/s

(600)² = 0² + 2 * (0.800) * (a)

360000 = (1.6) * (a)

a = [tex]\frac{360000}{1.6}[/tex]

a = 225000 m/s²

a = 2.25 x 10⁵ m/s²

Therefore the rate at which bullet accelerated while being in the barrel is 2.25 x 10⁵ m/s².

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A tuna jumps out of the water with an initial velocity of 44 feet per second (assume its starting height is 0 feet). Use the vertical motion model, h 16t²+uts where is the initial velocity in feet per second and is is the height in feet, to calculate the amount of time the tuna is in the air before it hits the ground again. Round your answer to the nearest tenth if necessary. ​

Answers

The time spent in air by the tuna before it hits the ground is determined as 1.375 seconds.

What is time spent in air by an object?

The time spent in air by an object or a projectile is the total time of motion of the object or projectile.

Final velocity of the tuna when it hits the ground

The time spent in air by the tuna can be determined by finding equation of the final velocity of the tuna when it hits the ground as follows;

h = 16t² + ut

where;

h is the height traveled by the tunau is the initial velocity of the tunet is the time of motion

Velocity of the tuna when it hits the ground,

v = dh/dt

dh/dt = 32t + u

when the tuna hits the ground, final velocity = 0

0 = 32t + u

0 = 32t + 44

t = -44/32

|t| = 1.375 seconds

Thus, the time spent in air by the tuna before it hits the ground is determined as 1.375 seconds.

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In his experiments on "cathode rays" during which he discovered the electron, J. J. Thomson showed that the same beam deflections resulted with tubes having cathodes made of different materials and containing various gases before evacuation.(b) When he applied various potential differences to the deflection plates and turned on the magnetic coils, alone or in combination with the deflection plates, Thomson observed that the fluorescent screen continued to show a single small glowing patch. Argue whether his observation is important.

Answers

The mobility of the light source depends on the interaction between the magnetic and electric forces.

What is a cathode ray?

Cathode ray refers to the electron beam that moves from the negatively charged cathode to the positively charged anode at the other end of the vacuum tube.

J. J. Thomson demonstrated that the identical beam deflections produced tubes with cathodes composed of varied materials and containing a range of gases prior to evacuation.

This discovery is significant because the interaction of the magnetic and electric forces determines the mobility of the light source.

Therefore, the interaction of the magnetic and electric forces determines how mobile the light source is.

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A police car is traveling east at 40.0 m/s along a straight road, overtaking a car ahead of it moving east at 30.0 m/s . The police car has a malfunctioning siren that is stuck at 1000 Hz. (c) What is it behind the police car?

Answers

λ=0.303m is behind the police car.

The frequency of the emission is [tex]f=1000\, \text{Hz}[/tex]

The police car is traveling at a speed of [tex]$v_s=+40.0 \frac{\mathrm{m}}{\mathrm{s}}$[/tex]

When the air is calm, the sound moves at a speed of [tex]$v=343 \frac{\mathrm{m}}{\mathrm{s}}$[/tex]

When travelling away from the source, the observer's (another car) speed is

[tex]$v_o=-30.0 \frac{\mathrm{m}}{\mathrm{s}}$[/tex]

(c) The sound's frequency is (for the observer at rest): in front of the police car.

[tex]f^{\prime}=\frac{v}{v-v_s} f[/tex]

consequently, the needed wavelength is:

[tex]\lambda=\frac{v}{f^{\prime}}=\frac{v}{\frac{v}{v-v_s} f}[/tex]

Rearranging and canceling v gives us:

[tex]\lambda=\frac{v-v_s}{f}=\frac{343 \frac{\mathrm{m}}{\mathrm{s}}-40 \frac{\mathrm{m}}{\mathrm{s}}}{1000 \mathrm{~Hz}}=0.303 \mathrm{~m}[/tex]

Hence, λ=0.303m  is behind the police car.

What do you mean by wave length?

A waveform signal's wavelength is defined as the separation between two identical points (adjacent crests) in adjacent cycles as the signal travels through space or along a wire.

This length is typically specified in meters (m), centimeters (cm), or millimeters (mm) in wireless systems (mm).

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Consider (a) an electron (b) a photon, and (c) a proton, all moving in vacuum. Choose all correct answers for each question. (i) Which of the three possess rest energy?

Answers

Electron and proton have rest energy which is option (a) and (c) .

An electron is a solid negatively charged factor of an atom. Electrons exist outdoor of and surrounding the atom nucleus. Each electron includes one unit of bad charge (1.602 x 10^-19 coulomb)A proton is a subatomic particle discovered with inside the nucleus of each atom.  The particle has a advantageous electric charge, identical and contrary to that of the electron A photon is a subatomic particle, having strength and momentum however no mass or electric powered charge, this is the quantum unit of electromagnetic radiation, together with light.

According to Einstein's quantum theory light propagates in the form of packets i.e. quanta of energy, which is called photon.

The rest mass of photons is being zero. So photos do not have rest energy they are never at rest .

As electrons and protons posses mass , therefore they have rest energy which is given by [tex]E=mc^2[/tex] .

Hence , only electron and proton have rest energy not photon.

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To win the game, a place kicker met kick a
football from a point 35 m (38.276 yd) from
the goal, and the ball must clear the crombar,
which is 3.05 m high. When kicked, the ball
leaves the ground with a speed of 20 m/s at
an angle of 45.9 from the horizontal.
The acceleration of gravity is 9.8 m/s²
By how much vertical distance does the ball
clear the crowbar?
Answer in units of m.

Answers

Answer:

2.08 m  (2 d.p.)

Explanation:

Constant Acceleration Equations (SUVAT)

[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}[/tex]

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

-------------------------------------------------------------------------------------

When a body is projected through the air with initial speed (u), at an angle of θ to the horizontal, it will move along a curved path.

Therefore, trigonometry can be used to resolve the body's initial velocity into its vertical and horizontal components:

Horizontal component of u (x) = u cos θVertical component of u (y) = u sin θ

Since the projectile is modeled as moving only under the influence of gravity, the only acceleration the projectile will experience will be acceleration due to gravity.

If the ball is kicked at an initial velocity of 20 m/s from flat ground at an angle of 45.9° then:

Horizontal component of u = 20 cos 45.9°Vertical component of u = 20 sin 45.9°

Resolving horizontally

The horizontal component of velocity is constant, as there is no acceleration horizontally.

Resolving horizontally, taking → as positive:

[tex]s=35 \quad u=20 \cos 45.9^{\circ} \quad v = 20 \cos 45.9^{\circ} \quad a=0[/tex]

[tex]\begin{aligned}\textsf{Using} \quad s&=\left(\dfrac{u+v}{2}\right)t\\\\35&=\left(\dfrac{20 \cos 45.9^{\circ}+20 \cos 45.9^{\circ}}{2}\right)t\\35&=(20 \cos 45.9^{\circ})t\\t&=\dfrac{35}{20 \cos 45.9^{\circ}}\\\implies t&=2.51468288...\; \sf s\end{aligned}[/tex]

Therefore, the time is takes for the ball to reach the crossbar is 2.51 s.

Resolving vertically

Acceleration due to gravity = 9.8 ms⁻²

Resolving vertically, taking ↑ as positive:

[tex]u=20 \sin 45.9^{\circ} \quad a=-9.8 \quad t=2.51468288...[/tex]

[tex]\begin{aligned}\textsf{Using} \quad s&=ut+\dfrac{1}{2}at^2\\\\s&=(20 \sin 45.9^{\circ})(2.51468288...)+\dfrac{1}{2}(-9.8)(2.51468288...)^2\\s&=36.1171982...-30.9857870...\\\implies s&=5.131411132...\; \sf m\end{aligned}[/tex]

Solution

To find by how much the ball clears the crossbar, subtract the height of the crossbar (3.05 m) from found the vertical height of the ball at 2.51 s:

[tex]\implies 5.13141132...-3.05=2.08\; \sf m\;(2 \: d.p.)[/tex]

Therefore, the ball clears the crossbar by 2.08 m (2 d.p.).

1ly/5.9 x 10^12 miles

Answers

Answer:

0.9964

Explanation:

I think that's the right answer but here you go always!

A circular loop of wire with a radius of 4.0cm is in a uniform magnetic field of magnitude 0.060T . The plane of the loop is perpendicular to the direction of the magnetic field. In a time interval of 0.50s , the magnetic field changes to the opposite direction with a magnitude of 0.040T . What is the magnitude of the average emf induced in the loop?(a) 0.20V(b) 0.025V(c) 5.0mV(d) 1.0mV(e) 0.20mV

Answers

E. The magnitude of the average emf induced in the loop is 0.20 mV.

Average emf induced in the circular loop

The magnitude of the average emf induced in the loop is calculated as follows;

emf = NA(B₁ - B₂)/t

where;

A is area of the loop N is number of turnsB1 is initial magnetic fieldB2 is final magnetic fieldt is time

A = πr² = π(0.04)² = 5.027 x 10⁻³ m²

emf = (1)(5.027 x 10⁻³)(0.06 - 0.04)/0.5

emf = 2.01 x 10⁻⁴ V

emf = 0.201 mV

Thus, the magnitude of the average emf induced in the loop is 0.20 mV.

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when approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately

Answers

When approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately 25 feet.

What impact, if any, would jet fuel and aviation gasoline have on a turbine engine?

Tetraethyl lead, which is present in gasoline, deposits itself on the turbine blades. Because jet fuel has a higher viscosity than aviation gasoline, it may retain impurities with greater ease.

Once the gasoline charge has been cleared, start the engine manually or with an electric starter while cutting the ignition and using the maximum throttle.

On the final approach, the aeroplane needs to be re-trimmed to account for the altered aerodynamic forces. A substantial nose-down tendency results from the airflow producing less lift on the wings and less downward force on the horizontal stabiliser due to the reduced power and slower velocity.

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if i apply newton's 1st law of motion to football, the player carrying or running with the ball will stay running in the same direction unless

Answers

The player would continue in the same direction unless a force acts on the player.

What is the Newton's first law?

The Newton's first law of motion states that an object will continue in its state of rest or uniform motion unless it is acted upon by an external force. This external force tries either make the object to move or to change its direction of motion accordingly.

Having said this, for a player that is running in the field with the ball, the player would continue in the same direction unless a force acts on the player such as when the player is pushed by another player.

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On the basis of the repulsive nature of the force between like charges and the freedom of motion of charge within a conductor, explain why excess charge on an isolated conductor must reside on its surface.

Answers

Excess charge on an isolated conductor must reside on its surface because excess charge moves away until they reach the surface of the body.

Charged conductors in electrostatic equilibrium exhibit a number of interesting properties. The electric field beneath the surface of a charged conductor is zero when it is in electrostatic equilibrium, which is one of its properties. The repulsive nature of force between similar charges and the liberty of motion of charges within a conductor allows for the establishment of equilibrium within charges present in the conductor. Due to this fact, excess charges carry like type of charge as contained within the conductor they are thus repelled away from the conductor till they reach the outer surface of the conductor.

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Consider this situation: A baseball travels upward and
rightward through the air. Of the forces listed, identify which
act upon the baseball.
Tap to select the forces that are present.

Answers

The forces acting on the baseball are air resistance, and force of gravity.

What are forces present during the upward motion of an object?

The forces which act on an object moving upward will either oppose the motion of the object or help to speed up the object.

The forces present during upward motion include;

force of air resistanceforce of gravityapplied force

The forces acting on a baseball thrown upwards which also travels rightward include the following;

force of air resistance which oppose the upward motion of the baseballforce of gravity acting downwards due to the weight of the ball.

Thus, the forces acting on the baseball are air resistance, and force of gravity.

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The complete question is below:

Consider this situation: A baseball travels upward and

rightward through the air. Of the forces listed, identify which

act upon the baseball. (force of air resistance, force of gravity, weight of the ball, frictional force).

A toy doll and a toy robot are standing on a frictionless surface facing each other. The doll has a mass of 0. 20 kg, and the robot has a mass of 0. 30 kg. The robot pushes on the doll with a force of 0. 30 n. The magnitude of the acceleration of the robot is.

Answers

The magnitude of the acceleration of the robot is 1m/s^2 .

Given ,

A toy doll and a toy robot are standing on a frictionless surface facing each other .

The mass of the doll = 0.20kg

The mass of the robot = 0.30kg

Force applied by the robot on the doll = 0.30N

We know the force and mass of the robot , so from this we can calculate the acceleration by applying Newton's laws of motion ,

Thus , F= ma

a = F/m = 0.30N / 0.30k

a = 1 m/s^2

Hence , the magnitude of the acceleration of the robot is 1m/s^2 .

What is acceleration ?

Acceleration is the rate of change in velocity.

Positive acceleration it occurs when an object speeds up. Negative acceleration it occurs when an object slows down.

Acceleration is a change of velocity over a period of time, expressed by the equation,

A=Δv/t

The unit of acceleration is m/s^2.

The dimensional formula of acceleration is [M^0 L^1 T^-2] .

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A tennis ball is dropped from 1.98 m above
the ground. It rebounds to a height of 1.05 m.
The acceleration of gravity is 9.8 m/s².
With what velocity does it leave the ground?

Answers

The ball will leave the ground with the velocity of 4.5 m/s.

What is Velocity ?

Velocity is the distance travelled in a specific direction per time taken. It is measured in m/s

Given that a tennis ball is dropped from 1.98 m above the ground. It rebounds to a height of 1.05 m. The acceleration of gravity is 9.8 m/s².

At maximum height, final velocity is zeroThe maximum height H = 1.05 mThe initial velocity u = 0

Using the formula

v² = u² - 2gH

0 = u² - 2 × 9.8 × 1.05

u² = 20.58

u = √20.58

u = 4.54 m/s

Therefore, the ball will leave the ground with the velocity of 4.5 m/s.

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The visible portion of the electromagnetic spectrum occurs for wavelengths between.

Answers

Answer:

380 to 700 nanometers

Brainliest please

Explanation:

Why is the following situation impossible? It is early on a Saturday morning, and much to your displeasure your next-door neighbor starts mowing his lawn. As you try to get back to sleep, your next-door neighbor on the other side of your house also begins to mow the lawn with an identical mower the same distance away. This situation annoys you greatly because the total sound now has twice the loudness it had when only one neighbor was mowing.

Answers

The situation is not possible here because "the total sound now has twice the loudness it had when only one neighbor was mowing" - this sentence can't be happened.

What are the levels of sound?

The scope of sounds estimated on the decibel scale is from 0 dB (the calmest sound) to 140 dB (the edge of agony). Sounds over 85 dB are viewed as by particular associations like NIOSH (the US Public Establishment for Word related Wellbeing and Wellbeing) to be risky to human hearing. Sound level alludes to different logarithmic estimations of discernible vibrations and may allude to. Sound openness level, proportion of the sound openness of a sound comparative with a reference esteem Sound power level, proportion of the rate at which sound energy is radiated, reflected, communicated or got, per unit time.

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apply concepts what two forces are acting on the blocks as they slide over the wet sand? which one is a noncontact force?

Answers

Gravity is the non contact force.

What is a Gravity?

Gravity is a fundamental interaction in physics that causes all objects with mass or energy to attract one another. The electromagnetic force, the weak interaction, and the strong interaction are all significantly stronger than gravity, which is by far the weakest of the four fundamental interactions. As a result, it has no appreciable impact on subatomic particle level phenomena. However, at the macroscopic level, gravity is the most important interaction between things and governs the motion of planets, stars, galaxies, and even light.

On Earth, gravitational pull lends weight to tangible objects, and the moon's gravitational pull generates ocean tides. In addition, gravity plays a significant role in many biological processes, including gravitropism, which directs plant growth, and the movement of fluids in multicellular organisms. Gravity may affect how the immune system and cells differentiate in the human body, according to research on the effects of weightlessness.

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S Review. Two identical particles, each having charge +q , are fixed in space and separated by a distance.d. A third particle with charge -Q is free to move and lies initially at rest. on the perpendicular bisector of the two fixed charges a distance x from the midpoint between those charges (Fig. P23.14). (b) Determine the period of that motion.

Answers

The period of motion is given as  ​π/2(√md³/KeqQ)

What is Coulombs law?

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

The period of motion from the instance given  ​π/2(√md³/KeqQ).

w²= (2π/T)²

T=2π/w

where w is gotten as 16KeqQ/md³

T=  ​π/2(√md³/KeqQ).

m is the mass of the body with charge-Q

In conclusion, the Coulomb's law equation provides an accurate description of the force between two objects whenever the objects act as point charges.

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Which formula is equivalent to D = m V ?

Answers

p = m/v p for density, m for mass, v for volume.

A coin is rolled in a straight line along a balcony edge at a steady speed of 0.46 m/s.
a) Calculate how far the coin rolls in 2.4 s.
b) Another coin is dropped from the balcony. It accelerates from rest and hits the ground after
8.0 seconds at a speed of 78.4 m/s. Calculate the acceleration of the coin during its fall

Answers

answer:

Use this equation (v=Δx/t); plug in velocity (0.500 m/s) and the time (2.4 s), then solve for displacement… (Δx=1.2 m)

If a coin is rolled in a straight line along a balcony edge at a steady speed of 0.46 m / s , then the distance traveled by this coin in 2.4 seconds would be 1.104 meters.

What are the three equations of motion?

There are three equations of motion given by Newton,

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

As given in the problem If a coin is rolled in a straight line along a balcony edge at a steady speed of 0.46 m / s ,

The distance traveled by the coin = 0.46 × 2.4

                                                         =1.104 meters per second

For another coin , by using the first equation of motion ,

v = u + a × t

78.8 = 0 + a × 8.0

a = 78.8 / 8.0

a = 9.85 m / s²

Thus,  the distance traveled by this coin in 2.4 seconds would be 1.104 meters .

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You wish to start a fire by reflecting sunlight from a mirror onto some paper under a pile of wood. Which would be the best choice for the typeof mirror? (a) flat (b) concave (c) convex

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You wish to start a fire by reflecting sunlight from a mirror onto some paper under a pile of wood which means the best choice of the type of mirror for this operation is a concave one and is denoted as option B.

What is Concave mirror?

This type of mirror usually has a reflective surface that is curved inward and away from the light source. They also reflect light inward to one focal point which is done by reflecting all the rays parallel to its axis and converge them at a single point.

For fire to occur, then we need a large concentration of light rays which will help reflect all the sunlight from a mirror onto some paper under a pile of wood thereby causing the fire.

This is therefore the reason why concave mirror was chosen as the most appropriate choice.

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A nonconducting wall carries charge with a uniform density of 8.60μC/cm²(b) Does your result change as the distance from the wall varies? Explain.

Answers

So long as the distance from the wall is small compared to the width and height of the wall, the distance does not affect the field.

Density, the mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre. For example, the density of water is 1 gram per cubic centimetre, and Earth’s density is 5.51 grams per cubic centimetre. Density can also be expressed as kilograms per cubic metre (in metre-kilogram-second or SI units).

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