Two positive point charges, each with charge q, separated by a distance d, repel each other with a force of magnitude 20 N. What is the magnitude of the force between two positive point charges of magnitude 11.11 q, separated by a distance 2.5 d in units of N

Answers

Answer 1

Answer:

The value is  [tex]F_2 = 395 \ N[/tex]

Explanation:

From the question we are told that

   The magnitude of the charge of each positive charge for the first case is [tex]q_1 = q_2 = q[/tex]

   The distance between the charges for the first case is  [tex]d[/tex]

   The  force between the charges for the first case  is  [tex]F = 20 \ N[/tex]

     The magnitude of the charge of each positive charge for the second case is [tex]q_1 = q_2=11.11q[/tex]

     The  distance between the charge for the second case is [tex]2.5d[/tex]

 

Generally for the first case the force between the charge is mathematically  represented as

       [tex]F_1 = \frac{k * q^2 }{d^2}[/tex]

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

  So  

     [tex]20= \frac{k * q^2 }{d^2}[/tex]

Generally for the second  case the force between the charge is mathematically  represented as

       [tex]F_2 = \frac{k * (11.11q)^2}{(2.5d)^2}[/tex]

       [tex]F_2 = \frac{k * 11.11^2 *q^2}{2.5^2d^2}[/tex]

=>      [tex]F_2 = \frac{11.11^2}{2.5^2} F_1[/tex]

=>      [tex]F_2 = \frac{11.11^2}{2.5^2} * 20[/tex]

=>      [tex]F_2 = 395 \ N[/tex]


Related Questions

If "38 %" of the light passes through this combination of filters, what is the angle between the transmission axes of the filters

Answers

Answer:

52°

Explanation:

The initial intensity [tex]I_{0}[/tex] = [tex]I[/tex]

The final intensity [tex]I_{f}[/tex] = 38% of [tex]I[/tex] = 0.38

From the polarizing equation

[tex]I_{f} = I_{0} cos^{2}[/tex]θ

substituting values, we have

[tex]0.38I = I cos^{2}[/tex]θ

0.38 = [tex]cos^2[/tex]θ

cosθ = [tex]\sqrt{0.38}[/tex]

cosθ = 0.6164

θ = [tex]cos^{-1}[/tex] 0.6164

θ = 51.9 ≅ 52°

All Houston Methodist buildings system wide have an emergency power generator that turns on to supply emergency power after normal power shuts down within:__________A. 5 secondsB. 30 secondsC. 60 secondsD. 10 seconds

Answers

Answer:

D. 10 seconds

Explanation:

All Houston Methodist Hospital buildings employs the use of EPSS(Emergency Power Supply System). This system employs the use of an emergency power generator that turns on to supply emergency power after normal power shuts down within 10 seconds during power outage.

Due to it being used in a hospital the importance of this emergency power system cannot be overemphasized as it is used to provide power supply to hospital equipment such as life support machines etc.

The highest that George can suck water up a very long straw is 2.0 m . (This is a typical value.) Part APart complete What is the lowest pressure that he can maintain in his mouth

Answers

Answer:

81,725 N/m^2

Explanation:

Given the following :

Height of water = 2m

Density of water = 1000kg/m^3

Atmospheric pressure (Pat) = 101325 N/m^2

Lowest pressure (Pl) can be obtained thus ;

Pat - Pl = density × height × acceleration due to gravity

101325 - Pl = 1000 × 2 × 9.8

101325 - Pl = 19600

101325 - 19600 = Pl

Lowest pressure (Pl) = 81,725 N/m^2

81,725 N/m^2 = (81,725/101325) atm

= 0.8065630 = 0.8065 atm

Ondrea could drive a Jetson's flying car at a constant speed of 540.0 km/hr across oceans and space, approximately how long would she take to drive from the Sun to Pluto in years? (Assume Pluto is at its average distance of 5.9 × 109 km from the Sun)

Answers

Answer:

The time taken in years is   [tex]x = 125 \  years[/tex]

Explanation:

From the question we are told that

   The  speed is  [tex]v  = 540.00 \ km /hr  =  \frac{540 *1000}{3600} =  150\  m/s[/tex]

    The distance from the sun to Pluto is  [tex]d =  5.9*10^{9} \  k m =  5.9*10^9 * 1000 =  5.9*10^{12} \  m[/tex]

Generally the time  taken is mathematically represented as

     [tex]t =  \frac{d}{v}[/tex]

=>   [tex]t = \frac{5.9*10^{11}}{150}[/tex]

=>   [tex]t =  3.933*10^{9}[/tex]

Converting to years

   [tex]1 year  \to  3.154*10^7 \  s[/tex]

    [tex]x \  years  \to 3.933*10^{9}[/tex]

=>  [tex]x = \frac{ 3.933*10^{9} * 1 }{ 3.154 *10^7}[/tex]

=>    [tex]x = 125 \  years[/tex]

An oil tanker took 7.2 hours to travel from city A to city B at an
average speed of 60 mph. If a passenger boat took 18 fewer
hours than the oil tanker to complete the same journey, what
was the average speed of the passenger boat?

Answers

Answer: 80

Explanation: 7.2 hours x 60mph = 432 m

7.2 hrs - 1.8 hrs= 5.4 hrs

432m/5.4hrs = 80

Answer: 80 mph

Explanation:

Just is

A certain automobile is 6.0 m long if at rest. If it is measured to be 4.8 m long while moving, its speed is:

Answers

Answer:

1.8*10^8m/s

Explanation:

Using

L= lo√1-v²/c²

So making v subject we have

V= c√1-4.8²/6²

V= 0.6*c

V= 0.6*3E8m/s

V= 1.8*10^8m/s

If an object is rolling without slipping, how does its linear speed compare to its rotational speed?

Answers

Answer:

v = rw

Explanation:

When an object is rolling continuously without slipping, then every angle it rotates through, is equal to a distance the perimeter has rotated.

If the object completes 10 revolutions and takes a particular time, let's say t to complete it. The angular distance would then be 20 π rad, while its angular velocity will be 20 π/t

The circumference will somehow translate to the distance it covers, which is 20πr, this means that the speed is 20πr/t

So, like the question asked, the linear speed compared to angular speed is

v : w

20πr/t : 20πt, which can be simplified to

r : 1

In essence, v = rw

While flying due east at 33 m/s, an airplane is also being carried due north at 12 m/s by the wind. What is the plane’s resultant velocity?

Answers

Answer:

35.11 m/s

Explanation:

While flying due east at 33 meters/second, an airplane is also being carried due north at 12 meters/second by the wind, then the plane’s resultant velocity would be 35.11 meters/second.

What is Velocity?

The total displacement covered by any object per unit of time is known as velocity.

the mathematical expression for velocity is given by

velocity = total displacement /total time

As given in the problem , while flying due east at 33 m/s, an airplane is also being carried due north at 12 m/s by the wind and we have to find the resultant velocity of the plane,

Resultant velocity = √( 33² + 12²)

                              = 35.11 meters/second

Thus, the plane’s resultant velocity would be 35.11 meters/second

Learn more about Velocity here, refer to the link;

brainly.com/question/18084516

#SPJ2

Refer to a long, straight wire carrying constant current I. What can be concluded about the magnitude of the magnetic field at distance a from the wire?

Answers

Answer:

"the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"

Explanation:

The magnitude of the magnetic field from a long straight wire (A approximately a finite length of wire at least for close points around the wire.) decreases with distance from the wire. It does not follow the inverse square rule as is the electric field from a point charge. We can then say that "the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"

From the Biot-Savart rule,

B = μI/2πR

where B is the magnitude of the magnetic field

I is the current through the wire

μ is the permeability of free space or vacuum

R is the distance between the point and the wire, in this case is = a

100 yards from his house. What kind of antenna should he use on the house and barn to get the best signal

Answers

Complete question:

a farmer wants to use a wireless camera he has installed in his barn which is about 100 yards from his house. what kind of antenna should he use on the house and barn to get the best signal.

Answer: Unidirectional antenna

Explanation: Based on the description given in the scenario above, the farmer should use a unidirectional antenna in other to get the best signal taking advantage of the fact that he knows where he wants the Transmission to come from. The good transmission signal attributed to unidirectional antennas stems from the fact that, the area covered is relatively small as signal transmission and radiofrequecy energy is focused in a particular direction, meaning that less area is covered using a unidirectional antenna,hiwevwr, it leverages this limitation to provide very good signal within the limited area covered.

the acceleration of a car is 10m/s².if the starts from rest , what will be its speed after 10sec.​

Answers

Answer:

v = 100 m/s

Explanation:

Given:

v₀ = 0 m/s

a = 10 m/s²

t = 10 s

Find: v

v = at + v₀

v = (10 m/s²) (10 s) + 0 m/s

v = 100 m/s

A flux density of 1.2Wb/m^2 is required in the 1 mm air gap of an electromagnet having an iron path of length 1.5 m. Calculate the mmf required. Given, relative permeability of iron is 1600. Neglect leakage. ​

Answers

Answer:

The mmf required is [tex]1.125[/tex]×[tex]10^{-3}[/tex] A

Explanation:

The Magnetomotive force (mmf) is given by the formula below

[tex]F_{M} = Hl\\[/tex]

where [tex]F_{M}[/tex] is the Magnetomotive force (mmf)

[tex]H[/tex] is the Magnetic field strength

[tex]l[/tex] is the magnetic length

The magnetic permeability μ is given by

μ = [tex]B / H[/tex]

Where [tex]B[/tex] is the Magnetic flux density

and [tex]H[/tex] is the Magnetic field strength

From the question,

[tex]B[/tex] = 1.2Wb/m^2

μ = 1600m

From μ = [tex]B / H[/tex]

∴[tex]H = B/[/tex]μ

[tex]H = 1.2 / 1600\\[/tex]

[tex]H = 7.5[/tex] × [tex]10^{-4}[/tex]A/m

Now, for the Magnetomotive force (mmf)

[tex]F_{M} = Hl\\[/tex]

From the question

[tex]l[/tex] = 1.5 m

∴ [tex]F_{M} = 7.5[/tex]×[tex]10^{-4}[/tex] × [tex]1.5[/tex]

[tex]F_{M} = 1.125[/tex]×[tex]10^{-3} A[/tex]

Hence, The mmf required is [tex]1.125[/tex]×[tex]10^{-3}[/tex] A

The electron beam inside an old television picture tube is 0.40 mm in diameter and carries a current of 50 μA. This electron beam impinges on the inside of the picture tube screen.
Part A. What is the current density in the electron beam?
Express your answer with the appropriate units.
Part B. The electrons move with a velocity of 3.8 ×107 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 mm?
Express your answer with the appropriate units.
Part C. Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the powerdelivered to the screen by the electron beam?
Express your answer with the appropriate units.

Answers

Answer:

A) J= 398 A/m²

B) E= 1.6×10⁶ N/C

C) P= ×10⁴ W

Explanation:

My work is in the attachment. Comment with questions or if something seems wrong with my work. (Honestly, they seem little high but it could just be the given numbers being unrealistic.) Below I have explanations of each part to match up with the image as well.

Part A:

Current density (J) is defined as the amount of current in a particular cross-sectional area. To get this, we simply need to divide the current (I) by the cross-sectional area of the electron beam tube (A).

Part B:

This one took the most work for me. I used a kinematic equation (yes they apply to electrons) to find the electric field (E). I used a modified form of the familiar: ∆d=V₀τ+aτ²/2

We can use the fact that τ= V/a, a=(qE/m), and V₀=0 here to rewrite the equation in terms of values we know and/or can look up. From there we solve for E and plug in the values.

Part C:

Power (P) is simply work (W) over time (τ). We know what τ is from before and can take W= mV²/2. Plugging these in and reducing some values gives us an equation for power as well.

If the first-order maximum for monochromatic light falling on a double slit is at an angle of 10.0∘, at what angle is the second-order maximum?

Answers

Answer:

The  value is  [tex]\theta_2 = 20.322^o[/tex]

Explanation:

From the question we are told that

  The angle of the first order maximum is  [tex]\theta _1 = 10.0^o[/tex]

Generally the condition for constructive interference is  

     [tex]dsin\theta = n \lambda[/tex]

Here  d is the separation between the slit ,

n  is the order of maxima  with values n  =  1, 2 , 3 ... for first , second , third ... order of maxima

    Now for first order of maximum

        [tex]dsin\theta_1 = \lambda \ \ ... \ \ ( 1)[/tex]

=>      [tex]dsin(10) = \lambda \ \ ... \ \ ( 1)[/tex]

    Now for second order of maximum

       [tex]dsin\theta = 2\lambda \ \ ... \ \ ( 2)[/tex]

dividing equation 1  by 2

      [tex]\frac{d sin (10)}{d sin (\theta_2 )} = \frac{\lambda}{2\lambda}[/tex]

     [tex]\frac{ sin (10)}{ sin (\theta_2 )} = \frac{1}{2}[/tex]

=>   [tex]2sin(10) = sin (\theta_2 )[/tex]

=>    [tex]0.3473 = sin(\theta_2)[/tex]

=>   [tex]\theta_2 = sin^{-1} [0.3473][/tex]

=>   [tex]\theta_2 = 20.322^o[/tex]

In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the slower runner when the faster runner finishes the race?

Answers

Answer:

The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

Explanation:

Given;

the speed of the slower runner, u₁ = 11.8 km/hr

the speed of the fastest runner, u₂ = 15 km/hr

distance, d = 8 km

The time when the fastest runner finishes the race is given by;

[tex]Time = \frac{Distance }{speed}\\\\Time = \frac{8}{15} \\\\Time = 0.533 \ hr[/tex]

The distance covered by the slower runner at this time is given by;

d₁ = u₁ x 0.533 hr

d₁ = 11.8 km/hr x 0.533 hr

d₁ = 6.29 km

Additional distance (x) the slower runner need to finish is given by;

6.29 km + x = 8km

x = 8 k m - 6.29 km

x = 1.71 km

Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, the distance to the cat appears to be

Answers

Answer:

The distance of the cat will appear farther than it really is

Explanation:

The index of refraction of water is more than that of air. Light rays from the cat  are refracted towards the normal at the surface and diverge outwards. This extends the virtual position of the cat, putting in a place farther than it really is.

what is the difference between temperature and heat

Answers

Explanation:

Heat is a form of energy that can transfer from hot body to cold body.Temperature is the degree of hotness and coldness of a body.

Heat is measured in joules, but temp. is measured in Kelvin

Answer:

temperature is the degree of hotness and coldness of the body or an object while heat is the energy transferred from hot to a cooler object as a result if differencein temperature ☺

A laser technician measures the wavelength of light from a new semiconductor laser. The wavelength is . What is the wavelength in nanometers

Answers

Answer:

The wavelength in nanometers is 849.81 nanometers (nm)

Explanation:

Here is the complete question:

A laser technician measures the wavelength of light from a new semiconductor laser. The wavelength is 8.4981 × 10⁻⁷m . What is the wavelength in nanometers. Write your answer as a decimal.

Explanation:

The given wavelength is 8.4981 × 10⁻⁷m.

To determine the wavelength in nanometers, we will convert the given wavelength from meters (m) to nanometers (nm).

1 nanometer = 1 × 10⁻⁹ meters

∴ 10⁻⁹ meters = 1 nanometer

To convert 8.4981 × 10⁻⁷m to nanometer (nm):

If 10⁻⁹ meters (m) = 1 nanometer (nm)

Then,  8.4981 × 10⁻⁷m = (8.4981 × 10⁻⁷m × 1 nm) / 10⁻⁹ m

8.4981 × 10⁻⁷m = 8.4981 × 10⁻⁷ × 10⁹ nm

= 8.4981 × 10² nm

= 849.81 nm

Hence, the wavelength in nanometers is 849.81 nm

he Himalayas now reach an elevation of 8.8 km, and radiometric dating suggests that their uplift began about 45 million years ago. Assuming a constant rate of uplift, how fast did the Himalayas rise?

Answers

Answer:

v = 6.20 10⁻¹² m / s,   v =  1.96 10⁻²  cm/year

Explanation:

This is a kinematics problem, specifically of uniform motion

        v = d / t

let's reduce the magnitudes to the SI system

      d = 8.8 km (1000 m / 1 km) = 8.8 10³ m

      t = 45 10⁶ years (365 day / 1 year) (24 h / 1 day) (3600 s / 1h)

       t = 1,419 10¹⁵ s

let's calculate the speed

        v = 8.8 10³ / 1.1419 10¹⁵

         v = 6.20 10⁻¹² m / s

This is the correct unit of the SI system, but it is more common to give the value in centimeters per year, so let's reduce the value

       v = 8.8 105 / 45 106

       v =  1.96 10⁻²  cm/year

     

what would be the behavior of an object that was immersed in a fluid if the object and the fluid had the same density

Answers

Answer:

The object will remain where it was placed on the fluid

Explanation:

If an object is immersed in a fluid, with both the object and the fluid having the same density, the force of buoyancy (the force acting upward) will be equal to the gravitational force (the weight pulling the object down) making the object remain in the position it was placed in the fluid. Meaning it will neither sink down or float through the surface.

For practical purpose, it should however be noted that the chances of this happening is very low.

NOTE: When an object is less dense than the liquid/fluid, it will float and when it is more dense than the liquid, it will sink

Why is it beneficial to perform focal point of a convex lens exercises in a dimly-lit room?
What is the focal length of a convex lens that produces an image on a screen 40 cm away with an object placed 10 cm from the lens? What is the magnification? Is the image inverted or upright? Show all calculations in your answer.

Answers

Explanation:

It is given that,

The object distance from the lens is 10 cm, u = -10 cm

Image is formed at a distance of 40 cm away from the lens, u = 40 cm

The lens formula is : [tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex], v is image distance

[tex]\dfrac{1}{40}-\dfrac{1}{-10}=\dfrac{1}{f}\\\\f=8\ cm[/tex]

The focal length is 8 cm

Magnification,

[tex]m=\dfrac{v}{u}\\\\m=\dfrac{40}{-10}\\\\m=-4[/tex]

The magnification is negative, it means that the formed image is inverted.

g a circular area with a radius of cm lies in the plane. What is the magnitude of 0.250 T in the

Answers

Complete Question

A circular area with a radius of 6.60 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field B = 0.250 T oriented in the following ways?

(a) in the +z-direction

Wb

(b) at an angle of 54

Answer:

a

[tex]\phi = 0.00342 \ Wb[/tex]

b

[tex]\phi = 0.00201 \ Wb[/tex]

Explanation:

From the question we are told that

   The  radius is  [tex]r = 6.60 \ cm = 0.066 \ m[/tex]

    The magnitude of the magnetic field is [tex]B = 0.250 \ T[/tex]

Generally the cross -sectional area is mathematically represented as

     [tex]A = \pi r^2[/tex]

     [tex]A = 3.142 * (0.066)^2[/tex]

     [tex]A = 0.01369 \ m^2[/tex]

Generally  when the magnetic field is oriented in the +z-direction the magnetic flux is mathematically represented as

      [tex]\phi = B* A cos(0)[/tex]

=>    [tex]\phi = 0.01369 * 0.250 * cos (0)[/tex]

=>     [tex]\phi = 0.00342 \ Wb[/tex]

  Now when the magnetic field is oriented  at an angle of 54°  the magnetic flux is mathematically represented as

      [tex]\phi = B* A cos(54)[/tex]

      [tex]\phi = 0.01369 * 0.250 * cos (54)[/tex]

     [tex]\phi = 0.00201 \ Wb[/tex]

As you did above for the gas, now estimate the number of molecules found in a cubic centimeter of a liquid at ordinary pressure and temperature.

Answers

Answer:

There are [tex]3.372 \times 10^{22}[/tex] molecules in a cubic centimeter of water at ordinary pressure and temperature.

Explanation:

Let suppose that liquid is water at a pressure of a atmosphere and a temperature of 25 ºC. Due to incompresibility of liquids, water density does not have any change of importance due to changes in pressure and temperature.

Density and molar mass of water are 1 gram per cubic centimeter and 18.015 grams per mole. The mass of water in a cubic centimeter ([tex]m[/tex]), measured in grams, is:

[tex]m = \rho\cdot V[/tex]

Where:

[tex]\rho[/tex] - Density, measured in grams per cubic meter.

[tex]V[/tex] - Volume of the sample, measured in cubic meters.

Given that [tex]\rho = 1\,\frac{g}{cm^{3}}[/tex] and [tex]V = 1\,cm^{3}[/tex], the mass of water is:

[tex]m = \left(1\,\frac{g}{cm^{3}} \right)\cdot (1\,cm^{2})[/tex]

[tex]m = 1\,g[/tex]

The amount of moles ([tex]n[/tex]) inside the sample is:

[tex]n = \frac{m}{M}[/tex]

Where:

[tex]m[/tex] - Mass of the sample, measured in grams.

[tex]M[/tex] - Molar mass, measured in grams per mole.

If [tex]m = 1\,g[/tex] and [tex]M = 18.015\,\frac{g}{mole}[/tex], then:

[tex]n = \frac{1\,g}{18.015\,\frac{g}{mole} }[/tex]

[tex]n = 0.056\,mole[/tex]

According to the Avogadro's Principle, there are [tex]6.022\times 10^{23}[/tex] molecules per mole. Hence, the number of molecules in a cubic centimeter of water at ordinary pressure and temperature is determined by simple rule of three:

[tex]x = (0.056\,mole)\times\left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)[/tex]

[tex]x = 3.372\times 10^{22}\,molecules[/tex]

There are [tex]3.372 \times 10^{22}[/tex] molecules in a cubic centimeter of water at ordinary pressure and temperature.

If you ride quickly down a hill on a bicycle your eardrums are pushed in before they pop back. Why is this?

Answers

Answer:

This is due to variation bin air pressure at the two different altitudes, so air rushes out through the eustachian tube to hit the eardrum and back when u reach the ground

Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a molecule

Answers

Complete Question

Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a molecule like [tex]N_2[/tex]?

Answer:

The  average volume per molecule is  

   [tex]\frac{V}{N} = 4.09 *10^{-26} \ m^3/molecule[/tex]

 The average distance between molecules

  [tex]d = 3.45 *10^{-9} \ m[/tex]

The size of  [tex]N_2[/tex] is 100 times smaller than the obtained value  

Explanation:

From the question we can deduce that we are considering an ideal

Generally the ideal gas equation is mathematically represented as

       [tex]PV = NkT[/tex]

Here  T  is the room temperature with value  T  =  300  \ K  

     k is the Boltzmann constant with value  [tex]k = 1.38 *10^{-23} \ J/K[/tex]  

      P  is the atmospheric pressure with value  [tex]P = 1.0 *10^{5} \ N/m^2[/tex]

     N is the number of molecules

Now  the  volume per molecule is mathematically deduced from the above equation as

        [tex]\frac{V}{N} = \frac{kT}{P}[/tex]

=>      [tex]\frac{V}{N} = \frac{ 1.381 *10^{-23} * 300}{ 1.0*10^{5}}[/tex]

=>      [tex]\frac{V}{N} = 4.09 *10^{-26} \ m^3/molecule[/tex]

Now the distance is mathematically evaluated as

      [tex]d = \sqrt{\frac{V}{N} }[/tex]

=>    [tex]d = \sqrt[3]{4.09*10^{-26}}[/tex]

=>     [tex]d = 3.45 *10^{-9} \ m[/tex]

Generally the size of  [tex]N_2[/tex]  is  115 pm which is  100 times smaller  than the  obtained value  

  Generally the size of  [tex]H_2O[/tex] is  [tex]95.84 \ pm[/tex] which is  10 times smaller than the obtained value  

Based on the ideal gas equation, the calculated values are as follows:

the average volume per molecule is 4.09 × 10^-23 m^3/moleculethe average distance between molecules is 3.45 nmthe N2 molecule is about 30 times smaller than the average distance between molecules.

How can the average volume per molecule be calculated?

The average volume per molecule is calculated using the given formula derived from the ideal gas equation:

V/n = kT/P

where:

V = gas volumen = number of molesk is Boltzmann constant = 1.38 × 10^-23 J/KT is temperature, and P is pressure

At room temperature and atmospheric pressure;

T = 300 k

P = 1.0 × 10^5 N/m^2

Substituting the values in the equation above to calculate the average volume per molecule, V/n:

V/n = 1.38 × 10^-23 × 300/1.0 × 10^5 N/m^2

V/n = 4.09 × 10^-23 m^3/molecule

Thus, the average volume per molecule is 4.09 × 10^-23 m^3/molecule

How can the average distance between molecules be determined?

The average distance between molecules can be determined by using the formula:

[tex]d = \sqrt[3]{ \frac{v}{n} } [/tex]

Substituting the values for V/n

[tex]d = \sqrt[3]{4.09 \times {10}^{ - 23} } [/tex]

d = 3.45 × 10^-9 m or 3.45 nm

Therefore, the average distance between molecules is 3.45 nm

A molecule of N2 has an average size of 115 pm or 0.115 nm.

Comparing the two values:

3.45/0.115 = 30

Therefore, the N2 molecule is about 30 times smaller than the average distance between molecules.

Learn more about distance between molecules ans Ideal gas equation at: https://brainly.com/question/14375674

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Objects fall near the surface of the earth with a constant downward acceleration of 10 m/s2 . Suppose a falling object is moving downward at 10 m/s at a certain instant. How fast is it falling 2 sec later

Answers

Answer:

The final velocity of the object after 2 seconds is 30 m/s

Explanation:

Given;

constant downward acceleration, a =  10 m/s²

initial velocity of the object falling down, v = 10 m/s

time of fall, t = 2 s

The final velocity of the object is given by;

v = u + at

where;

v is the final velocity

v = 10 + (10)(2)

v = 10 + 20

v = 30 m/s

Therefore, the final velocity of the object after 2 seconds is 30 m/s

A worker is pushing a crate of tools up a ramp into a truck. The crate has a mass of 120 kg and is accelerating at a rate of 1.05 m/s^2. Once he finishes pushing that crate up, he then takes another crate that is 3 times as massive and pushes it up the ramp with an acceleration of 0.71 m/s^2. What is the ratio of the force the worker used on the on the 120 kg crate and the more massive crate? a. 0.218 b. 0.493 c. 1.68 d. 1.07

Answers

Answer:

b. 0.493

Explanation:

the density of mercury is 13.6 g/ml. what is its density in lbs/L

Answers

Answer

density of mercury is 13.6 g/ml and that 1 lb = 0.45 kg.

A particle is confined to a one-dimensional box that is 50 pm long. What is the smallest possible uncertainty in momentum for the particle?

Answers

Answer:

The smallest possibility is 0.01E-22kgm/s

Explanation:

Using

Momentum= h/4πx

= 6.6x 10^-34Js/ 4(3.142* 50*10-12m)

= 0.01*10^-22kgm/s

1. Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish really is

Answers

Answer: I believe farther. Correct me if I am wrong

Explanation:

Answer:

Closer.

Explanation:

Due to the refraction of light passing through water and air.

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