How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners of an equilateral triangle)?

Answer:

2.568 × 10⁻³ kg

Explanation:

The amount of sodium to be taken by an adult is measured in terms of a limit, as given here.

That limit has been set as 2568 mg , when measured in milligrams.

The basic conversion from milligrams to g is done by dividing with 1000. Then 2568 milligrams will be 2.568 grams.

Now 100 grams are present in 1 kilogram.

So 2.568 grams are divided with 1000 to get the specified mass in kg.

The gives it as 2.568 × 10⁻³.

Answer:

Normal Force is usually perpendicular to the movement and static friction usually means that there is no movement.

Explanation:

The work donde by any force on an object is equal to the displacement of the object multiplied by the component of the force that is in the direction of the displacement.

Normal force is usually perpendicular to the movement, so there is no component in the direction of the displacement. This is why it is zero in most circumstances.

Static friction on the other hand, usually means that there is no movement at all (it's static). It means that there is no displacement between the object and ground (in most cases). If there is no displacement, there is no work.

The normal force and the force of static friction do no work on an object for different reasons. The normal force acts at right angles to the displacement of the object, while the force of static friction only acts when there is no relative motion between the object and the surface.

The normal force and the force of static friction do no work on an object in most circumstances, but for different reasons.

The normal force is a force exerted by a surface that is perpendicular to the surface. It does no work because it acts at right angles to the displacement of the object.

The force of static friction is a force that opposes the motion of an object on a surface. It does no work because it only acts when there is no relative motion between the object and the surface.

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Answer:

a) To the nearest tenth the average value of the measures is 217.4cm

b) The standard deviation for the four measurements is 0.415

Explanation:

The average is the result of adding all the values and dividing by the number of measures, in this case 4 then

[tex]Average = \frac{217.6 + 217.2 + 216.8 + 217.9}{4} \\[/tex]

[tex]Average = \frac{869.5}{4} \\[/tex]

[tex]Average = 217.375 [/tex]

a) To the nearest tenth the average value of the measures is 217.4cm

The standard deviation equals to the square root of the variance, and the variance is the adition of all the square of the difference between the measure and the average for each value divided by the number of measurements

So first, we must calculate the variance:

[tex]Variance = \frac{(217.6-217.4)^2+(217.2-217.4)^2+(216.8-217.4)^2+(217.9-217.4)^2}{4}[/tex].

[tex]Variance = \frac{(0.2)^2+(-0.2)^2+(-0,6)^2+(0.5)^2}{4}[/tex].

[tex]Variance = \frac{0.04+0.04+0.36+0.25}{4}[/tex].

[tex]Variance = \frac{0.69}{4}[/tex].

[tex]Variance = 0.1725[/tex].

This represent the difference between the average and the measurements.

Now calculate the standard deviation

Standard deviation  = [tex]\sqrt{Variance}[/tex]

Standard deviation  = [tex]\sqrt{0.1725}[/tex]

Standard deviation  = 0.415

b) The standard deviation for the four measurements is 0.415

This measure represent a standard way to know what is normal in this sample. so the differences between the average should be of ±0.415

The average value of the four door height measurements is 217.4 cm, and the standard deviation of these measurements is approximately 0.4 cm.

To find the average value of the four measurements: 217.6 cm, 217.2 cm, 216.8 cm, and 217.9 cm, you add them up and then divide by the number of measurements, which is four:

Average = (217.6 + 217.2 + 216.8 + 217.9) / 4 = 217.375

Rounded to the nearest tenths place, the average is 217.4 cm.

For the standard deviation, we first calculate the variance. First, find the difference of each measurement from the mean, square that difference, and then find the average of those squared differences. Finally, take the square root of that average to find the standard deviation:

Substituting the values and performing the calculations gives us a standard deviation of approximately 0.4 cm.

Answer:

The density of nuclear matter is [tex]10^{6}\ Mg/\mu L[/tex]

Explanation:

Given that,

Density [tex]\rho= 10^{18}\ kg/m^3[/tex]

Using unit conversation,

[tex]\Rightarrow \dfrac{10^{18}\ kg}{1\ m^3}\times\dfrac{1 Mg}{10^{6}g}\times\dfrac{1000\ g}{1\ kg}\times\dfrac{1\ m^3}{10^6\ cm^3}\times\dfrac{1\ cm^3}{1\ mL}\times\dfrac{1000\ mL}{1\ L}\times\dfrac{10^{-6}\ L}{1\ \muL}[/tex]

[tex]\Rightarrow \dfrac{10^{18}\times10^{3}\times10^{3}\times10^{-6}}{10^{6}\times10^{6}\times10^{6}}[/tex]

The density of nuclear matter is

[tex]\rho=10^{6}\ Mg/\mu L[/tex]

Hence, The density of nuclear matter is [tex]10^{6}\ Mg/\mu L[/tex]

Answer:

shearing force is [tex]3.40\times 10^{-4} lb[/tex]

Explanation:

we know that

force can be determined [tex]F = \tau \times A[/tex]

Area can be determine as

[tex]A  = \frac{\pi}{4} d^2 = \frac{\pi}{4} [\frac{0.2}{12}]^2 =   2.18\times 10^{-4} ft^2[/tex]

linear velocity can be determines as

[tex]\tau = \mu_{air} \frac{U}{b}[/tex]

dynamic viscosity of air [tex]\mu_{air} = 3.74\times 10^{-7} lb-s/ft^2[/tex]

veolcity of disc

[tex]U =\omega R[/tex]

[tex]U = \frac{2\pi N}{60} \times R = \frac{2\pi 10,000}{60} \times \frac{2}{12}[/tex]

U = 174.5 ft/s

so

[tex]\tau = 3.74\times 10^{-7} \times \frac{174.5}{\frac{0.0005}{12}}[/tex]

[tex]\tau = 1.56 lb/ft^2[/tex]

[tex]F = 1.56\times 2.18\times 10^{-4} = 3.40\times 10^{-4} lb[/tex]

shearing force is [tex]3.40\times 10^{-4} lb[/tex]

Answer:

[tex]v_{ft} = 21.11 \frac{m}{s}[/tex]   :  Speed of the truck immediately after the collision , to the east.

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction and direction as velocity) and its magnitude is calculated like this:

P=m*v

where

P:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀=Pf   Formula (1)

P₀ :Initial  linear momentum quantity

Pf : nitial  linear momentum quantity

Nomenclature and data

mc: car mass= 1.43*10³ kg  = 1430kg

V₀c: initial car speed,  = 25.0 m/s

Vfc: final car speed = 18.0 m/s

mt: truck mass =  9000 kg  

V₀t: initial truck speed, = 20.0 m/s

Vft: final truck speed

Problem development

For this problem the collision is inelastic because after the collision the objects are deformed .

Because the known speeds go east they are positive, we assume that the truck continues moving east after the collision and its speed will also be positive:

We apply formula (1)

P₀=Pf

mc*V₀c+mt*V₀t=mc*Vfc+mt*Vft

1430*25+9000*20=1430*18+9000*Vft

215750=25740+9000*Vft

[tex]v_{ft} =\frac{215750-25740}{9000} = 21.11 \frac{m}{s}[/tex]

[tex]v_{ft} = 21.11 \frac{m}{s}[/tex]

Because the response was positive the truck moves east after the collision

Rate of change of mass is given by,

[tex]\frac{dm}{dt}=18.816\,g/s[/tex]

In the question,

We have the Base Area of the vertical container = 14 cm x 16 cm

Now,

Let us take the height of the container = h

Rate of change of height with time, dh/dt = 0.21 cm/s = 2.1 mm/s

So,

Volume of the container = Base Area x Height

So,

V = 14 x 16 x h

V = 140 x 160 x h (because, 1 cm = 10 mm)

V = (22400)h

Now,

Volume of one of the candy = 50 mm³

Mass of the candy = 0.0200 g

So,

Density of the candy = Mass/ Volume

So,

[tex]Density=\frac{0.0200}{50}\\Density=0.0004[/tex]

Now,

V = (22400)h

On differentiating with respect to time, t, we get,

[tex]\frac{dV}{dt}=\frac{22400h}{dt}\\\frac{d}{dt}(\frac{mass}{density})=22400.\frac{dh}{dt}\\\frac{1}{density}.\frac{dm}{dt}=22400.\frac{dh}{dt}\\[/tex]

Therefore, on putting the value of density in the equation and also the value of rate of change of height with time, we get,

[tex]\frac{1}{density}.\frac{dm}{dt}=22400.\frac{dh}{dt}\\\frac{1}{0.0004}.\frac{dm}{dt}=22400(2.1)\\\frac{dm}{dt}=18.816\,g/s[/tex]

Therefore, the rate of change of mass with respect to time is given by,

[tex]\frac{dm}{dt}=18.816\,g/s[/tex]

The distance of the image is 60 cm and the focal length of this concave mirror is 30 cm.

The focal length of the lens is length of the distance between the middle of the lens to the focal point.

It can be find out using the following formula as,

[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

Here, (v)is the distance of the image, (u) is the distance of the object, and (f) is the focal length of the lens.

Here, the concave mirror produces a real image that is three times as large as the object. The object is 20 cm in front of the mirror, and the concave mirror produces a real image that is three times as large as the object.

Hence, the value of magnification of the mirror is 3.

The object distance is 20 cm thus the image distance, using the magnification formula, can be given as,

[tex]m=\dfrac{v}{u}\\3=\dfrac{v}{20}\\v=60\rm cm[/tex]

Put the values in the lens formula as,

[tex]\dfrac{1}{60}+\dfrac{1}{20}=\dfrac{1}{f}\\f=30 \rm \; cm[/tex]

Hence, the distance of the image is 60 cm and the focal length of this concave mirror is 30 cm.

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Final answer:

Using the mirror equation and magnification formula, the image distance is found to be -60 cm and the focal length of the concave mirror is calculated to be 30 cm.

Explanation:

The question involves determining the image distance and the focal length of a concave mirror that produces a real image three times larger than the object. The object is placed 20 cm in front of the mirror. To find these values, we can use the mirror equation which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.

Given that the magnification (m) is -3 (negative sign since the image is real and inverted), and magnification is also equal to -di/do, we can express di as -3do. We are told that do is 20 cm, so di is -3(20 cm) = -60 cm (negative because the image is on the same side as the object).

Now, we can plug these values into the mirror equation to find the focal length (f):

Therefore, the image distance is -60 cm and the focal length of the concave mirror is 30 cm.

Answer:

Part a)

[tex]y_m = 0.157 mm[/tex]

part b)

[tex]k = 101.8 rad/m[/tex]

Part c)

[tex]\omega = 579.3 rad/s[/tex]

Part d)

here since wave is moving in negative direction so the sign of [tex]\omega[/tex] must be positive

Explanation:

As we know that the speed of wave in string is given by

[tex]v = \sqrt{\frac{T}{m/L}}[/tex]

so we have

[tex]T = 17.5 N[/tex]

[tex]m/L = 5.4 g/cm = 0.54 kg/m[/tex]

now we have

[tex]v = \sqrt{\frac{17.5}{0.54}}[/tex]

[tex]v = 5.69 m/s[/tex]

now we have

Part a)

[tex]y_m [/tex] = amplitude of wave

[tex]y_m = 0.157 mm[/tex]

part b)

[tex]k = \frac{\omega}{v}[/tex]

here we know that

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(92.2) = 579.3 rad/s[/tex]

so we  have

[tex]k = \frac{579.3}{5.69}[/tex]

[tex]k = 101.8 rad/m[/tex]

Part c)

[tex]\omega = 579.3 rad/s[/tex]

Part d)

here since wave is moving in negative direction so the sign of [tex]\omega[/tex] must be positive

Answer:

5.05 m

Explanation:

Given:

According to the work-energy theorem, work done by the gravitational force will be equal to the kinetic energy change in the vaulter.

[tex]\therefore -mg(h)=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow -gh=\dfrac{1}{2}(v^2-u^2)\\\Rightarrow h=\dfrac{1}{-2g}(v^2-u^2)\\\Rightarrow h=\dfrac{1}{-2\times 9.8}((1)^2-(10)^2)\\\Rightarrow h=\dfrac{-99}{-19.6}\\\Rightarrow h=5.05[/tex]

Hence, the height of the vaulter is 5.05 m.

Answer:

Surely Achilles will catch the Tortoise, in 400 seconds

Explanation:

The problem itself reduces the interval of time many times, almost reaching zero. However, if we assume the interval constant, then it is clear that in two hours Achilles already has surpassed the Tortoise (20 miles while the Tortoise only 3).

To calculate the time, we use kinematic expression for constant speed:

[tex]x_{final}=x_{initial}+t_{tor}v_{tor}=1+t_{tor}\\x_{final}=x_{initial}+t_{ach}v_{ach}=10t_{ach}[/tex]

The moment that Achilles catch the tortoise is found by setting the same final position for both (and same time as well, since both start at the same time):

[tex]1+t=10t\\t=1/9 hour=0.11 hours[/tex]

To calculate average speed, divide the total distance by the total time taken. In this case, the average speed is 30.0 km/h.

To calculate average speed, we divide the total distance traveled by the total time taken. In this case, you drove 10.0 km in 20.0 minutes. To convert minutes to hours, we divide by 60. The average speed in kilometers per hour is calculated as:

Average speed = Total distance / Total time = 10.0 km / (20.0 min / 60 min/h) = 30.0 km/h

Answer:

the square of the average atomic velocity.

Explanation:

From the formulas for kinetic energy and temperature for a monoatomic gas, which has three translational degrees of freedom, the relationship between root mean square velocity and temperature is as follows:

[tex]v_{rms}=\sqrt{\frac{3RT}{M}}[/tex] (1)

Where  [tex]v_{rms}[/tex] is the root mean square velocity, M is the molar mass of the gas, R is the universal constant of the ideal gases and T is the temperature.

The root mean square velocity is a measure of the velocity of the particles in a gas. It is defined as the square root of the mean square velocity of the gas molecules:

[tex]v_{rms}=\sqrt{<v^2>}[/tex] (2)

substituting 2 in 1, we find the relationship between mean square speed and temperature:

[tex]\sqrt{<v^2>}=\sqrt{\frac{3RT}{M}}\\T=\frac{M<v^2>}{3R}\\\\T\sim  <v^2>[/tex]

Final answer:

The temperature of a monatomic ideal gas is directly proportional to the average atomic velocity.

Explanation:

The temperature of a monatomic ideal gas is directly proportional to the average atomic velocity. As the temperature increases, the average atomic velocity also increases. This relationship is a result of the fact that temperature is a measure of the kinetic energy of the gas particles, and the average velocity is related to the kinetic energy. Therefore, temperature and average atomic velocity are directly proportional in a monatomic ideal gas.

Answer:

Option (D)

Explanation:

The formula for the potential energy between the two charges is given by

[tex]U=\frac{KQq}{r}[/tex]

where, r is the distance between the two charges.

In first case the distance between the two charges is r1.

The potential energy is

[tex]U_{1}=\frac{KQq}{r_{1}}[/tex]

In first case the distance between the two charges is r2.

The potential energy is

[tex]U_{2}=\frac{KQq}{r_{2}}[/tex]

The change in potential energy is

[tex]\Delta U = U_{2}-U_{1}[/tex]

[tex]\Delta U=\frac{KQq}{r_{2}}-\frac{KQq}{r_{1}[/tex]

[tex]\Delta U=KQq \times \left ( \frac{1}{r_{2}} -\frac{1}{r_{2}} \right )[/tex]

The change in the potential energy of the charge +q during this process is [tex]\mathbf{\Delta U = KQq \Big (\dfrac{1}{r_2}- \dfrac{1}{r_1} \Big) }[/tex]

Option D is correct.

In an electrical circuit, the electric potential energy is the entire potential energy of a unit charge will have if it is positioned at any point in space.

The electric potential generated by a charge at any point in space is directly proportional to its magnitude and also varies inversely proportional to the distance out from the point charge.

Mathematically, it can be expressed as:

[tex]\mathbf{U = \dfrac{KQq}{r}}[/tex]

here;

So,

The potential energy for the first scenario can be expressed as:

[tex]\mathbf{U_1 = \dfrac{KQq}{r_1}}[/tex]

The potential energy for the second scenario can be expressed as:

[tex]\mathbf{U_2= \dfrac{KQq}{r_2}}[/tex]

Therefore, the change in the potential energy is:

[tex]\mathbf{\Delta U =U_2 -U1}[/tex]

[tex]\mathbf{\Delta U = \dfrac{KQq}{r_2}- \dfrac{KQq}{r_1}}[/tex]

[tex]\mathbf{\Delta U = KQq \Big (\dfrac{1}{r_2}- \dfrac{1}{r_1} \Big) }[/tex]

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Answer :

(a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is [tex]2.38\times10^{-38}\ m[/tex]

Explanation :

Given that,

Speed = 100 km/hr

Mass of car = 1 ton

(a). We need to calculate the wavelength of electron

Using formula of wavelength

[tex]\lambda_{e}=\dfrac{h}{p}[/tex]

[tex]\lambda_{e}=\dfrac{h}{mv}[/tex]

Put the value into the formula

[tex]\lambda_{e}=\dfrac{6.63\times10^{-34}}{9.1\times10^{-31}\times100\times\dfrac{5}{18}}[/tex]

[tex]\lambda=0.00002622[/tex]

[tex]\lambda=26.22\times10^{-6}\ m[/tex]

[tex]\lambda=26.22\ \mu m[/tex]

(II).  We need to calculate the wavelength of car

Using formula of wavelength again

[tex]\lambda_{e}=\dfrac{6.63\times10^{-34}}{1000\times100\times\dfrac{5}{18}}[/tex]

[tex]\lambda=2.38\times10^{-38}\ m[/tex]

The wavelength of the electron is greater than the dimension of electron and the wavelength of car is less than the dimension of car.

Therefore, electron is quantum particle and car is classical.

Hence, (a). The wavelength of electron is 26.22 μm.

(b).The wavelength of car is [tex]2.38\times10^{-38}\ m[/tex].

Answer:0.10283 J

Explanation:

Given

mass of block(m)=0.0475 kg

radius of track (r)=0.425 m

when the Block is at Bottom Normal has a magnitude of 3.95 N

Force acting on block at bottom

[tex]N-mg=\frac{mu^2}{r}[/tex]

[tex]N=mg+\frac{mu^2}{r}[/tex]

[tex]3.95=0.0475\timess 9.81+\frac{0.0475\times u^2}{0.425}[/tex]

[tex]u^2=31.172[/tex]

[tex]u=\sqrt{31.172}[/tex]

u=5.583 m/s

At top point

[tex]N+mg=\frac{mv^2}{r}[/tex]

[tex]0.670+0.0475\timess 9.81=\frac{mv^2}{r}[/tex]

[tex]v^2=10.1639[/tex]

v=3.188 m/s

Using Energy conservation

[tex]\frac{mu^2}{2}=\frac{mv^2}{2}+mg\left ( 2r\right )+W_f[/tex]

[tex]W_f=0.4989-0.39607[/tex]

[tex]W_f=0.10283 J[/tex]

i.e. 0.10283  J of energy is wasted while moving up.

The work done on the block by friction during the motion of the block from point A to point B is 0.102 J.

The velocity of the block at the top of the vertical circle is calculated as follows;

[tex]W + mg = \frac{mv^2}{r} \\\\0.67 + 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\1.136 = 0.112 v^2\\\\v^2 = 10.14\\\\v = 3.18 \ m/s[/tex]

The velocity of the block at the bottom of the vertical circle is calculated as follows;

[tex]W - mg = \frac{mv^2}{r} \\\\3.95 - 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\3.485 = 0.112 v^2\\\\v^2 = 31.12\\\\v = 5.58 \ m/s[/tex]

The work done by friction is the change in the kinetic energy of the block.

[tex]W _f =P.E_f - \Delta K.E \\\\W_f = mgh - \frac{1}{2} m(v_f^2 - v_i^2)\\\\Wf = 0.0475\times 9.8(2 \times 0.425) - \frac{1}{2} \times 0.0475(5.58^2 - 3.18^2)\\\\W_f = 0.396 - 0.498\\\\W_f = -0.102 \ J[/tex]

Thus, the work done on the block by friction during the motion of the block from point A to point B is 0.102 J.

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Answer:

potential energy at origin is [tex]2.57*10^{6} volt[/tex]

Explanation:

given data:

electric field E = 5*10^{6} N/C

at x = 43 cm, y = 28 cm

distance btween E and origin

[tex]\Delta r = \sqrt{43^2 +28^2}[/tex]

[tex]\Delta r = 51.313 cm[/tex]

potential energy per unit charge [tex]\Delta V = - Edr[/tex]

[tex]\Delta V = 5*10^6*51.313*10^{-2} J/C[/tex]

[tex]\Delta V  =  2.57*10^{6} volt[/tex]

potential energy at origin is 2.57*10^{6} volt

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × [tex]10^{7}[/tex] m/s

uniform electric field E = 1.18 × [tex]10^{4}[/tex] N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × [tex]10^{-31}[/tex] kg ×a = 1.18 × [tex]10^{4}[/tex] × 1.602 × [tex]10^{-19}[/tex]

a = 20.75 × [tex]10^{14}[/tex] m/s²

so acceleration is 20.75 × [tex]10^{14}[/tex] m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × [tex]10^{7}[/tex] = 0 + 20.75 × [tex]10^{14}[/tex] (t)

t = 11.80 × [tex]10^{-9}[/tex] s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × [tex]10^{-9}[/tex]

time is 23.6 × [tex]10^{-9}[/tex] s

time will elapse before it return to  its staring point is 23.6 ns

Answer:

28.8 cm

Explanation:

Magnification in a microscope is:

M = Mo * Me

Where

Mo: magnification of the objective,

Me: Magnification of the eyepiece.

The magnification of the objective if:

Me = npd/fe

Where:

npd: near point distance

fe: focal length of the eyepiece

The magnification of the objective:

Mo = d/fo

Where

d: the distance between lenses

fo: focal length of the objective

Then

M = npd/fe * d/fo

d = M * fe * fo / npd

d = 12 * 5 * 12 / 25 = 28.8 cm

Answer:

Frequency of the wave will be 31.84 Hz

Explanation:

We have given the equation [tex]y=0.02sin(30x-200t)[/tex]-----eqn 1

The standard equation of sine wave is given by [tex]y=Asin(kx-\omega t)[/tex]----eqn 2

On comparing eqn 1 and eqn 2

[tex]\omega =200[/tex]

Angular frequency [tex]\omega[/tex] is given by [tex]\omega =2\pi f[/tex]

So [tex]200=2\times 3.14\times f[/tex]

[tex]f=31.84Hz[/tex]

Answer:

The answer is 9 m.

Explanation:

Using the kinematic equation for an object in free fall:

[tex]y = y_o - v_o-\frac{1}{2}gt^{2}[/tex]

In this case:

[tex]v_o = \textrm{Initial velocity} = 1.5[m/s]\\t = \textrm{air time} =  1.2 [s][/tex][tex]y_o = 0[/tex]

[tex]g = \textrm{gravity} = 9.8 [m/s^{2} ][/tex]

Plugging those values into the previous equation:

[tex]y = 0 - 1.5*1.2-\frac{1}{2}*9.8*1.2^{2} \\y = -8.85 [m] \approx -9 [m][/tex]

The negative sign is because the reference taken. If I see everything from the rescuer point of view.

Answer:

a)

3.43*10^{14} Hz

8.75*10^{-7} m

b)

2.70*10^{14} Hz

1.10*10^{-6} m

Explanation:

GIVEN DATA:

a)

i)we know that

E = h\nu

where E is energy

h = plank's constant = 6.625* 10^{-34} j-s

[tex]\nu = frequency[/tex]

[tex]\nu = \frac{E}{h} = \frac{1.42*1.6*10^{-19} v}{6.625*10^{-34}}[/tex]

[tex]\nu = 3.43*10^{14} Hz[/tex]

ii)wavelength  is given as

[tex]\lambda = \frac{c}{\nu}[/tex]

            [tex]= \frac{3*10^8}{3.43*10^{14}} = 8.75*10^{-7} m[/tex]

b) i) i)we know that

[tex]E = h\nu[/tex]

where E is energy

h = plank's constant = 6.625* 10^{-34} j-s

[tex]\nu = frequency[/tex]

[tex]\nu = \frac{E}{h} = \frac{1.12*1.6*10^{-19} v}{6.625*10^{-34}}[/tex]

[tex]\nu =2.70*10^{14} Hz[/tex]

ii)wavelength  is given as

[tex]\lambda = \frac{c}{\nu}[/tex]

[tex]= \frac{3*10^8}{2.70*10^{14}} = 1.10*10^{-6} m[/tex]

Answer:

3.122×10¹⁴ g/cm³

Explanation:

Diameter of neutron star = 23 km = 2300000 cm

Radius of neutron star = 2300000/2 = 1150000 cm = r

Mass of neutron star = 1.989 × 10³⁰ kg = 1.989 × 10³³ g = m

Volume of neutron star

[tex]v=\frac{4}{3}\pi r^3\\\Rightarrow v=\frac{4}{3}\pi 1150000^3[/tex]

Density = Mass / Volume

[tex]\rho=\frac{m}{v}\\\Rightarrow \rho=\frac{1.989\times 10^{33}}{\frac{4}{3}\pi 1150000^3}\\\Rightarrow \rho=3.122\times 10^{14}\ g/cm^3[/tex]

∴ Density of neutron star is 3.122×10¹⁴ g/cm³

Final answer:

The typical density of a neutron star is about 10¹⁴g/cm³. This is calculated by dividing the mass of the star in grams by its volume in cubic centimeters, and taking into account that a neutron star's mass is typically 1.4 solar masses and its diameter is about 20 kilometers.

Explanation:

To calculate the density of a neutron star, we consider it as a sphere with a typical mass of 1.4 solar masses and a diameter of about 20 kilometers. The formula for density (d) is mass (m) divided by volume (V), and the volume of a sphere is given by the formula V = ⅔πr3, where r is the radius of the sphere.

First, we convert the solar mass to kilograms (1 solar mass = 1.99 × 1030 kg) and 1.4 solar masses to kg gives us 2.786 × 1030 kg. Next, we convert the diameter to radius in centimeters (10,000 cm), then calculate the volume. Now we can find the density:

Mass: 2.786 × 1030 kg
Volume: ⅔π(105 cm)³= 4.18879 × 1015 cm³
Density = Mass/Volume
Density = 2.786 × 1030 kg / 4.18879 × 1015 cm³

The density, in grams per cubic centimeter (g/cm³), is calculated by converting the mass from kilograms to grams. This gives us a typical neutron star density of about 1014 g/cm³, which is exceedingly high compared to materials we experience on Earth.

Answer:

for -12db

 [tex]\frac{V1}{V2}=0.251\\\\\frac{P1}{P2}=0.0631[/tex]

for 3db

[tex]\frac{V1}{V2}=\sqrt{2}\\\\\frac{P1}{P2}=2[/tex]

for 10db

[tex]\frac{V1}{V2}=3.16\\\\\frac{P1}{P2}=10[/tex]

for 0db

[tex]\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1[/tex]

Explanation:

The decibel is a logaritmic value given by:

[tex]db=10*log(\frac{P1}{P2})=20*log(\frac{V1}{V2})[/tex]

we use 10 for power values and 20 for other values such voltages or currents.

[tex]\frac{V1}{V2}=10^{\frac{db}{10}}\\\\\frac{P1}{P2}=10^{\frac{db}{20}}[/tex]

for -12db

[tex]\frac{V1}{V2}=10^{\frac{-12}{10}}=0.251\\\\\frac{P1}{P2}=10^{\frac{-12}{20}}=0.0631[/tex]

for 3db

[tex]\frac{V1}{V2}=10^{\frac{3}{10}}=\sqrt{2}\\\\\frac{P1}{P2}=10^{\frac{3}{20}}=2[/tex]

for 10db

[tex]\frac{V1}{V2}=10^{\frac{10}{10}}=3.16\\\\\frac{P1}{P2}=10^{\frac{10}{20}}=10[/tex]

for 0db

[tex]\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1[/tex]

Explanation:

Let [tex]q_1[/tex] is the first charge and [tex]q_2[/tex] is the second charge.

Force between them, F = 0.12 N

Distance between charges, d = 0.46 m

(a) Force acting between two point charges is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

[tex]q_1q_2=\dfrac{Fd^2}{k}[/tex]

[tex]q_1q_2=\dfrac{0.12\times (0.46)^2}{9\times 10^9}[/tex]

[tex]q_1q_2=2.82\times 10^{-12}[/tex]..............(1)

Also,

[tex]q_1+q_2=-8.9\ \mu C=-8.9\times 10^{-6}\ C[/tex]............(2) (both charges are negative)

On solving equation (1) and (2) :

[tex]q_1=-8.571\ C[/tex]

and

[tex]q_2=-0.329\ C[/tex]

(b) If the force is attractive, F = -0.12 N

[tex]q_1q_2=\dfrac{Fd^2}{k}[/tex]

[tex]q_1q_2=\dfrac{-0.12\times (0.46)^2}{9\times 10^9}[/tex]

[tex]q_1q_2=-2.82\times 10^{-12}[/tex]..............(3)

[tex]q_1+q_2=-8.9\ \mu C=-8.9\times 10^{-6}\ C[/tex]............(4)    

Solving equation (3) and (4) we get :

[tex]q_1=-0.306\ C[/tex]

[tex]q_2=9.206\ C[/tex]      

Hence, this is the required solution.                                  

Using Coulomb's law, set up a system of equations with q1 and q2 as unknowns and solve to find the values of the individual charges for the case of repulsive force. Repeat the process for attractive forces, keeping in mind that charges will have the opposite sign.

This problem can be solved using the formula for Coulomb's law, which states that the force between two charges is equal to the product of the charges divided by the distance squared, times the Coulomb constant: F = k*q1*q2/r², where F is the force, k is the Coulomb constant (8.99 * 10^9 N.m²/C²), q1 and q2 are the charges, and r is the distance between them.

For repulsive force, both charges have the same sign. But in this particular problem, we are given that the total charge is 8.90 μC, so let's take q1 and q2 as unknowns. Now q1 + q2 = 8.90 μC and using the above formula we get another equation. Now you have two equations to solve the unknown charges. Same procedure applies for the attractive force, but know that charges are of opposite sign for attractive force.

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The tension in the rope must equal the weight of the supported mass. For the given situation, the tension can be calculated by multiplying the mass by the acceleration due to gravity.

Tension in the rope must equal the weight of the supported mass, as we can prove using Newton's second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus Fnet = 0. The only external forces acting on the mass are its weight w and the tension T supplied by the rope. Thus, T - w = 0.

Thus, for a 5.00-kg mass (neglecting the mass of the rope), we can find that T = mg = (5.00 kg) (9.80 m/s²) = 49.0 N.

Thus, Tension in the rope equals the weight of the supported mass (5.00 kg) due to Newton's second law, making T = mg = 49.0 N.

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Answer:

Distance, d = 99990 meters

Explanation:

It is given that,

Speed of the train, v = 200 km/h = 55.55 m/s

Time taken, t = 1800 s

Let d is the distance covered by the train. We know that the speed of an object is given by total distance covered divided by total time taken. Mathematically, it is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]d=v\times t[/tex]

[tex]d=55.55\times 1800[/tex]

d = 99990 m

So, the distance covered by the train is 99990 meters. Hence, this is the required solution.

To find the charge per unit length on the copper rod, use the formula E = k * λ / r. Plug in the given values to find the charge per unit length. The electric flux through the cube can be calculated by multiplying the electric field by the area of one face of the cube.

(a) To find the charge per unit length on the copper rod, we can use the formula for an electric field (E = k * λ / r), where E is the electric field, k is the electrostatic constant, λ is the charge per unit length, and r is the distance from the center of the rod. Rearranging the formula, we can solve for λ: (λ = E * r / k). Plugging in the given values, we have: λ = (4 N/C) * (0.04 m) / (9 x 10^9 Nm^2/C^2) = 1.78 x 10^-10 C/m.

(b) The electric flux through a surface is given by the formula (Φ = E * A), where Φ is the electric flux, E is the electric field, and A is the area of the surface. In this case, the electric flux through the cube can be calculated by multiplying the electric field (4 N/C) by the area of one face of the cube (3 cm)^2 = (0.03 m)^2 = 9 x 10^-4 m^2. Therefore, the electric flux is Φ = (4 N/C) * (9 x 10^-4 m^2) = 3.6 x 10^-3 Nm^2/C.

(a) The charge per unit length on the copper rod is [tex]\( {8.90 \times 10^{-10} \, \text{C/m}} \).[/tex]

(b) The electric flux through the cube is [tex]\( {0.0036 \, \text{N} \cdot \text{m}^2/\text{C}} \).[/tex]

(a): Charge per Unit Length on the Copper Rod

The electric field ( E ) at a distance ( r ) from the axis of a long charged rod is given by:

[tex]\[E = \frac{2k\lambda}{r}\][/tex]

From the given data:

[tex]\[4 = \frac{2 \cdot 8.99 \times 10^9 \cdot \lambda}{0.04}\][/tex]

Solving for [tex]\( \lambda \):[/tex]

[tex]\[\lambda = \frac{4 \cdot 0.04}{2 \cdot 8.99 \times 10^9}\][/tex]

[tex]\[\lambda = \frac{0.16}{1.798 \times 10^9}\][/tex]

[tex]\[\lambda \approx 8.90 \times 10^{-10} \, \text{C/m}\][/tex]

(b): Electric Flux through a Cube

To find the electric flux [tex]\( \Phi_E \)[/tex] through the cube, we use Gauss's law, which states:

[tex]\[\Phi_E = \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{q_{\text{enclosed}}}{\epsilon_0}\][/tex]

Since the rod is long and the cube is oriented such that the rod passes through opposite sides perpendicularly, the electric flux [tex]\( \Phi_E \)[/tex] through the cube is:

[tex]\[\Phi_E = E \cdot A\][/tex]

Calculate ( A ):

[tex]\[A = a^2 = (0.03)^2 = 0.0009 \, \text{m}^2\][/tex]

Now, calculate [tex]\( \Phi_E \):[/tex]

[tex]\[\Phi_E = 4 \times 0.0009\][/tex]

[tex]\[\Phi_E = 0.0036 \, \text{N} \cdot \text{m}^2/\text{C}\][/tex]

Answer:

(a) [tex]2.31\times10^{-8}\ N[/tex]

(b) [tex]1.44\times 10^{-19}\ eV[/tex]

Explanation:

Given:

*p = charge on proton = [tex]1.602\times 10^{-19}\ C[/tex]

*e = magnitude of charge on an electron = [tex]1.602\times 10^{-19}\ C[/tex]

*r = distance between the proton and the electron in the Rutherford's atom = [tex]1.0\times 10^{-10}\ m[/tex]

Part (a):

Since two unlike charges attract each other.

According to Coulomb's law:

[tex]F=\dfrac{kqe}{r^2}\\\Rightarrow F = \dfrac{9.0\times 10^{9}\times 1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(1.0\times 10^{-10})^2}\\\Rightarrow F = 2.31\times 10^{-8}\ N[/tex]

Hence, the attractive electrostatic force of attraction acting between an electron and a proton of Rutherford's atom is  [tex]2.31\times 10^{-8}\ N[/tex].

Part (b):

Potential energy between two charges separated by a distance r is given by:

[tex]U= \dfrac{kqQ}{r}[/tex]

So, the potential energy between the electron and the proton of the Rutherford's atom is given by:

[tex]U = \dfrac{kqe}{r}\\\Rightarrow U = \dfrac{9\times 10^{9}\times1.602\times 10^{-19}\times e}{1.0\times 10^{-10}}\\\Rightarrow U = 1.44\times 10^{-19}\ eV[/tex]

Hence, the electrostatic potential energy of the atom is  [tex]1.44\times 10^{-19}\ eV[/tex].

The attractive electrostatic force at the atomic size in Rutherford's model is about 2.3 x 10^-9 N. The electrostatic potential energy at this distance is approximately 14.4 eV.

The question asks for the attractive electrostatic force between an electron and a proton at a distance of 1.0 x 10-10 m (Rutherford's atomic size) and the electrostatic potential energy in eV. We use Coulomb's Law to find the force and potential energy equations in physics.

(a) The attractive electrostatic force (F) is given by the equation: F = (k * e^2) / r^2, where k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), e is the charge of an electron/proton (1.6 x 10^-19 C), and r is the distance (1.0 x 10^-10 m). Plugging in these values, we get F = (8.99 x 10^9 N m^2/C^2 * (1.6 x 10^-19 C)^2) / (1.0 x 10^-10 m)^2 ≈ 2.3 x 10^-9 N.

(b) The electrostatic potential energy (U) is given by the equation: U = (k * e^2) / r, which results in U = (8.99 x 10^9 J m/C^2 * (1.6 x 10^-19 C)^2) / (1.0 x 10^-10 m) ≈ 2.3 x 10^-18 J. Because 1 J = 6.242 x 10^18 eV, we convert this to eV to get approximately U ≈ 14.4 eV.

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Answer:

[tex]Density=6.51*10^{14}g/cm^{3[/tex]

Explanation:

Sun mass:

Ms=1.989 × 10^30 kg

Neutron star has the same mass.

Radius Neutron:

R=9Km (because diameter is 18Km)

R=9*10^3m

Density neutron star:

[tex]D=Mn/Vol=Ms/(4/3*\pi*R^3)=1.989*10^{30} /(4/3*\pi*(9*10^{3})^{3})=6.51*10^{17}kg/m^{3}[/tex]

[tex]D=6.51*10^{17}kg/m^{3}*(1000g/1Kg)*(1m^{3}/1000000cm^{3})=6.51*10^{14}g/cm^{3[/tex]

The density of a neutron star is about 10¹⁴ grams / cm³.

Any member of the class of extremely dense, compact stars known as neutron stars is assumed to be predominantly made of neutrons. The average diameter of a neutron star is 18 km (12 miles). Their masses range from 1.18 to 1.97 times those of the Sun, with the majority being 1.35 times the Sun. They have exceptionally high mean densities, almost 10¹⁴ times that of water.

This is similar to the density of an atomic nucleus, and one may think of a neutron star as a massive nucleus. Where the pressure is greatest, at the centre of the star, it is not known for sure what is there; hypotheses include hyperons, kaons, and pions. Most of the particles in the intermediate layers are neutrons, which are likely in a "superfluid" condition.

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