the charge density in an insulateed solid sphere of radius find the electric field at a distance of from the center of the solid

Answer:

The AC generator has one unbroken slip ring

Explanation:

In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.

An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.

We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.

It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.

Explanation:

Given that,

Initial speed of the rock, u = 30 m/s

The acceleration due to gravity at the surface of the moon is 1.62 m/s².

We need to find the time when the rock is ascending at a height of 180 m.

The rock is projected from the surface of the moon. The equation of motion in this case is given by :

[tex]h=ut-\dfrac{1}{2}gt^2\\\\180=30t-\dfrac{1}{2}\times 1.62t^2[/tex]

It is a quadratic equation, after solving whose solution is given by:

t = 7.53 s

or

t = 8 seconds

(e)If it is decending, v = -20 m/s

Now t' is the time of descending. So,

[tex]v=-u+gt\\\\t=\dfrac{v+u}{g}\\\\t=\dfrac{20+30}{1.62}\\\\t=30.86\ s[/tex]

Let h' is the height of the rock at this time. So,

[tex]h'=ut-\dfrac{1}{2}gt^2\\\\h'=30\times 30.86-\dfrac{1}{2}\times 1.62\times 30.86^2\\\\h'=154.40\ m[/tex]

or

h' = 155 m

Answer:

The average power dissipated is 72 W.

Explanation:

Given;

peak voltage of the AC circuit, V₀ = 120 V

resistance of the resistor, R = 100 -ohm

The average power dissipated by the resistor is given by;

[tex]P_{avg} = \frac{1}{2} I_oV_o= I_{rms}V_{rms} = \frac{V_{rms}^2}{R}[/tex]

where;

[tex]V_{rms}[/tex] is the root-mean-square-voltage

[tex]V_{rms} = \frac{V_o}{\sqrt{2}} \\\\V_{rms} = \frac{120}{\sqrt{2}}\\\\V_{rms} = 84.853 \ V[/tex]

The average power dissipated by the resistor is calculated as;

[tex]P_{avg} = \frac{V_{rms}^2}{R}\\\\P_{avg} = \frac{84.853^2}{100}\\\\P_{avg} = 72 \ W[/tex]

Therefore, the average power dissipated is 72 W.

Answer:

t = T / 2 all energy is stored in the inductor

Explanation:

The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is

          w = √1/LC

          2π / T =√1 / LC

          T = 2π  √LC

The energy is constant and for the initial instant it is completely stored in the capacitor

         Uc = Q₀² / 2C

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor

        U = L I² / 2

in the intermediate instant the energy is stored in the two elements.

Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor

After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation  (t = T/2).

The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,

[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]

here, T is the period of oscillation.

Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,

[tex]U =\dfrac{Q^{2}}{2C}[/tex]

here, C is the capacitance.

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,

[tex]U'=\dfrac{L I^{2}}{2}[/tex]

here, L is the inductance and I is the current.

Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.

Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation  (t = T/2).

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Answer:

The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].

Explanation:

First, each vector is determined in terms of absolute coordinates:

6-meter vector with direction: 30º north of east.

[tex]\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)[/tex]

[tex]\vec A = 5.196\,i + 3\,j[/tex]

4-meter vector with direction: 30º east of north.

[tex]\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)[/tex]

[tex]\vec B = 2\,i + 3.464\,j[/tex]

The resultant vector is obtaining by sum of components:

[tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex]

The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].

Answer:

Explanation:

Equation of refraction of a lens is expression according to the formula given below;

[tex]\dfrac{n_2}{v} = \dfrac{n_1}{u}= \dfrac{n_2-n_1}{R}[/tex]

R is the radius of curvature of the convex refracting surface = 12cm

v is the image distance from the refracting surface

u  is the object distance from the refracting surface

n₁ and n₂ are the refractive indices of air and the medium respectively

Given parameters

R = 12 cm

u = [tex]\infty[/tex] (since light incident is parallel to the axis)

n₁  = 1

n₂  = 1.5

Required

focus point of the light that is incident and parallel to the central axis (v)

Substituting this values into the given formula we will have;

[tex]\dfrac{1.5}{v} - \dfrac{1}{\infty}= \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0= \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}= \dfrac{0.5}{12}\\\\[/tex]

Cross multiply

[tex]1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm[/tex]

Hence  Light incident parallel to the central axis is focused at a point 36cm from the surface

Answer:

1. Berenice  = 0.67 D

2. Avishka  = 1.50 D

3. Francesca  = 2.00 D

4. Edouard = 2.75 D

Explanation:

The farsighted people are those with near point greater than 25 cm.

They include Avishka, Berenice, Edouard and Francisca.

A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative

From the lens formula, 1/f = 1/v  1/u; but v is negative

Therefore, 1/f = 1/u - 1/v

But, power of a lens = 1/f in meters.

Therefore,  P =  1/u - 1/v

For Avishka, u = 25 cm or 0.25 m, v = 0.4 m

P = 1/ 0.25 - 1/0.4 = 1.5 D

For Berenice, u = 0.25 m, v = 0.3 m

P = 1/0.25 - 1/0.30 =0.7 D

For Edouard, u = 0.25 m, v = 0.80 m

P = 1/0.25 - 1/0.80 =2.75 D

For Francesca, u = 0.25 m, v = 0.50 m

P = 1/0.25 - 1/0.5 = 2.0 D

When The farsighted people are those with a near point greater than 25cm.

Berenice is = 0.67 D

Avishka is = 1.50 D

Francesca is = 2.00 D

Edouard is = 2.75 D

When The farsighted people are those with a near point greater than 25 cm. Then, They include Avishka, Berenice, Edouard, and also Francisca.

Also, A converging lens is needed to correct farsightedness, or hyperopia, thus, When the focal length, f, is positive. Also, The image formed is virtual and also on the same side of the lens. hence the image distance is negative.

Also, From the lens formula, 1/f = 1/v 1/u; but v is negative

Thus, 1/f = 1/u - 1/v

But, when the power of a lens = 1/f in meters.

Thus, P = 1/u - 1/v

For Avishka, u = 25 cm or 0.25 m, v = 0.4 m

After that, P = 1/ 0.25 - 1/0.4 = 1.5 D

Then, For Berenice, u = 0.25 m, v = 0.3 m

Now, P = 1/0.25 - 1/0.30 =0.7 D

For Edouard, u = 0.25 m, v is = 0.80 m

Then, P = 1/0.25 - 1/0.80 is =2.75 D

Now, For Francesca, u = 0.25 m, v is = 0.50 m

Therefore, P = 1/0.25 - 1/0.5 = 2.0 D

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Answer:

93.3x10^-3s

Explanation:

If

Resistance = 1.8 kΩ

Current = 50 mA

Capacitor = 120 μF

Voltage = 140 V

to calculate the discharge current

Applying the formula of discharge current

io=vo/R

io= 140/ 1.8x 10³

= 0.078A

to calculate the time

Applying the formula of current

io= vo/R e-t/RC

50= 140/1800e-t/RC

0.649= e-t/RC

-t/RC= ln( 0.649)

t = 0.432x 120x10^-6x 1800

t=93.3 x 10^-3seconds

Answer:

The magnetic field is 0.0857 T.

Explanation:

The electrons orbit the magnetic field with a centripetal force equal to

F = [tex]\frac{mv^{2} }{r}[/tex]

also, the force on an electron in a magnetic field is gotten as

F = Bqv

equating this two equations give

[tex]\frac{mv^{2} }{r}[/tex] = Bqv

mv/r = Bq

where m is the mass of the electron = 9.11 x 10^-31 kg

v is the the linear speed of the electron

B is the magnetic field on the electron

r is the radius of the orbital movement

q is the charge on an electron = 1.602 x 10^-19 C

but, the linear speed v = ωr

where ω is the angular speed of the electron

substituting into equation above, we have

mωr/r = Bq

which reduces to

mω = Bq

finally, w know that the angular speed is related to the frequency of the electron by

ω = 2πf

we then finally have

2mπf = Bq

where f is the frequency emitted by the electron = 2.4 GHz = 2.4 x 10^9 Hz

substituting values into the equation, we have

2 x 9.11 x 10^-31 x 3.142 x 2.4 x 10^9 = B x 1.602 x 10^-19

B = (1.3734 x 10^-20)/(1.602 x 10^-19) = 0.0857 T

= 85.7 mT

Answer:

The speed of the proton is 4059.39 m/s

Explanation:

The centripetal force on the particle is given by;

[tex]F = \frac{mv^2}{r}[/tex]

The magnetic force on the particle is given by;

[tex]F = qvB[/tex]

The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.

[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]

where;

r is the radius of the circular path moved by both electron and proton;

⇒For electron;

[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]

⇒For proton

The speed of the proton is given by;

[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]

Therefore, the speed of the proton is 4059.39 m/s

Answer:

The magnitude of the magnetic flux through the loop is 0.0982 T.m²

Explanation:

Given;

magnitude of magnetic field, B = 0.5 T

radius of the loop, r = 0.25 m

Area of the loop is given by;

A = πr²

A = 3.142 x (0.25)²

A = 0.1964 m²

The magnitude of the magnetic flux through the loop is given by;

Ф = BA

Where;

B is the magnitude of the magnetic field

A is area of the field

Ф = 0.5 x 0.1964

Ф = 0.0982 T.m²

Therefore, the magnitude of the magnetic flux through the loop is 0.0982 T.m²

Answer:

95.7v

Explanation

Using Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop

So it is given as

E= Ndစ/dt

E= N BA-0/ deta t

Given that

N = 58turns

B = 1.10T

A = 0.150m^²

Deta t= 0.1s

now we have

E = 58(1.10x0.150)/0.1

= 95.7v

Magnetic flux is decreasing, so the direction of the current will be to aid the decreasing flux $decrease= CLOCKWISE

Explanation:

The researchers need to compare those who contracted the disease to those who did not.

Answer:

B. mineral oil – density of 0.910 g/mL.

Explanation:

Hello,

In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.

In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.

Best regards.

Image is missing, so i have attached it

Answer:

19.04 × 10⁻⁴ T in the +x direction

Explanation:

We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.

Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)

We are given;

I = 2.4 A

R = 5 cm = 0.05 m

μ₀ is a constant = 4π × 10⁻⁷ H/m

B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)

B = 9.6 × 10⁻⁴ T

To get the direction of the field from each wire, we will use Flemings right hand rule.

From the diagram attached:

We can say the field at P from the top wire will point up/right

Also, the field at P from the bottom wire will point down/right

Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.

If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.

Let the angle be θ

Thus;

sin(θ) = (1.3/2)/5

θ = sin⁻¹(0.13) = 7.47⁰

The x component of each field would be:

9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T

Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction

The magnitude of the magnetic field at the point P will be "9.6 × 10⁻⁴ T".

The region of the environment close to something like a magnetic entity or a current-carrying body wherein this same magnetic forces caused by the body as well as a current might well be sensed.

According to the question,

Current, I = 2.4 A

Radius, R = 5 cm or,

                = 0.05 m

Constant, μ₀ = 4π × 10⁻⁷ H/m

We know the relation,

The magnetic field, B = [tex]\frac{\mu_0 I}{2 \pi R}[/tex]

By substituting the values in the above relation, we get

                                    = [tex]\frac{4 \pi\times 10^{-7}\times 2.4}{2 \pi\times 0.05}[/tex]

                                    = 9.6 × 10⁻⁴ T

Thus the above answer is appropriate.

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Answer:

The film thickness is 4.32 * 10^-6 m

Explanation:

Here in this question, we are interested in calculating the thickness of the film.

Mathematically;

The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;

ΔN = (2L/λ) (n-1)

where λ is the wavelength of the light used

Let’s make L the subject of the formula

(λ * ΔN)/2(n-1) = L

From the question ΔN = 8 , λ = 540 nm, n = 1.5

Plugging these values, we have

L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m

The correct answer is C. The inner planets are also called terrestrial planets.

Explanation:

Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".

Answer:

The answer is c.) The inner planets are also called terrestrial planets.

Explanation:

The question is incomplete. Here is the complete question.

The isotope of Plutonium 238Pu is used to make thermoeletric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 3.5 kg of 238Pu.

(a) If the half-life of 238Pu is 87.7 yr, write a function of the form [tex]Q(t)=Q_{0}e^{-kt}[/tex] to model the quantity Q(t) of 238Pu  left after t years. Round ythe value of k to 3 decimal places. Do not round intermediate calculations.

(b) If 1.7kg of 238Pu is required to power the spacecraft's data transmitter, for low long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.

Answer: (a) [tex]Q(t)=3.5e^{-0.0079t}[/tex]

              (b) 91 years.

Explanation:

(a) Half-life is time it takes a substance to decrease to half of itself, i.e.:

Q(t) = [tex]0.5Q_{0}[/tex]

[tex]0.5Q_{0}=Q_{0}e^{-87.7k}[/tex]

[tex]0.5=e^{-87.7k}[/tex]

[tex]ln(0.5)=ln(e^{-87.7k})[/tex]

[tex]ln(0.5)=-87.7k[/tex]

[tex]k = \frac{ln(0.5)}{-87.7}[/tex]

k = 0.0079

Knowing k and [tex]Q_{0}[/tex]=3.5kg, function is [tex]Q(t)=3.5e^{-0.0079t}[/tex]

(b) Using function:

[tex]Q(t)=3.5e^{-0.0079t}[/tex]

[tex]1.7=3.5e^{-0.0079t}[/tex]

[tex]e^{-0.0079t}=\frac{1.7}{3.5}[/tex]

[tex]e^{-0.0079t}=0.4857[/tex]

[tex]ln(e^{-0.0079t})=ln(0.4857)[/tex]

[tex]-0.0079t=-0.7221[/tex]

[tex]t = \frac{-0.7221}{-0.0079}[/tex]

t = 91.41

t ≈ 91 years

Scientists will be able to receive data for approximately 91 years.

Answer:

2)The Cell with OER 7

Explanation:

OER is the acronym for Oxygen Enhancement Ratio. It is the measure of the  enhancement of the effect of ionizing radiation due to the presence of oxygen. The ionization effect can be detrimental or therapeutic (use in cancer treatment). OER is the ratio of radiation dose during the lack of oxygen (hypoxia), to the dosage in the presence of oxygen (air is used as a reference). From the definition, one can see that the higher the OER the higher the sensitivity of the cell.

The color that is reflected when a red card is illuminated by red light is white.

The color an object is perceived to have, depends on the frequency of light it reflects.

If white light incidents on a red filter, red is transmitted while blue and green are absorbed.

Consequently, when a red card is illuminated by red light, the red card will  reflect back almost all the incident light on it, causing it to appear brighter which creates an  illusion of white color to the eyes.

Thus, we can conclude the color that is reflected when a red card is illuminated by red light is white.

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Answer:

In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.

Answer:

0.6 m/s

Explanation:

2Hz = 2^-1 = 2 /s

30cm = .3m

Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.

(.3m)*(2 /s) = 0.6 m/s

Answer:

Explanation:

B. There is a counterclockwise induced current in the loop

Explanation:

This in line with the right hand grip rule,

The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of F.

Answer:

The torque is 11.27 N m

Explanation:

Recall that torque is the vector product of the force times the distance to the pivoting point. So in our case, the distance to the pivoting point is the radius of the cylinder (0.05 m), and the force is given by the weight of the bucket full of water (W = 9.8 * 23 N = 225.4 N)

Then the torque is: 0.05 * 225.4 N m = 11.27 N m

Answer:

Explanation:

The phase angle of an RLC circuit  ϕ is expressed as shoen below;

ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]

Xc is the capacitive reactance = 1/2πfC

Xl is the inductive reactance = 2πfL

R is the resistance = 25.0Ω

Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz

Xl = 2π * 70*0.0940

Xl = 41.32Ω

For the capacitive reactance;

Xc = 1/2π * 70*35.5*10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]

ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]

[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]

Since tan is negative in the 2nd quadrant;

[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]

Hence the phase angle ϕ of the circuit in degrees is 137.69°

The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°

Given that:

capacitance C = 35.5 μF,

Inductance L = 0.0940 H,

The resistance R = 25.0Ω

and frequency f = 70.0Hz

The capacitive reactance is given by:

Xc = 1/2πfC

Xc = 1/2π × 70 × 35.5× 10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

The inductive reactance is given by:

Xl = 2πfL

Xl = 2π × 70 × 0.0940

Xl = 41.32Ω

The phase angle of an RLC circuit ϕ  is given by:

[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]

Ф = -42.31°

Since tan is negative in the 2nd quadrant, thus:

ϕ = 180° - 42.31°

ϕ = 137.69°

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Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

Answer:middle

Explanation:

Because it will make the seasaw balanced

Answer:

Cardiovascular exercise is the activity use that aerobic metabolism and cellular reaction.

Explanation:

Answer:

1.34 mm

Explanation:

A double slit experiment is conducted with a light which has a wavelength of 620 nm

The fringes are separated 2.3 mm apart

The light is changed to a wavelength length of 360 nm

Let x represent the fringe spacing as a result of the change in wavelength

Therefore,the fringe spacing can be calculated as follows

2.3mm/x= 620nm/360nm

Multiply both sides

x × 620= 2.3×360

620x= 828

x= 828/620

x= 1.34 mm

The correct answer is D. An average city

Explanation:

A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.