As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:_______.
a. approaches zero.
b. approaches infinity.
c. approaches unity.
d. none of the above.

Answer:

Well, yes.

We can have an isolated light wave that is defined by only one frequency (and one wavelenght). But this is not a really common situation, most of the light that we can see in nature, is actually a composition of different waves with different frequencies.

Even if we have, for example, a red laser, the actual frequency of the light that comes from the laser may be in a range of frequencies, so the actual wave is a composition of different waves with really close frequencies.

An example of a light wave defined by only one frequency can be, for example, the photon that comes out of a change in energy of an electron.

Here we have a single photon, with a single frequency, that is modeled as a single frequency wave.

Answer:

Objects that are closer together have a stronger force of gravity between them.

Explanation:

For example, the moon is closer to Earth than it is to the more massive sun, so the force of gravity is greater between the moon and Earth than between the moon and the sun.

Answer:

Explanation:

This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.

If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).

But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).

A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".

Answer:

Star A has a higher surface temperature than star B.

Explanation:

The effective temperature of a star can be determined by means of its spectrum and Wien's displacement law:

[tex]T = \frac{2.898x10^{-3} m. K}{\lambda max}[/tex] (1)

Where T is the effective temperature of the star and [tex]\lambda_{max}[/tex] is the maximum peak of emission.  

A body that is hot enough emits light as a consequence of its temperature. For example, if an iron bar is put in contact with fire, it will start to change colors as the temperature increase, until it gets to a blue color, that scenario is known as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase and higher wavelengths as the temperature decreases.

Therefore, star A has a higher surface temperature than star B, as it is shown in equation 1 since T and [tex]\lambda max[/tex] are inversely proportional.

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

Answer:

The volume of the block is 306 cm³

The density of the block is 7.07 g/cm³

Explanation:

Given;

weight of block in air, [tex]W_a[/tex] = 21.2 N

Weight of block in water, [tex]W_w[/tex] = 18.2 N

Mass of the block in air;

[tex]W_a = mg[/tex]

21.2 = m x 9.8

m = 21.2 / 9.8

m = 2.163 kg

mass of the block in water;

[tex]W_w = mg[/tex]

18.2 = m x 9.8

m = 18.2 / 9.8

m = 1.857 kg

Apply Archimedes principle

Mass of object in air  - mass of object in water = density of water   x  volume                  of object

2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block

0.306 kg = 1000 kg/m³ x Volume of block

Volume of the block = [tex]\frac{0.306 \ kg}{1000 \ kg/m^3}[/tex]

Volume of the block = 3.06 x 10⁻⁴ m³

Volume of the block = 306 cm³

Determine the density of the block

[tex]Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3[/tex]

Answer:

198 ms-1

Explanation:

According to the Heisenberg uncertainty principle; it is not possible to simultaneously measure the momentum and position of a particle with precision.

The uncertainty associated with each measurement is given by;

∆x∆p≥h/4π

Where;

∆x = uncertainty in the measurement of position

∆p = uncertainty in the measurement of momentum

h= Plank's constant

But ∆p= mΔv

And;

m= 1.770×10^-27 kg

∆x = 0.15 nm

Making ∆v the subject of the formula;

∆v≥h/m∆x4π

∆v≥ 6.6 ×10^-34/1.770×10^-27 × 1.5×10^-10 ×4×3.142

∆v≥198 ms-1

Answer:

A. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

B. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

C. You can't say anything about the forces required until we know about the time frames required for each one to stop. So If they all stopped in the same time interval, then the rankings are the same.

Answer:

The correct choice is B & E.  

Explanation:

A solenoid is a coil of wire (usually copper) which is used as an electromagnet. Solenoids are used to convert electrical energy to mechanical energy. When this type of device is created it is also called a solenoid. One can increase the energy density within the solenoid or the coil by upping the electric current in the coil.

Cheers!

Answer:

The wavelength is 2.16 m.

Explanation:

Given the speed of the sound = 343 m/s

Trombone generate the frequency = 159 Hz

Now we have to find the wavelength of the sound. Here, we can find the wavelength by dividing the speed of the sound with frequency.

The wavelength of the sound = Speed of sound/frequency

Wavelength of the sound = 343 / 159 = 2.16 m

Answer:

[tex]\frac{di}{dt} = 7.31 \ A/s[/tex]

Explanation:

From the question we are told that  

     The  number of turns is  [tex]N = 450 \ turns[/tex]

      The  radius is  [tex]r = 1.17 \ cm = 0.0117 \ m[/tex]

       The  position from the center consider is  x =  3.45 cm  =  0.0345 m

       The  induced emf is  [tex]e = 8.20 *10^{-6} \ V/m[/tex]

Generally according to Gauss law

        [tex]\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A[/tex]

=>    [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A[/tex]

Where A is the  cross-sectional area of the solenoid which is mathematically represented as

                [tex]A = \pi r ^2[/tex]

=>      [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2[/tex]

=>       [tex]\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}[/tex]ggl;

Here  [tex]\mu_o[/tex] is the permeability of free space with value

          [tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]

=>     [tex]\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}[/tex]

=>      [tex]\frac{di}{dt} = 7.31 \ A/s[/tex]

The value of di/dt from the given values of the solenoid electric field is;

di/dt = 7.415 A/s

We are given;

Number of turns; N = 450 per m

Radius; r = 1.17 cm = 0.0117 m

Electric Field; E = 8.2 × 10⁻⁶ V/m

Position of electric field; r' = 3.45 cm = 0.0345 m

According to Gauss's law of electric field;

∫| E*dl | = |-d∅/dt |

Now, ∅ = BA = μ₀niA

where;

n is number of turns

i is current

A is Area

μ₀ = 4π × 10⁻⁷ H/m

Thus;

E(2πr') = (d/dt)(μ₀niA)  (negative sign is gone from the right hand side because we are dealing with magnitude)

Since we are looking for di/dt, then we have;

E(2πr') = (di/dt)(μ₀nA)

Making di/dt the subject of the formula gives;

di/dt = E(2πr')/(μ₀nA)

Plugging in the relevant values gives us;

di/dt = (8.2 × 10⁻⁶ × 2 × π × 0.0345)/(4π × 10⁻⁷ × 450 × π × 0.0117²)

di/dt = 7.415 A/s

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Answer:

145060km

Explanation: Given that

speed = dx/dt = 2t^2 +1

integrate

x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)

given that at t=0, x = 1000

so 1000 = 2/3 X (0)^3 + 0 + c

or c = 1000

So x = 2/3t^3 + t + 1000

for t = 1 min = 60s

x = 2/3 X 60^3 + 60 + 1000

x = 2/3×216000+ 1060

x = 144000+1060

= 145060km

At one minute, it will be 145060km far from the earth

Answer: 156.02 metre.

Explanation:

Give that a certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied.

Let us use third equation of motion,

V^2 = U^2 + 2as

Since the car is decelerating, V = 0

And acceleration a will be negative.

U = 33 mph

S = 39 m

Substitute both into the formula

0 = 33^2 - 2 × a × 39

0 = 1089 - 78a

78a = 1089

a = 1089 / 78

a = 13.96 m/h^2

If we assume that the car decelerate at the same rate.

the distance the car will stop had it been going 66.0mph will be achieved by using the same formula

V^2 = U^2 + 2as

0 = 66^2 - 2 × 13.96 × S

4356 = 27.92S

S = 4356 / 27.92

S = 156.02 m

Therefore, the car would stop at

156.02 m

Answer:

μ = 1

F = P√2

Explanation:

The parabola equation is: y = ½ x².

The slope of the tangent is dy/dx = x.

The angle between the tangent and the x-axis is θ = tan⁻¹(x).

At x = 1, θ = 45°.

Draw a free body diagram of the block.  There are three forces:

Weight force P pulling down,

Normal force N pushing perpendicular to the surface,

and friction force Nμ pushing up tangential to the surface.

Sum of forces in the perpendicular direction:

∑F = ma

N − P cos 45° = 0

N = P cos 45°

Sum of forces in the tangential direction:

∑F = ma

Nμ − P sin 45° = 0

Nμ = P sin 45°

μ = P sin 45° / N

μ = tan 45°

μ = 1

Draw a new free body diagram.  This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.

Sum of forces in the tangential direction:

∑F = ma

F − Nμ − P sin 45° = 0

F = Nμ + P sin 45°

F = (P cos 45°) μ + P sin 45°

F = P√2

Answer:

T = 3.14 hours

Explanation:

We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.

We know that the radius of Mars is 3,389.5 km.

So, r = 1,787 + 3,389.5 = 5176.5 km

Using Kepler's law,

[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]

M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]

So,

[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]

or

T = 3.14 hours

So, the orbital period is 3.14 hours

Answer:

a

   [tex]\lambda = 1.667 nm[/tex]

b

     [tex]\theta = 0.8681^o[/tex]

Explanation:

From the question we are told that

   The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]

    The  is distance of the screen from the slit is  [tex]D = 2.60 \ m[/tex]

    The distance between the central bright fringe and either of the adjacent bright   [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]

Generally  the condition for constructive interference is  

      [tex]d sin \tha(\theta ) = n \lambda[/tex]

From the question we are told that small-angle approximation is valid here.

So    [tex]sin (\theta ) = \theta[/tex]

=>        [tex]d \theta = n \lambda[/tex]

=>        [tex]\theta = \frac{n * \lambda }{d }[/tex]

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         [tex]y = D * sin (\theta )[/tex]

From the question we are told that small-angle approximation is valid here.

So

       [tex]y = D * \theta[/tex]

=>   [tex]\theta = \frac{ y}{D}[/tex]

So

     [tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]

     [tex]\lambda =\frac{d * y }{n * D}[/tex]

substituting values

       [tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]

        [tex]\lambda = 1.667 *10^{-6}[/tex]

        [tex]\lambda = 1.667 nm[/tex]

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

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

    [tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]

    [tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]

    [tex]\theta = 0.8681^o[/tex]

Answer:

The magnetic force would be:

[tex]F\approx 2.625\,\,10^{-5}\,\,N[/tex]

Explanation:

Recall that the magnetic force on a charged particle (of charge q) moving with velocity (v) in a magnetic field B, is given by the vector product:

F = q v x B

(where the bold represents vectors)

the vector product involves the sine of the angle ([tex]\theta[/tex]) between the vectors, so we can write the relationship between the magnitudes of these quantities as:

[tex]F=q\,v\,B\,sin(\theta)[/tex]

Therefore replacing the known quantities for the first case:

[tex]F=q\,v\,B\,sin(\theta)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\,sin(90^o)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\\q\,\,B=\frac{2.98\,\,10^{-4} }{25.7} \,\frac{N\,\,s}{m}[/tex]

Now, for the second case, we can find the force by using this expression for the product of the particle's charge times the magnetic field, and the new velocity and angle:

[tex]F=q\,v\,B\,sin(\theta)\\F=q\,(4.64\,\,m/s)\,B\,sin(29.2^o)\\F=q\,B(4.64\,\,m/s)\,\,sin(29.2^o)\\F=\frac{2.98\,\,10^{-4} }{25.7} \,(4.64\,\,m/s)\,\,sin(29.2^o)\\F\approx 2.625\,\,10^{-5}\,\,N[/tex]

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

Answer:

The magnification is  [tex]m = -2.15[/tex]

Explanation:

From the question we are told that

   The  radius is  [tex]r = 14.0 \ m[/tex]

    The  focal length eyepiece is  [tex]f_e = 3.25 \ m[/tex]

Generally the objective focal length is mathematically represented as

        [tex]f_o = \frac{r}{2}[/tex]

=>     [tex]f_o = \frac{14}{2}[/tex]

=>     [tex]f_o = 7 \ m[/tex]

The  magnification is mathematically represented as

      [tex]m = - \frac{f_o }{f_e }[/tex]

=>    [tex]m = - \frac{7 }{ 3.25 }[/tex]

=>   [tex]m = -2.15[/tex]

Answer:

The intensity of light from the first polarizer  is [tex]I_1 = 26 W/m^2[/tex]

Explanation:

  The intensity of the unpolarized light is  [tex]I_o = 52.0 \ W/m^2[/tex]

Generally the intensity of light that emerges from the first polarized light is

            [tex]I_1 = \frac{I_o}{2 }[/tex]

 substituting values

             [tex]I_1 = \frac{52. 0}{2 }[/tex]

             [tex]I_1 = 26 W/m^2[/tex]

Answer:

[tex]v_B/v_A=\sqrt{3}[/tex]

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]

Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:

[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]

Answer:

The wavelength is [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]

Explanation:

From the question we are told that

   The wavelength of the first light is  [tex]\lambda _ 1 = 587 \ nm[/tex]

    The order of the first light that is being considered is  [tex]m_1 = 10[/tex]

     The order of the second light that is being considered is  [tex]m_2 = 11[/tex]

Generally the distance between the fringes for the first light is mathematically represented as

      [tex]y_1 = \frac{ m_1 * \lambda_1 * D}{d}[/tex]

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         [tex]y_2 = \frac{ m_2 * \lambda_2 * D}{d}[/tex]

Now given that both of the light are passed through the same double slit

       [tex]\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1[/tex]

=>    [tex]\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1[/tex]

=>     [tex]\lambda_2 = \frac{m_1 * \lambda_1}{m_2}[/tex]

=>    [tex]\lambda_2 = \frac{10 * 587 *10^{-9}}{11}[/tex]

=>   [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]

Answer:

This is because the same electron removed from the positively charged plate is what is taken to the negatively charged plate, maintaining the same amount of electron according to the conservation of charge in an electric circuit.

Explanation:

In any circuit, electrons are neither created nor destroyed according to the laws of conservation of charge, but are transferred from one point to another on the circuit. When the plates of a capacitor are connected to battery, the battery pushes the electron to move due to its potential difference. Electrons are then moved from the positive plate, at a steady rate, to the negative plate. The removal of electrons from the positive plate is what leaves it positively charged from deficiency of electrons, and the addition of electrons at the negatively charged plate is what leaves the plate negatively charge from excess of electrons. From this, we can see that the same electrons removed from the positively charged plate are  taken to the negatively charged plate.

Answer:

The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.

Explanation:

Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.

Answer:

Chemical changes are recognized when a substance changes its properties permanently and it cannot be the same substance as before.

Instead the physical changes implies that if you can return to the same substance through a reverse process.

Explanation:

A chemical change is, by example, a combustion, if a sheet of paper burns, its result is ashes, the ashes cannot go back to being a sheet of paper because its properties changed, heat energy was generated that changed matter permanently.

A physical change, by example, is that of freezing water, the water becomes ice, but this can easily become water again if the temperature is increased, its properties do not change and the chemistry of the substance does not change.

A chemical change takes place when a chemical reaction takes place, while when a matter changes forms but not the chemical identity then a physical change takes place.  

• A product or a new compound formation takes place from a chemical change as the rearrangement of atoms takes place to produce novel chemical bonds.  

• In a chemical change always a chemical reaction takes place.  

• Some of the chemical changes examples are souring milk, burning wood, digesting food, mixing acid and base, cooking food, etc.  

• In a physical change no new chemical species form.  

• The changing of the state of a pure substance between liquid, gas, or solid is a physical change as there is no change in the identity of the matter.  

• Some of the physical changes are melting of ice, tempering of steel, breaking a bottle, crumpling a sheet of aluminum foil, boiling water, and shredding paper.  

Thus, a new substance is formed during a chemical change, while a physical change does not give rise to a new substance.  

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Answer:

(a)106.4C

b)0.5676mm

Explanation:

(a)To get the charge that have passed through the starter then The current will be multiplied by the duration

I= current

t= time taken

Q= required charge

Q= I*t = 140*0.760 = 106.C

(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)

diameter of the conductor is 4.20 mm

But Radius= diameter/2= 4.20/2=

The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then

radius of the conductor is 0.0021m.

We can now calculate the area of the conductor which is

A = π*r^2

= π*(0.0021)^2 = 13.85*10^-6 m^2

We can proceed to calculate the current density below

J = 140/13.85*10^-6 = 10108303A/m

According to the listed reference:

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s

Therefore , the distance traveled is:

x = v*t = 0.7468 * 0.760 = 0.5676mm

(a) The charge passes through the starter motor is 106.4C.

(b) An electron travel along the wire while the starter motor is on 0.5676mm.

Answer (a)

I= current

t= time taken

Q= required charge

Q= I*t

Q= 140*0.760

Q= 106.C

Answer (b)

The n electron travel along the wire while the starter motor is on:

Diameter of the conductor is 4.20 mm

Radius= diameter/2= 4.20/2

Radius =2.1mm

Radius of the conductor is 0.0021m.

A = π*r^2

A= π*(0.0021)^2

A= 13.85*10^-6 m^2

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 )

Vd  =0.0007468m/s

Vd =0 .7468 mm/s

The distance traveled is:

x = v*t

x= 0.7468 * 0.760

x = 0.5676mm

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CHECK COMPLETE QUESTION BELOW

you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?

A) four times more than you do on Earth.

B) two times less than you do on Earth.

C) the same as you do on Earth

D) two times more than you do on Earth

Answer:

OPTION C is correct

The same as you do on Earth

Explanation :

According to law of gravitation :

F=GMm/R^2......(a)

F= mg.....(b)

M= mass of earth

m = mass of the person

R = radius of the earth

From law of motion

Put equation b into equation a

mg=GMm/R^2

g=GMm/R^2

g=GM/R^2

We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have

m= 4M

r=(2R)^2=4R^2

g= G4M/4R^2

Then, 4in the denominator will cancel out the numerator we have

g= GM/R^2

Therefore, g remain the same

Answer:

The distance is  [tex]D = 0.000712 \ m[/tex]

Explanation:

From the question we are told that

    The wavelength of  the  light source is  [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]

     The distance from a pin hole is  [tex]x = 9\ m[/tex]

       The  diameter of the pin  hole is  [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

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

substituting values

             [tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]

             [tex]D = 0.000712 \ m[/tex]

Answer:

a) the distance between the interference fringes is reduced by half

b) the distance between stripes is doubled

Explanation:

Interference experiments constructive interference is described by the expression

          d sin θ = m λ

let's use trigonometry to find the distance between the interference fringes

              tan θ=  y / L

dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles

            tan θ = sin θ / cos θ = sin θ

we substitute

             sin θ = y / L

            d y / L = m  λ

           y = m λ / d L

a) it asks us when the screen doubles its distance

           L ’= 2 L

subtitute in the equation

           y ’= m λ / (d 2L)

           y ’=( m λ / d L) /2

           y ’= y / 2

the distance between the interference fringes is reduced by half

b) the density of the network doubles

      if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half

            d ’= d / 2

we substitute

          y ’= m λ (L d / 2)

          y ’= m λ / (L d) 2

          y ’= y 2

the distance between stripes is doubled

Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, [tex]N_P[/tex] = 50 turns

number of turns in the secondary winding, [tex]N_S[/tex] = 10 turns

the secondary load resistance, [tex]R_S[/tex] = 250 Ω

Determine the turns ratio;

[tex]K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5[/tex]

Now, determine the reflected resistance in the primary winding;

[tex]\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms[/tex]

Therefore, the reflected resistance in the primary winding is 6250 Ω