When a magnet moves above a conducting ladder, the currents induced in the ladder produces a magnetic field. This field interacts with the magnetic field of the magnet to produce a force that A) pulls the ladder in the direction opposite to the direction of the moving magnet. B) pulls the ladder in the same direction as the moving magnet. C) brakes the ladder. D) None of the above.

Answer:

Visible light intensity at the surface of the bulb (I) = 331 W/m²

Explanation:

Given:

Energy = 75 W

Radius = 6 /2 = 3 cm = 3 × 10⁻² m

Energy goes to visible light = 5% = 0.05

Find:

Visible light intensity at the surface of the bulb (I)

Computation:

Visible light intensity at the surface of the bulb (I) = P / 4A

Visible light intensity at the surface of the bulb (I) = (0.05)(75) / 4π(3 × 10⁻²)²

Visible light intensity at the surface of the bulb (I) = 3.75 / 4π(9 × 10⁻⁴)

Visible light intensity at the surface of the bulb (I) = 331 W/m²

Answer:

the final angular velocity of the particle is approximately 38.18  Rad/s

Explanation:

To start with, let's make sure that units of angle measure are the same, converting everything into radians:

[tex]4500^o\, \frac{\pi}{180^o}= 25\,\pi[/tex]

And now we can use the kinematic formulas for rotational motion:

[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2[/tex]

Therefore we can find the initial angular velocity [tex]\omega_0[/tex]  of the particle:

[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2\\25\,\pi=\omega_0\,(3)+\frac{1}{2} (8)\,(3)^2\\25\,\pi-36=\omega_0\,(3)\\\omega_0=\frac{25\,\pi-36}{3} \\\omega_0\approx 14.18\,\,\,rad/s[/tex]

and now we can estimate the final angular velocity using the kinematic equation for angular velocity;

[tex]\omega=\omega_0\,+\alpha\,t\\\omega=14.18+8\,(3)\\\omega=38.18\,\,\,rad/s[/tex]

Answer:

The number of windings is 1.

Explanation:

The radius of the solenoid = 8.0 cm = 0.08 m

Length of the solenoid = 45.0 cm = 0.45 m

number of turn = ?

circumference of each winding = 2πr = 2 x 3.142 x 0.08 = 0.503 m

The number of windings = (Length of the solenoid)/(circumference of each winding)

==> 0.45/0.503 = 0.89 ≅ 1

Answer:

Explanation:

a) To get a rainbow from a prism arrangement, we will need

First, you cut a slit in one end of the cardboard with the pen knife.

Next you open up a space on top of the cardboard through which you can observe the experiment and its result.

Next, you place the triangular prism with its slant face facing the the cut slit.

Finally, position the slit to face the light from the open window, and adjust the prism till the projected bands of colored light (rainbow) is very much obvious on the other end of the box, opposite the slit.

b) For a light to yield rainbow, it most be composed of different component colors of light. The colors of light is due to the difference in wavelength, and dispersion is due to the different in the wavelengths of the component light. So to get rainbow from a light source, the light must not be monochromatic. This means that only light composed of component light of different colors can produce rainbow. Light from the sun for example is composed of 7 distinct colors of light, and white light can be created with just three colors; blue, green, and red light.

Answer:

 t = 8.98 10⁻⁷ m

Explanation:

This is an exercise in interference by reflection, let's analyze what happens on each surface of the film.

* When the light ray shifts from a medium with a lower refractive index to a medium with a higher refractive index, the reflected ray has a reflection of 180

* The beam when passing to the middle its wavelength changes

           λ = λ₀ / n

if we take this into account, the constructive interference equation for normal incidence is

            2t = (m + ½) λ₀ / n

let's apply this equation to our case

for λ₀ = 479 nm = 479 10⁻⁹ m

             t = (m + ½) 479 10⁻⁹ / 1.33

             (m + ½) = 1.33 t / 479 10⁻⁹

for λ₀ = 798 nm = 798 10⁻⁹ m

             t = (m' + ½) 798 10⁻⁹ /1.33

            (m' + ½) = 1.33 t / 798 10⁻⁹

as they tell us that no other constructive interference occurs between the two wavelengths, the order of interference must be consecutive, let's write the two equat⁻ions

               (m + ½) = 1.33 t / 479 10⁻⁹

             ((m-1) + ½) = 1.33 t / 798 10⁻⁹

             (m + ½) = 1.33 t / 798 10⁻⁹ +1

resolve

             1.33 t / 479 10⁻⁹ = 1.33 t / 798 10⁻⁹ +1

             1.33 t / 479 10⁻⁹ = (1.33t + 798 10⁻⁹) / 798 10⁻⁹

             1.33t = (1 .33t + 798 10⁻⁹) 479/798

             1.33t = (1 .33t + 798 10⁻⁹) 0.6

             1.33 t = 0.7983 t + 477.6 10⁻⁹

             t (1.33 - 0.7983) = 477.6 10⁻⁹

             t = 477.6 10⁻⁹ /0.5315

             t = 8.98 10⁻⁷ m

Answer:

2500 kg/m³

Explanation:

P = P

ρgh = ρgh

ρh = ρh

(1000 kg/m³) (8.9 cm) = ρ (3.5 cm)

ρ ≈ 2500 kg/m³

Answer:

The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]

Explanation:

From the question we are told that

    The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]

     The  speed is  [tex]v_1 = 785 \ m/s[/tex]

     The mass of the water is  [tex]m_w = 13.5 \ kg[/tex]

     The velocity it emerged with is  [tex]v_2 = 534 \ m/s[/tex]

Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then  

the change in kinetic energy of the bullet =  the heat gained by the water

 So

 The change in kinetic energy of the water is  

          [tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]

substituting values  

        [tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]

        [tex]\Delta KE = 2814.1 \ J[/tex]

Now the heat gained by the water is

     [tex]Q = m_w* c_w * \Delta T[/tex]

Here [tex]c_w[/tex] is the specific heat of water which has a value  [tex]c_w = 4190 J/kg \cdot K[/tex]

So  since   [tex]\Delta KE = Q[/tex]  

we have that

          [tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]

          [tex]\Delta T = 0.0497 \ ^oC[/tex]

Answer:

51. 7m/s

Explanation:

Take speed of sound in air = 340 m/s

fp = fs (V + Vp)/(V + Vs)

128 = 123 (340 + Vp)/(340 + 36.4)

Vp = 51.7m/s

Explanation:

Answer:

R = 36.885 km

Explanation:

In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other

The diffraction equation for slits is

            a sin θ = m λ

the first minimum occurs for m = 1

             sin θ = λ a

as the diffraction experiments the angles are very small, we approximate

             sin θ = θ

             θ = λ / a

This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form

            θ = 1.22 λ / a

In this problem they give us the frequency, let's find the wavelength with the relation

           c = λ f

           λ = c / f

           θ = 1.22 c/ f a

since they ask us for the distance between the planes, we can use the definition of radians

          θ = s / R

if we assume that the distance is large, we can approximate the arc to the horizontal distance

          s = x

we substitute

             x / R = 1.22 c / fa

             R = x f a / 1.22c

Let's reduce the magnitudes to the SI system

            f = 9000 MHz = 9 109 Hz

            a = 15 m

           x = 100 m

let's calculate

            R = 100 10⁹ 15 / (1.22 3 108)

            R = 3.6885 10⁴ m

let's reduce to km

            R = 3.6885 10¹ km

            R = 36.885 km

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

The correct option is option (A)

the frequencies produced by the pipes are (A) 110Hz,220Hz, 330 Hz

(I) For an organ pipe open at open both ends the frequency of different modes is given by:

F =  nv/2L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = v/2L

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

The first overtone corresponds to n = 2, the second overtone corresponds to n = 3, and so on...

F₁ =2v/2L

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

The difference between successive overtones is F₀

(II) For an organ pipe open at one end the frequency of different modes is given by:

F =  nv/4L

where

F is the frequency

L is the length of the organ pipe

v is the speed of the wave

and, n is the mode of frequency

the fundamental frequency corresponds to n = 1, given by:

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

For an organ pipe open at one end, only those overtones are present which correspond to odd n, that is n = 3,5,...so:

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

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Answer:

Stay the same

Explanation:

Since, friction is negligible:

Initial Momentum = Final Momentum

Initial KE = Final KE

m1 * v1 = m2 * v2

When m increases v decreases.

The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.

Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.

Given:

The paperclips into an open cart rolling along a straight horizontal track with negligible friction,

Calculate the momentum, Since friction is negligible,

Initial Momentum = Final Momentum

Initial Kinetic Energy = Final Kinetic Energy

m₁ × v₁ = m₁  × v₂

When m increases, v decreases,

Thus, momentum will remain the same.

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Answer:

Kidney

Explanation:

One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.

Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.

Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.

Answer:

Length of pipe organ(L) = 0647 m (Approx)

Explanation:

Given:

Temperature (T) = 14°C

Fundamental frequency (F) = 262 Hz

Speed of sound (v) = 331 + 0.60(T) m/s

Find:

Length of pipe organ(L)

Computation:

Speed of sound (v) = 331 + 0.60(14) m/s

Speed of sound (v) = 339.4

Length of pipe organ(L) = Speed of sound (v) / 2(Fundamental frequency)

Length of pipe organ(L) = 339.4 / 2 (262)

Length of pipe organ(L) = 0647 m (Approx)

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.

Answer:

d. unchanged.

Explanation:

The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.

In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from

v = fλ

that the frequency is tied to the wave, and does not change throughout the waveform.

where v is the speed of the sound wave

f is the frequency

λ is the wavelength of the sound wave.

Answer:

2/5 m/s

Explanation:

There are two vectors  v and w . Let θ be angle b/w the two vector.

[tex]cos\theta =\frac{\overleftarrow{v}\cdot \overleftarrow{w}}{\left | v \right |\left | w \right |}\\=\frac{6-4}{\sqrt(2^2+1^2)\sqrt(3^2+4^2)} =\frac{2}{5\sqrt(5)}[/tex]

velocity of the ball in direction of the the wind

[tex]\left | vcos\theta \right |\\\left | v \right |cos\theta\\\sqrt(2^2+1^2)\frac{2}{5\sqrt(5)} = \frac{2}{5}[/tex]

The size of the velocity of the ball in the direction of the wind is 2/5 ms.

Since there are two vectors v and w

Also, here we assume θ be angle b/w the two vector.

So

Cos θ = 6-4 / √(2^2 + 1^2) √(3^2 + 4^2)

= 2/5√5

Now the velocity of the ball should be

= √(2^2 + 1^2) 2 ÷ 5√(5)

= 2 /5

hence, The size of the velocity of the ball in the direction of the wind is 2/5 ms.

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Given that,

Energy [tex]H=2.7\times10^{31}\ W[/tex]

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

[tex]H=Ae\sigma T^4[/tex]

[tex]A=\dfrac{H}{e\sigma T^4}[/tex]

[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]

[tex]R=5.0\times10^{10}\ m[/tex]

(b). Given that,

Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]

[tex]R=5.42\times10^{6}\ m[/tex]

Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]

(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]

Answer:

The water pressure on the upper pipe is 92.5 kPa.

Explanation:

Given that,

Pressure in lower pipe= 120 kPa

Speed of water in lower pipe= 1 m/s

Acceleration due to gravity = 10 m/s²

Density of water = 1000 kg/m³

Radius of lower pipe = 12 m

Radius of uppes pipe = 6 m

Height of upper pipe = 2 m

We need to calculate the velocity in upper pipe

Using continuity equation

[tex]A_{1}v_{1}=A_{2}v_{1}[/tex]

[tex]\pi r_{1}^2\times v_{1}=\pi r_{2}^2\times v_{2}[/tex]

[tex]v_{2}=\dfrac{r_{1}^2\times v_{1}}{r_{2}^2}[/tex]

Put the value into the formula

[tex]v_{2}=\dfrac{12^2\times1}{6^2}[/tex]

[tex]v_{2}=4\ m/s[/tex]

We need to calculate the water pressure on the upper pipe

Using bernoulli equation

[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]

Put the value into the formula

[tex]120\times10^{3}+\dfrac{1}{2}\times1000\times1^2+1000\times10\times0=P_{2}+\dfrac{1}{2}\times1000\times(4)^2+1000\times10\times2[/tex]

[tex]120500=P_{2}+28000[/tex]

[tex]P_{2}=120500-28000[/tex]

[tex]P_{2}=92500\ Pa[/tex]

[tex]P_{2}=92.5\ kPa[/tex]

Hence, The water pressure on the upper pipe is 92.5 kPa.

Answer:

The values is  [tex]m_{max} = 8001 \ bright \ spots[/tex]

Explanation:

From the question we are told that

    The slit distance is  [tex]d = 2 \ mm = 2*10^{-3} \ m[/tex]

    The  wavelength is  [tex]\lambda = 500 \ nm = 500 *10^{-9} \ m[/tex]

At the first half of the screen from the central maxima

   The number of bright spot according to the condition for constructive interference is  

          [tex]n = \frac{d * sin (\theta )}{\lambda}[/tex]

For maximum number of spot [tex]\theta = 90^o[/tex]

So  

       [tex]n = \frac{2*10^{-3} * sin (90 )}{500 *10^{-9}}[/tex]

        [tex]n =4000[/tex]

Now for the both sides plus the central maxima  we have

      [tex]m_{max} = 2 * n + 1[/tex]

substituting values

       [tex]m_{max} = 2 * 4000 + 1[/tex]

       [tex]m_{max} = 8001 \ bright \ spots[/tex]

Answer:

1.1micro meter

Explanation:

Given that

Constructive interference is

ma = alpha x sin theta

Alpha = 1 x 594 x10^ -9/ sin 32.8°

= 1.1 x 10^ -6m

Explanation:

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

Answer:

A) E(r) = 1.3957 × 10^(5) N/C

B) E(r) = 9.8864 × 10⁴ N/C

C) E(r) = 1.13 × 10^(5) N/C

Explanation:

We are given;

q = 6 nc = 6 × 10^(-9) C

L = 10 cm = 0.1 m

d = 4.4 cm = 0.044 m

r1 = 1 cm = 0.01 m

r2 = 2 cm = 0.02 m

r3 = 3 cm = 0.03 m

Formula for the electric field strength in this question is given as;

E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)

When factorized, we have;

E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]

Plugging in the relevant values for q/(2π(ε_o)L)

We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m

Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53

Thus;

E(r) = 1078.52 [1/r + 1/(d - r)]

A) E1 is at r = 1 cm = 0.01m

Thus;

E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))

E(r) = 1.3957 × 10^(5) N/C

B) E2 is at r = 2 cm = 0.02 m

Thus;

E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))

E(r) = 9.8864 × 10⁴ N/C

C) E2 is at r = 3 cm = 0.03 m

Thus;

E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))

E(r) = 1.13 × 10^(5) N/C

Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

Answer:

10m

Explanation:

Since Given frequency f= 1/t

and velocity ν=10 m/s

We know ν=λf

λ= ν/f

​ = 10/1/0.5

=5m

Since both the waves are similar but moves in opposite direction its total wavelength of the wave will be 10 m

Explanation:

1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.

2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.

Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.

We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.

These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.

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Answer:

The induced emf in the coil is 7.843 V

Explanation:

Given;

number of turns of the coil, N = 1300 turn

diameter of the coil, d = 2.4 cm = 0.024 m

initial magnetic field, B₁ = 0 T

final magnetic field, B₂ = 0.12 T

change in time, dt = 9.0 ms = 9 x 10⁻³ s

Area of the coil is given by;

A = πr²

radius of the coil, r = 0.024 / 2

radius of the coil, r = 0.012 m

A = π(0.012)²

A = 4.525 x 10⁻⁴ m²

The induced emf in the coil is given by;

E = NA(dB/dt)

E =  NA [(B₂ - B₁) /dt]

E = 1300 x 4.525 x 10⁻⁴ (0.12 - 0) / (9 x 10⁻³)

E = 7.843 V

Therefore, the induced emf in the coil is 7.843 V

Answer:

The complete question is

NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26  W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.

a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.

For totally reflective, F = (2I/c)A    ....1

for totally reflective, F = (I/c)A       ....2

where I is the intensity of the light

c is the speed of light = 3 x 10^8 m/s

A is the area the sail

b) The intensity of the light from the sun = power/area

==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]

where r is the distance from the sun and the sail

The Force from the sail from equation 1  is therefore

[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]

gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]

where

G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.

m is the mass of the sail = 10000 kg

M is the mass of the sun = 1.99 x 10^30 kg.

==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]

Equating the forces, we have

[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]  =  [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]

the distance cancels out

A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2

==> 6428.2 km^2

c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation  is inversely proportional to the square of the distance, so they both cancel out.

Answer:

3x10^5m/s

Explanation:

See attached file

Explanation:

The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.

Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.

[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]

Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.

[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]

[tex]v = 3\times 10^5\;\rm m/s[/tex]

Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

To know more about the doppler effect, follow the link given below.

https://brainly.com/question/1330077.

Answer:

Explanation:

Find:

Computation:

F = (9/5)C + 32

3 times that of the Celsius

If C = x

So F = 3x

So,

3x = (9/5)x + 32

15x = 9x +160

6x = 160

x = 26.67

So, C = 26.67° and F = 80°

1/5 times that of the Celsius

If C = x

So F = x/5

So,

x/5 = (9/5)x + 32

x = 9x + 160

x = -20

So, C = -20° and F = -4°