Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 120 years into the future. According to special relativity, how fast must you travel

Complete Question    

The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by

[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex]    in SI units.

Answer:

The  value  is  [tex]f = 1.98918*10^{5}\ Hz[/tex]

Explanation:

From the question we are told that

   The magnetic field is    [tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex]

 This above  equation can be modeled as

       [tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ] \equiv A sin ( kz -wt )[/tex]

So  

       [tex]w = \frac{10^7}{8}[/tex]

Generally the frequency is mathematically represented as

       [tex]f = \frac{w}{2 \pi}[/tex]

=>    [tex]f = \frac{ \frac{10^7}{8} }{2 \pi}[/tex]

=>    [tex]f = 1.98918*10^{5}\ Hz[/tex]

Answer:

Explanation:

Current in the same direction

 When current flow through to parallel conductors of a given length, when the current flows in the same direction

1. A force of attraction between the wires occurs and this tends to draw the wires inward

2. A magnetic field in the same direction is produced.

Current in opposite direction

when the current is in opposite direction

1. Force of repulsion between the two wires occurs, draws the wire outward

2. A magnetic field in opposite direction occurs

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

Answer:

The average density of the rod is 1.605 kg/m.

Explanation:

The average density of the rod is given by:

[tex] \rho = \frac{m}{l} [/tex]    

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:

[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex]   (1)

Using u = x+1  →  du = dx  → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]

[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]

Therefore, the average density of the rod is 1.605 kg/m.  

I hope it helps you!    

The average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

Given data:

The length of rod is, L = 3 m.

The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3,  The expression for the average density is given as,

[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)

Using u = x+1  

du = dx

u₁= x₁+1 = 0+1 = 1

and

u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]

Thus, we can conclude that the average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

Learn more about the average density here:

https://brainly.com/question/1371999

Answer:

The amplitude [tex]A(5) = 1 \ m[/tex]

Explanation:

From the question we are told that

     The  spring constant is  [tex]k = 100 \ N/m[/tex]

      The  damping constant is  [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]

       The mass is  [tex]m = 0.050 \ kg[/tex]

       The  maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]

       The  time  considered is  t =  5.0 s

Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as

           [tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]

substituting values

         [tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]

         [tex]A(5) = 1 \ m[/tex]

Answer:

The frequency will change by a factor of 0.4

Explanation:

T = 2(pi)*sqrt(L/g)

Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.

Answer:

A) 2

Explanation:

Given;

magnetic field of the coil, B = 1.6 T

frequency of the coil, f = 75 Hz

maximum emf developed in the coil, E = 56.9 V

area of the coil, A = 0.15 m x 0.25 m = 0.0375 m²

The maximum emf in the coil is given by;

E = NBAω

Where;

N is the number of turns

ω is the angular velocity = 2πf = 2 x 3.142 x 75 = 471.3 rad/s

N = E / BAω

N = 56.9 / (1.6 x 0.0375 x 471.3)

N = 2 turns

Therefore, the value of N is 2

A) 2

Answer:

The correct option is  D

Explanation:

From the question we are told that

   The  refractive index of oil film is [tex]k = 1.48[/tex]

   The  thickness is [tex]t = 290 \ nm = 290*10^{-9} \ m[/tex]

Generally for constructive interference

      [tex]2t = [m + \frac{1}{2} ]* \frac{\lambda}{k}[/tex]

For reflection of a bright fringe m =  1

 =>   [tex]2 * (290*10^{-9}) = [1 + \frac{1}{2} ]* \frac{\lambda}{1.48}[/tex]

=>     [tex]\lambda = 5.723 *10^{-7} \ m[/tex]

This wavelength fall in the range of a yellow light

Answer:

Rp = 3.04×10² Ω.

Explanation:

From the question given:

1/Rp = 1/4.5×10² Ω + 1/ 9.4×10² Ω

Rp =?

We can obtain the value of Rp as follow:

1/Rp = 1/4.5×10² + 1/ 9.4×10²

Find the least common multiple (lcm) of 4.5×10² and 9.4×10².

The result is 4.5×10² × 9.4×10²

Next, divide the result of the lcm by each denominator and multiply the result obtained with the numerator as shown below:

1/Rp = (9.4×10² + 4.5×10²) /(4.5×10²) (9.4×10²)

1/Rp = 13.9×10²/4.23×10⁵

Cross multiply

Rp × 13.9×10² = 4.23×10⁵

Divide both side by 13.9×10²

Rp = 4.23×10⁵ / 13.9×10²

Rp = 3.04×10² Ω.

Answer:

200cm

Explanation:

Answer:

100cm

Explanation:

Using

F= ( N/2L)(√T/u)

F1 will now be (0.5*2)( √600/0.015)

=> L( wavelength)= 200/2cm = 100cm

Answer: Astro 1 will have a 10x larger field of view than the Hubble.

Explanation: The hubble will also be extremely light weight that way it can go further into space and the mission will be able to last a longer amount of time.

Answer:

The  period is [tex]T = 0.00255 \ s[/tex]

Explanation:

From the question we are told that

  The  frequency is  [tex]f = 392 \ Hz[/tex]

Generally the period is mathematically represented as  

           [tex]T = \frac{1}{f}[/tex]

=>       [tex]T = \frac{1}{ 392}[/tex]

=>       [tex]T = 0.00255 \ s[/tex]

Answer:

a)step-down" transformer

b) 1440 turns

Explanation:

There are two types of transformers; step up transformers and step down transformers. A step down transformer converts a higher voltage to a lower voltage.

In a stepdown transformer, there are more turns in the primary coil than in the secondary coil, the turns ratio Ns/Np is less than 1 for a stepdown transformer.

If

Number of turns in primary coil Np= 2760

Number of turns in secondary coil Ns= unknown

Voltage in primary coil Vp= 230 V

Voltage in secondary coil Vs= 120 V

Ns/Np= Vs/VP

NsVp= NpVs

Ns= NpVs/VP = 2760 × 120/230

Ns= 1440 turns

Answer:

Distance = 13 units

Explanation:

The overall path covered by an object during its journey is called distance covered.

In this problem, a person starts at position zero, walks to position 8, then walks to position 5.

We need to find the person's distance traveled. It can be calculated simply by adding all the positions i.e.

Distance = 0+8+5

Distance = 13

Hence, the distance covered by the person is 13 units.

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      [tex]Em_{f}[/tex] = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

          K / K₀ = 4

B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

Answer:

m = 9.1 x 10⁶ kg

Explanation:

First, we need to find the rate of heat transfer through the box to the ice. For this purpose, we use Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

Q = Rate Of Heat Transfer = ?

K = Thermal Conductivity = 0.16 KW/m.°C = 160 W/m.°C

A = Area = 1.5 m²

ΔT = Difference in Temperature = 29°C - 0°C = 29°C

L = Thickness of wall = 2 cm = 0.002 m

Therefore,

Q = (160 W/m °C)(1.5 m²)(29°C)/(0.002 m)

Q = 3.48 x 10⁶ W

Now, we find the amount of heat transferred in one day to the ice:

q = Qt

where,

q = amount of heat = ?

t = time = (1 day)(24 h/1 day)(3600 s/1 h) = 86400 s

Therefore,

q = (3.48 x 10⁶ W)(8.64 x 10⁴ s)

q = 3 x 10¹¹ J

Now, for mass of ice melted in a day:

q = m H

m = q/H

where,

m = mass of ice melted in a day = ?

H = latent heat of fusion of ice = 3.3 x 10⁵ J/kg

Therefore,

m = (3 x 10¹¹ J)/(3.3 x 10⁵ J/kg)

m = 9.1 x 10⁶ kg

Answer:

Explanation:

To convert from °C to °F , the formula is

( F-32 ) / 9 = C / 5

F is reading fahrenheit scale and C is in centigrade scale .

F = 205 , C = ?

(205 - 32) / 9 = C / 5

C = 96°C approx .

Let us calculate the heat required .

Total heat required = heat required to heat up the ice at - 20 °C  to 0°C  + heat required to melt the ice + heat required to heat up the water at  0°C to

96°C.

=  5 x 2.04 x (20-0) +  5 x 336 + 5 x ( 96-0 ) x 4.2  kJ .

= 204 + 1680 + 2016

= 3900 kJ .

Complete question:

A charge of 15C is moving with velocity of 6.2 x 10³ m/s which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.

a. 0.06 T

b. 0.08 T

c. 0.07 T

d. 0.05 T

Answer:

The magnitude of the magnetic field is 0.07 T.

Explanation:

Given;

magnitude of the charge, q = 15C

velocity of the charge, v = 6.2 x 10³ m/s

angle between the charge and the magnetic field, θ = 48°

the force on the particle, F = 4838 N

The magnitude of the magnetic field can be calculated by applying Lorentz force formula;

F = qvBsinθ

where;

B is the magnitude of the magnetic field

B = F / vqsinθ

B = (4838) / (6.2 x 10³ x 15 x sin48)

B = 0.07 T

Therefore, the magnitude of the magnetic field is 0.07 T.

Answer:

A) it transforms a small force acting over a large distance into a large force acting over a small distance.

Explanation:

The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.

Pressure transmitted P = F/A

where F is the force applied

and A is the area over which the force is applied.

This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.

F/A = f/a

where the left side of the equation is for the output, and the right side is for the input.

The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that

volume V = (area A) x (distance d)

this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.

From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.

Answer:

The  speed of the water wave is [tex]v = 88 \ m/s[/tex]

Explanation:

From the question we are told that

      The  width of the tube is  [tex]L = 55 \ m[/tex]

     The fundamental  frequency is  [tex]f = 0.80 \ Hz[/tex]

Generally the fundamental frequency is mathematically represented as

      [tex]f = \frac{v}{2 * L }[/tex]

=>    [tex]v = f * 2 * L[/tex]

substituting values

       [tex]v = 0.8 * 2 * 55[/tex]

       [tex]v = 88 \ m/s[/tex]

The speed of the water wave will be 88 m/s.

Given information:

When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center.

The frequency of the standing wave is [tex]f=0.8[/tex] Hz.

The width of the tub is [tex]w=55[/tex] m.

Let v be the speed of the standing wave.

The speed of the wave can be calculated as,

[tex]v=2wf\\v=2\times 55\times 0.8\\v=88\rm\; m/s[/tex]

Therefore, the speed of the water wave will be 88 m/s.

For more details, refer to the link:

https://brainly.com/question/1967686

Answer:

The maximum height reached in the second trial is 16times the maximum height reached in the first trial.

Explanation:

The following data were obtained from the question:

First trial

Initial speed (u) = v

Final speed (v) = 0

Second trial

Initial speed (u) = 4v

Final speed (v) = 0

Next, we shall obtain the expression for the maximum height reached in each case.

This is illustrated below:

First trial:

Initial speed (u) = v

Final speed (v) = 0

Acceleration due to gravity (g) = 9.8 m/s²

Height (h₁) =.?

v² = u² – 2gh₁ (going against gravity)

0 = (v)² – 2 × 9.8 × h₁

0 = v² – 19.6 × h₁

Rearrange

19.6 × h₁ = v²

Divide both side by 19.6

h₁ = v²/19.6

Second trial

Initial speed (u) = 4v

Final speed (v) = 0

Acceleration due to gravity (g) = 9.8 m/s²

Height (h₂) =.?

v² = u² – 2gh₂ (going against gravity)

0 = (4v)² – 2 × 9.8 × h₂

0 = 16v² – 19.6 × h₂

Rearrange

19.6 × h₂ =16v²

Divide both side by 19.6

h₂ = 16v²/19.6

Now, we shall determine the ratio of the maximum height reached in the second trial to that of the first trial.

This is illustrated below:

Second trial:

h₂ = 16v²/19.6

First trial:

h₁ = v²/19.6

Second trial : First trial

h₂ : h₁

h₂ / h₁ = 16v²/19.6 ÷ v²/19.6

h₂ / h₁ = 16v²/19.6 × 19.6/v²

h₂ / h₁ = 16

h₂ = 16 × h₁

From the above illustrations, we can see that the maximum height reached in the second trial is 16times the maximum height reached in the first trial.

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:

External force    W₁ = F L

Friction force    W₂ = - fr L

weight component   W₃ = - mg sin θ L

Y Axis   Force      W=0

Explanation:

When the block rises up the plane with constant velocity, it implies that the sum of the forces is zero.

For these exercises it is indicated to create a reference system with the x axis parallel to the plane and the y axis perpendicular

let's write the equations of translational equilibrium in given exercise

X axis

        F - fr -Wₓ = 0

        F = fr + Wₓ

the components of the weight can be found using trigonometry

         Wₓ = W sin θ

         [tex]W_{y}[/tex] = W cos θ

let's look for the work of these three forces

          W = F x cos θ

External force

          W₁ = F L

since the displacement and the force have the same direction

Friction force

          W₂ = - fr L

since the friction force is in the opposite direction to the displacement

For the weight component

          W₃ = - mg sin θ L

because the weight component is contrary to displacement

Y Axis  

          N- Wy = 0

in this case the forces are perpendicular to the displacement, the angle is 90º and the cosine 90 = 0

therefore work is worth zero

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:

Inertia! I hope this helps!

Answer:

inertia

Explanation

Inertia.

Answer:

1. the sphere of the radius a

Explanation:

Because the charge distribution for each case is spherically symmetric, we can choose a spher- ical Gaussian surface of radius r , concentric with the sphere in question.

So E = k (Q /r 2) (for r ≥ R ) , where R is the radius of the sphere being considered, either a or b .

With the choice of potential at r = ∞ being zero, the electric potential at any distance r from the center of the sphere can be expressed as V = - integraldisplay r ∞ E dr = k /Q r

(for r ≥ R ) .

On the spheres of radii a and b , we have V a = k (Q/ a)and V b = k (Q/ b), respectively.

So Since b > a , the sphere of radius a will have the higher potential.

Also recall Because E = 0 inside a conductor, the potential

Answer:

24

Explanation:

The mass = 3 kg

at point O all the mechanical energy of the system is due to its potential energy PE. The body is at rest.

PE = mgh

but ME = PE + KE = constant   (law of energy conservation)

KE is the kinetic energy

since KE is zero at this point, then,

ME = mgh

where m is the mass

g is acceleration due to gravity = 9.81 m/s^2

h is the height = O

ME = 3 x 9.81 x O

ME = 29.43-O

At point A the total ME is due to its PE and its kE

PE at this point = mgh = 3 x 9.81 x A = 29.43-A

KE = [tex]\frac{1}{2}mv^{2}[/tex]

velocity = v at this point, therefore,

KE = [tex]\frac{1}{2}*3*v^{2}[/tex] = [tex]\frac{3}{2} }v^{2}[/tex]

therefore,

ME = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]

Equating ME for the points O and A, we have

29.43-O = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]

29.43-O - 29.43-A = [tex]\frac{3}{2} }v^{2}[/tex]

(O - A)29.43 = [tex]\frac{3}{2} }v^{2}[/tex]

O - A = 0.051[tex]v^{2}[/tex]   this is the distance between point O and A

For point B

PE = 29.43-B

KE = [tex]\frac{25}{2}*3*v^{2}[/tex] = 37.5[tex]v^{2}[/tex]        (velocity is equal to [tex]5v[/tex] at this point)

therefore,

ME = 29.43-B + 37.5[tex]v^{2}[/tex]

Equating the ME for points A and B, we have

29.43-A + [tex]\frac{3}{2} }v^{2}[/tex] = 29.43-B + 37.5[tex]v^{2}[/tex]

29.43-A - 29.43-B = 37.5[tex]v^{2}[/tex] - [tex]\frac{3}{2} }v^{2}[/tex]

(A - B)29.43 = 36[tex]v^{2}[/tex]

A - B = 1.22[tex]v^{2}[/tex]    this is the distance between points A and B

The distance between points A & B divided by the distance between points 0 & A will be

1.22[tex]v^{2}[/tex]/0.051 = 23.9 ≅ 24

Answer:

A)a red giant star

Answer:

The induced emf  is  [tex]\epsilon = 0.1041 \ V[/tex]  

Explanation:

From the question we are told that

   The  radius of the circular loop is  [tex]r = 9.50 \ cm = 0.095 \ m[/tex]

     The  intensity of the wave is  [tex]I = 0.0215 \ W/m^2[/tex]

      The wavelength is  [tex]\lambda = 6.90\ m[/tex]

Generally the intensity is mathematically represented as

         [tex]I = \frac{ c * B^2 }{ 2 * \mu_o }[/tex]

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

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

B is the magnetic field which can be mathematically represented from the equation as

          [tex]B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }[/tex]

substituting values

          [tex]B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }[/tex]

          [tex]B = 1.342 *10^{-8} \ T[/tex]

The  area is mathematically represented as

       [tex]A = \pi r^2[/tex]

substituting values

       [tex]A = 3.142 * (0.095)^2[/tex]

       [tex]A = 0.0284[/tex]

The angular velocity is mathematically represented as

        [tex]w = 2 * \pi * \frac{c}{\lambda }[/tex]

substituting values          

       [tex]w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }[/tex]  

        [tex]w = 2.732 *10^{8} rad \ s^{-1}[/tex]  

Generally the induced emf is mathematically represented as

        [tex]\epsilon = N * B * A * w * sin (wt )[/tex]

At maximum induced emf  [tex]sin (wt) = 1[/tex]

    So

         [tex]\epsilon = N * B * A * w[/tex]

substituting values

         [tex]\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}[/tex]  

         [tex]\epsilon = 0.1041 \ V[/tex]  

Answer: d₂ = 170 mGya

Explanation:

the relationship between absonbed 'd' and exposure 'E' is given as;

D(Gv) = F . x (AS/xB)

F is a conversion coefficient depending on medium

so we can simply write

d₁/d₂ = x₁/x₂

Given that;

our x₁ = 60 mAs, x₂ = 120 mAs,  d₁ = 85 mGya, d₂ = ?

from the given formula,

d₂ = (x₂d₁ / x₁)

now we substitute

d₂ = (120 × 85) / 60

d₂ = 170 mGya

∴ if 120 mAa is used,  the new exposure will be 170 mGya