Red and orange stars are found evenly spread throughout the galactic disk, but blue stars are typically found

Answer:

Explanation:

Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/

α = 26.56°

Laser angle with Normal = 90 - 26.56 = 63.44 °

Assuming a red laser, refractive index in water is 1.331.

Angle of refraction in water is given by:

Ref Ind = Sin i / Sin r

1.331 = Sin 63.44 / Sin r

Sin r = 0.8945 / 1.331 = 0.6721

Angle r = 42.22°

For the path in water:

Tan 42.22 = x / 3.2

x = 2.9m where x is the lateral displacement of the laser ince it hits the water

So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool

Answer:

space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.

Answer:

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Explanation:

Given that:

the time period T = 88.5 min = 88.5 × 60  = 5310 sec

The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg

if  the radius of orbit is r,

Then,

[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]

[tex]{V^2} = \dfrac{GM_e }{r}[/tex]

[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Similarly :

[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]

where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]

Then:

[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]

[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]

[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]

[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]

[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]

[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]

[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]

[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]

[tex]r= 2565.38^2[/tex]

r = 6579225 m

The  circumference of the satellites  orbit can now be determined by using the formula:

 circumference = 2π r

 circumference = 2π  × 6579225 m

 circumference = 41338489.85 m

 circumference of the satellite orbit  = 4.13 × 10⁷ m

Answer:

A. 4.82 cm

B. 24.66 cm

Explanation:

The depth of water = 19.6 cm

Distance of fish  = 6.40 cm

Index of refraction of water = 1.33

(A). Now use the below formula to compute the apparent depth.

[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.[/tex]

(B). the depth of the fish in the mirror.

[tex]d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm[/tex]

Now find the depth of reflection of the fish in the bottom of the tank.

[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8 = 24.66\\[/tex]

1. Which example best describes a restoring force?

B) the force applied to restore a spring to its original length

2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?

C) The spring exerts a restoring force to the left and returns to its equilibrium position.

3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

D) 1 m

4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?

D)It is a vector quantity.

5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?

A) It decreases in magnatude.

Hope this Helps!! Sorry its late

Answer:

A = 6.8 km²

Explanation:

A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.

B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;

F_rad = 2IA/c

I is given by the formula;

I = P/(4πr²)

Thus;

F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²

Where;

A is the area of the sail

r is the distance of the sail from the sun

c is the speed of light = 3 × 10^(8) m/s

P is total power output of the sun = 3.90 × 10^(26) W

Now,F_rad = F_g

Where F_g is gravitational force.

Thus;

PA/2cπr² = G•m•M_sun/r²

r² will cancel out to givw;

PA/2cπ = G•m•M_sun

Making A the subject, we have;

A = (2•c•π•G•m•M_sun)/P

Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg

G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²

Thus;

A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))

A = 6.8 × 10^(6) m² = 6.8 km²

Answer:

a. 0.058°

b.  0.117°

Explanation:

a. The angular position of the first-order is:

[tex] d*sin(\theta) = m\lambda [/tex]

[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{1* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.058 ^{\circ} [/tex]

Hence, the angular position of the first-order, two-slit, interference maxima is 0.058°.

b. The angular position of the second-order is:

[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{2* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.12 ^{\circ} [/tex]

Therefore, the angular position of the second-order, two-slit, interference maxima is 0.117°.

I hope it helps you!

Answer:

d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

Explanation:

Let us take the momentum of a photon unit as u

we know that the rate of change of momentum is proportional to the force exerted.

For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that

F = (u - 0)/t = u/t

for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,

F = u

For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to

F = (u - (-u))/t = 2u/t

just as the we did above, it becomes

F = 2u.

From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.

Answer

3.340J

Explanation;

Using the relation. Energy stored in capacitor = U = 7.72 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.

Initial separation between the plates =d1= 3.30mm .

Final separation = d2 = 1.45 mm

(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.72*1.45/3.30

(A) Final energy = Uf = 3.340J

Answer:

objetive: a converging lens for large diameter lenses

eyepiece you must select a lens with a small focal length and the diameter is not important

The selected lenses should decrease chromatic aberration.

Explanation:

A telescope is an instrument that collects light from very distant objects, therefore very weak.

Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.

For the eyepiece you must select a lens with a small focal length and the diameter is not important

the telescope magnification is

                 m = f_objective / F_ocular

The selected lenses should decrease chromatic aberration.

In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.

Answer: this the real answer try it objetive: a converging lens for large diameter lenseseyepiece you must select a lens with a small focal length and the diameter is not importantThe selected lenses should decrease chromatic aberration.Explanation:A telescope is an instrument that collects light from very distant objects, therefore very weak.Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.For the eyepiece you must select a lens with a small focal length and the diameter is not importantthe telescope magnification is                 m = f_objective / F_ocularThe selected lenses should decrease chromatic aberration.In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.

Explanation:

Answer:

C $0.75 my friend I wish it is right answer

Answer:

30.93 cm

Explanation:

Given that:

A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses

The power of the old pair of lens p = 2.25 diopters

The focal point length = 1/p

The focal point length =  1/2.25

The focal point length = 0.444 m

The focal point length = 44.4 cm

The near point of the person from the glass = (85 -2)cm , This is because the glasses are usually 2 cm from the lens

The near point of the person from the glass = 83 cm

Let consider s' to be the image on the same sides of the lens,

∴ s' = -83 cm

We known that:

the focal length of a mirror image 1/f =1/u +1/v

Assume the near point is at an excellent distance s from the glass where the person wears the corrective glasses.

Then:

1/f = 1/s + 1/s'

1/s = 1/f - 1/s'

1/s = (s' -f)/fs'

s = fs'/(s'-f)

s =( 44.4× -83)/(-83 - 44.4)

s = - 3685.2 / - 127.4

s = 28.93 cm

Thus , the near distance point measured from the eye wearing the old glasses, if they rest 2.0 cm in front of the eye = (28.93 +2.0)cm

= 30.93 cm

Answer:

Explanation:

Charge on an electron (q) = 1.6 * 10 ^ -19 C

Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec

We know that, Force exerted on moving particle moving through a magnetic field :

[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]

1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B

B =  0.08573 T

Answer:

f. 80 and 90

Explanation:

1 x 10⁻¹² W/m² sound intensity falls within 0 sound level

1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level

1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level

1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level

1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level

1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level

1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level

1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level

1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level

1 x 10⁻³ W/m² sound intensity falls within 90 sound level

Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.

f. 80 and 90

Answer:In the case of two proton beams the protons repel one another because they have the same sign of electrical charge. There is also an attractive magnetic force between the protons, but in the proton frame of reference this force must be zero! Clearly then the attractive magnetic force that reduces the net force between protons in the two beams as seen in our frame of reference is relativistic. In particular the apparent magnetic forces or fields are relativistic modifications of the electrical forces or fields. As such modifications, they cannot be stronger than the electrical forces and fields that produce them. This follows from the fact that switching frames of reference can reduce forces, but it can’t turn what is attractive in one frame into a repulsive force in another frame.

In the case of wires the net charges in two wires are zero everywhere along the wires. That makes the net electrical forces between the wires very nearly zero. Yet the relativistic magnetic forces and fields will be of the same sort as in the case of two beams of charges of a single sign. This is true even in the frame of reference of what we think as the moving charges, that is, the electrons. In the frame of reference moving at the drift velocity of these current-carrying electrons, it is the protons or positively charged ions that are moving in the other direction. Consequently in any frame of reference for current-carrying wires in parallel, the net electrical force will be essentially zero, and there will be a net attractive magnetic force

Explanation:                                                                              

Explanation:

Particles with similar charges (both positive or both negative) will always repel each other, regardless of their speed or direction.

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:

1.125×10⁻⁹ J

Explanation:

Applying,

E = 1/2CV²................... Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.

Given; C = 1.0 nF,  = 1.0×10⁻⁹ F, V = 1.5 V

Substitute into equation 1

E = 1/2(1.0×10⁻⁹×1.5²)

E = 1.125×10⁻⁹ J

Hence the energy stored by the capacitor is 1.125×10⁻⁹ J

Answer:

The correct answer is 500 N

Explanation:

This is an exercise in Newton's third law or law of action and reaction

The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.

Which force we call action does not matter, the analysis and conclusions are the same

The correct answer is 500N

Answer:

The charge is  [tex]q = 1.50 *10^{-5} \ C[/tex]

Explanation:

From the question we are told that

   The electric field strength is  [tex]E = 1860 \ N/C[/tex]

    The force is  [tex]F = 0.02796 \ N[/tex]

Generally the charge on this particle is mathematically represented as

     [tex]q = \frac{F}{E}[/tex]

=>   [tex]q = \frac{0.02796}{ 1860}[/tex]

=>   [tex]q = 1.50 *10^{-5} \ C[/tex]

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

Answer:

Vrel= 0.75c

Explanation:

See attached file

Answer:

Explanation:

Let the race be of a fixed distance x

[tex]Average Speed = \frac{Total Distance}{Total Time}[/tex]

Troy's Average speed = 3 miles/hr = x / 0.2 hr

x = 0.6 miles

Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr

Answer:

this raise the temperature is x = 3

Explanation:

Heat capacity is the relationship between heat and temperature change

          C = Q / ΔT

if the heat in the system is given by the change in energy and we carry this differential formulas

          [tex]c_{v}[/tex] = dE / dT

In this problem we are told that the energy of thermal radiation is

        E = A V T⁴

Let's look for the specific heat

        c_{v} = AV 4 T³

the power to which this raise the temperature is x = 3

Answer:

16 times.

Explanation:

The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr

Time consumed, t = 40 hr

The magnitude of Q.F for the neutrons, Q.F = 2

Thus the effective radiation dose is:

[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]

Thus, the effective dose allowable yearly = 16 times

Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

[tex]B = \frac{N\mu_o I}{2R}[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Answer:

Explanation:

Let B= bead

Q = rod

the electric field at the glass bead pocation is

(Gauss theorem)

E = Q / (2 π d L εo)

the force is

F = q E = q Q / (2 π d L εo)

then

Q = 2 π d L εo F / q

Q = 2*3.14*4x10^-2*10^-1*8.85x10^-12*910x10^-4 / 5x10^-9 = 2.87x10^-8 C = 40.5 nC

Answer:

Explanation:

1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;

M ∝ V

M = kV

For every proportionality sign, there will always be a proportionality constant 'k'

Since the proportionality constant is the density (D) of the metal, the equation will become;

M = DV

Given the density to be 11.3 g/cm3, the equation will become;

M = 11.3V

Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V

2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;

M = 11.3V

M = 11.3 * 17.3

M = 195.49 grams

Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.

Answer:

A:  magnitude and direction

B: Force that the field exerts on a test charge

C: its magnitude and direction is the same.

D: electrostatic machine

two rollers that are driven by a motor that generates a rotation

Explanation:

Answer:

The moment of inertia is  [tex]I= 312.09 \ kg \cdot m^2[/tex]

Explanation:

From the question we are told that

    The  mass of the platform is   m =  137 kg

     The radius is  r  =  1.53 m

    The mass of the person is  [tex]m_p = 68.7 \ kg[/tex]

    The distance of the person from the center is  [tex]d_c =1.19 \ m[/tex]

    The mass of the dog is  [tex]m_d = 25.9 \ kg[/tex]

     The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]

Generally the moment of inertia of the system is mathematically represented as

      [tex]I = I_1 + I_2 + I_3[/tex]

Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as

          [tex]I_1 = \frac{m * r^2}{2}[/tex]

substituting values

           [tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]

           [tex]I_1 = 160.35 \ kg\cdot m^2[/tex]

Also  [tex]I_2[/tex]  is the moment of inertia of the person about the axis which is mathematically represented as

          [tex]I_2 = m_p * d_c^2[/tex]

substituting values

          [tex]I_2 = 68.7 * 1.19^2[/tex]

          [tex]I_2 = 97.29 \ kg \cdot m^2[/tex]

Also  [tex]I_3[/tex]  is the moment of inertia of the dog about the axis which is mathematically represented as

          [tex]I_3 = m_d * d_d^2[/tex]

substituting values

          [tex]I_3 = 25.9 * 1.45^2[/tex]

          [tex]I_3 = 54.45 \ kg \cdot m^2[/tex]

Thus  

        [tex]I= 160.35 + 97.29 + 54.45[/tex]

        [tex]I= 312.09 \ kg \cdot m^2[/tex]

Answer:

(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV

(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV

Explanation:

Given;

radius of the circular loop, r = 31.0 cm = 0.31 m

initial magnetic field, B₁ = 0.7 T

final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T

duration of change in the field, t = 29

(a) The magnitude of induced emf in the loop while the magnetic field is increasing.

[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]

[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]

Where;

A is the area of the circular loop

A = πr²

A = π(0.31)² = 0.302 m²

[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]

(b) the magnitude of the induced voltage at a constant magnetic field

E = A x B/t

E = (0.302 x 1.61) / 3.9

E = 0.1247 V

E = 124.7 mV

Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV