Unpolarized light passes through two ideal Polaroid sheets. The axis of the first is vertical, and the axis of the second is at 30.0∘ to the vertical. What fraction of the incident light is transmitted?

A source emits monochromatic light of wavelength 549 nm in air. when the light passes through a liquid, its wavelength reduces to 433 nm, the liquid's index of refraction is approximately 1.269.

The index of refraction (n) of a medium can be calculated using the formula

n = λair / λmedium

Where λair is the wavelength of light in air and λmedium is the wavelength of light in the medium.

Given:

λair = 549 nm

λmedium = 433 nm

Substituting these values into the formula, we get

n = 549 nm / 433 nm

Simplifying the calculation

n = 1.269

Therefore, the liquid's index of refraction is approximately 1.269.

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Answer:

3 m/s²

Explanation:

Initial Velocity, u = 18

Final velocity, v = 36

Time, t = 6 seconds

Acceleration is the change in velocity of a body with time. It obtained using the relation :

Acceleration = (v - u) / t

Acceleration = (36 - 18) / 6

Acceleration = 18 / 6

Acceleration = 3m/s²

Hence, acceleration of the skier is 3m/s²

Answer:

11 cm and concave

Explanation:

edge 2021

Answer:

it is -21, The top part is correct it is only the second part this is the first part.

Explanation:

plants

i hope its right and have a nice day

Answer:

number 1

Explanation:

The change in potential energy of the spring system is 15.6 x 10³J.

Mass of the disc, m = 20 kg

Velocity of the disc, v = 3 m/s

Spring constant of the spring, k = 200 N/m

Angular displacement of the disc, x = 2 revolutions = 2 x 2π = 4π radians

The potential energy that is stored when an elastic object is stretched or compressed by an external force, such as the stretching of a spring, is known as elastic potential energy. It is equivalent to the effort required to extend the spring, which is dependent on both the length of the stretch and the spring constant k.

The expression for the elastic potential energy of the spring is given by,

PE = 1/2 kx²

PE = 1/2 x 200 x (4π)²

PE = 1.57 x 10⁴J

The kinetic energy of the disc is given by,

KE = 1/2 mv²

KE = 1/2 x 20 x 3²

KE = 90 J

Therefore, the change in potential energy is,

E = 1.57 x 10⁴- 90

E = 15.6 x 10³J

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Answer:

The moon doesn’t have light of its own, the moon lights up because of the sun. So at night, as the light of the sun doesn't reach the grass directly because of the moon, it doesnt reflect any color off the grass, and so our eyes detect grass as black.

Explanation:

Sun appears white, but it is made up of the colors: red, orange, yellow, green, blue, indigo, and violet. When white light hits an object, it absorbs some colors and reflect the others. Grass appears green because it absorbs all the wavelengths except green. Green is reflected off the grass, so we see grass as green.

The groups aren't well formatted ;

The groups are ;

(-3, -1, 0, 2, 7) ; (9, 7, 6, -5, -4) ; (8, -6, 5, -4, 1) ; (-3, -4, -7, -8, -9)

Answer:

(-3, -4, -7, -8, -9)

Explanation:

Given the following group of numbers :

Evaluating each group of values for which is correctly arranged from greatest to least.

(-3, -1, 0, 2, 7) : - 1 is greater than - 3 (the group isn't arranged from greatest to least)

(9, 7, 6, -5, -4) : - 4 is greater than - 5 ; hence, the group isn't arranged from greatest to least

(8, -6, 5, -4, 1) : 5 is greater than - 6 ; hence, the group isn't arranged from greatest to least

(-3, -4, -7, -8, -9) ; the group of numbers here is arranged from greatest to least ;

(-3 > -4 > -7 > -8 > -9) ; hence, the correct group

The line, 100 mm in length, is parallel to and 40 mm above the h.p. Its ends, positioned 25 mm and 50 mm in front of the v.p., are connected to form projections. The inclination with the v.p. is 48.59°.

To draw the projections of the line, we start by drawing the plan view (H₁) and the front view (V₁).

In the plan view (H₁), we draw a horizontal line of 100 mm length. The line is parallel to the horizontal plane (h.p.), so it remains at the same height as the h.p.

In the front view (V₁), we draw a vertical line to represent the height above the h.p. Since the line is 40 mm above the h.p., we draw a line 40 mm long above the ground line. Then, we draw the line segments representing the ends of the line.

The first end is 25 mm in front of the vertical plane (v.p.), and the second end is 50 mm in front of the v.p. We connect these ends to the top of the vertical line to complete the front view.

To find the inclination of the line with the v.p., we use the right triangle formed by the height of the line (40 mm) and the distance from the v.p. to the second end of the line (50 mm). We can calculate the inclination angle using the tangent function:

tan(θ) = opposite/adjacent = 40/50

Solving for θ, we find:

θ = tan^(-1)(40/50) = 48.59°

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Another bumblebee could detect the presence of the charged bumblebee from a distance of approximately 3.5 meters.

To determine the distance at which another bumblebee could detect the presence of the charged bumblebee, we can use Coulomb's law, which states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for the electric force between two charges is given by:

F = (k * q1 * q2) / r^2

where F is the electric force, k is the electrostatic constant (approximately 9 × 10^9 N·m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.

Given that the electric field detected by the bumblebee is 60 N/C, we can relate the electric field to the electric force using the equation:

E = F / q

where E is the electric field and q is the charge.

Rewriting the equation to solve for the electric force:

F = E * q

Substituting the given values:

F = (60 N/C) * (21 × 10^-12 C)

Simplifying:

F = 1.26 × 10^-9 N

Rearranging the Coulomb's law equation to solve for the distance:

r = sqrt((k * q1 * q2) / F)

Substituting the values into the equation:

r = sqrt((9 × 10^9 N·m^2/C^2 * (21 × 10^-12 C)^2) / (1.26 × 10^-9 N))

Simplifying:

r ≈ 3.5 meters

Therefore, another bumblebee could detect the presence of the charged bumblebee from a distance of approximately 3.5 meters.

Another bumblebee could detect the presence of the charged bumblebee from a distance of approximately 3.5 meters. This is based on the ability of bumblebees to sense electric fields and the known electric field strength and charge of the bumblebee in question.

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Answer:

0.0217 m

Explanation:

Ftom the question,

Applying Coulomb's law

F = kqq'/r²............... Equation 1

Where F = Electric Force, q and q' = First and second charge respectively, r = distance between the charges, k = coulomb's constant.

make r the subeject of the equation

r = √[(kqq')/F]............. Equation 2

Given: q = +3.6×10⁻⁸ C, q' = -3.2×10⁻⁸ C, F = -2.2×10⁻² N

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation

r = √(+3.6×10⁻⁸×-3.2×10⁻⁸×8.98×10⁹/-2.2×10⁻²)

r = √(-103.4489×10⁻⁷/-2.2×10⁻²)

r = √(47.02×10⁻⁵)

r = 21.68×10⁻³

r ≈ 21.7×10⁻³ m.

r ≈ 0.0217 m

Hence the seperation between the two balloons is 0.0217 m

Answer:

Nepal is one of the least developed countries. Solar power is well-founded than electricity. It's better to use solar power because it's a clean resource, the sun provides more energy than we will ever need.

Explanation:

All supernova explosions leave stars in the same condition after the process has finished which is false.

Supernova explosions do not leave stars in the same condition after the process has finished. Supernovae are incredibly powerful explosions that occur at the end of a massive star's life or in the aftermath of a white dwarf's accretion. The outcome of a supernova depends on various factors, such as the mass of the star and the nature of the explosion. There are two main types of supernovae: Type I and Type II.

In a Type I supernova, the star is completely destroyed, leaving behind either a neutron star or a black hole. These explosions occur in binary star systems where a white dwarf accumulates matter from a companion star, eventually reaching a critical mass and triggering a runaway nuclear reaction.

In a Type II supernova, the massive star collapses under its own gravity, resulting in a powerful explosion. After the explosion, what remains can vary. It could leave behind a neutron star or even a black hole. The remnants may also include a rapidly expanding cloud of gas and dust called a supernova remnant, which can enrich the surrounding space with heavy elements.

Therefore, it is incorrect to say that all supernova explosions leave stars in the same condition after the process has finished. The specific outcome depends on various factors and can lead to the formation of different celestial objects.

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The work done on the hoop to stop it is 0.344 Joules.

What is work done?

Work is defined as the transfer of energy that occurs when a force is applied to an object, causing it to move in the direction of the force. Work is the product of the force applied to an object and the displacement of the object in the direction of the force.

Given:

Mass of the hoop (M) = 100 kg

Speed of the center of mass (v) = 0.0830 m/s

Radius of the hoop (R) = ?

To find the radius (R) of the hoop, we can use the relationship between linear and angular velocity:

v = R * ω

Rearranging the equation, we get:

R = v / ω

We need to find the angular velocity (ω) of the hoop. Using the relationship between linear velocity and angular velocity, we have:

ω = v / R

Substituting the given values, we can calculate the angular velocity:

ω = 0.0830 m/s / R

Next, we can calculate the moment of inertia (I) of the hoop:

I = MR^2

Substituting the mass and radius, we get:

I = 100 kg * R^2

Now, we can calculate the initial kinetic energy (KE_initial) of the hoop:

KE_initial = (1/2) * I * ω^2

Substituting the values, we have:

KE_initial = (1/2) * (100 kg * R^2) * (0.0830 m/s / R)^2

Simplifying the equation, we get:

KE_initial = (1/2) * 100 kg * 0.0830^2 m^2/s^2

Finally, the work done on the hoop to stop it is equal to the initial kinetic energy:

Work = KE_initial

Substituting the values, we have:

Work = (1/2) * 100 kg * 0.0830^2 m^2/s^2

Calculating the value, we find:

Work ≈ 0.344 J

Therefore, the work done on the hoop to stop it is 0.344 Joules.

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The additional volume of ethanol that can be put into the tank is 0.0136 m³.

The given informations are,

Part A

When the tank and its contents have cooled to 20.0 °C, the volume of ethanol decreases due to the decrease in temperature.

Let's assume the volume of ethanol at 33.0 °C be V1 and at 20.0 °C be V2 and coefficient of cubical expansion of ethanol be α.

From the temperature coefficients of cubical expansion, we can say that the volume of ethanol decreases with decrease in temperature.

So, the additional volume of ethanol which can be put into the tank is,

Additional volume = V1 - V2

The volume of ethanol changes due to the change in temperature,

V2 = V1 / [1 + α (T2 - T1)]

where T1 and T2 are the initial and final temperatures of the ethanol in degree Celsius.

The coefficient of cubical expansion of ethanol, α = 1.12 × 10^-3 / °C.

Now, let's substitute the given values in the above equation:

V2 = 1.60 m³ / [1 + (1.12 × 10^-3 / °C) × (33.0 °C - 20.0 °C)]

V2 = 1.5864 m³

Therefore, the additional volume of ethanol that can be put into the tank when the tank and its contents have cooled to 20.0 °C is,

Additional volume = V1 - V2= 1.60 m³ - 1.5864 m³

                              = 0.0136 m³

Hence, the additional volume of ethanol that can be put into the tank is 0.0136 m³.

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Answer:  TRUE /   IT IS TRUE

Explanation:

Answer:

[tex]\frac{60kgm}{s}[/tex]

Explanation:

momentum = mass * velocity

= 20kg * 3m/s

= 60kgm/s

Force is push or pull....

thank you

(a) The thickness of the magnesium fluoride film should be approximately 120 nm to suppress the reflection of 565 nm light.

(b) To suppress the reflection of light with a higher frequency, the coating of magnesium fluoride should be made thinner.

(a) The condition for suppressing reflection is given by the equation:

2nt = mλ

where n is the refractive index of the film, t is the thickness of the film, m is an integer representing the order of the interference, and λ is the wavelength of light.

For reflection suppression of 565 nm light, we can substitute the given values:

2(1.38)t = λ

2(1.38)t = 565 nm

t = (565 nm) / (2(1.38))

t ≈ 120 nm

Therefore, the thickness of the magnesium fluoride film should be approximately 120 nm to suppress the reflection of 565 nm light.

(b) To suppress the reflection of light with a higher frequency, the coating of magnesium fluoride should be made thinner. This is because higher frequencies correspond to shorter wavelengths. As the wavelength decreases, the required thickness of the film for interference suppression decreases. So, a thinner coating of magnesium fluoride would be needed to achieve reflection suppression for higher-frequency light.

(a) To suppress the reflection of 565 nm light, the magnesium fluoride film should have a thickness of approximately 120 nm.

(b) To suppress the reflection of light with a higher frequency, the coating of magnesium fluoride should be made thinner. This is because higher frequencies correspond to shorter wavelengths, requiring a smaller film thickness for interference suppression.

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To find the electric potential at the center of one of the rings, we can use the formula for the electric potential due to a charged ring: V = k * Q / r

where V is the electric potential, k is the Coulomb's constant (8.99 × 10^9 N m²/C²), Q is the charge of the ring, and r is the distance from the center of the ring. In this case, the charge of each ring is +3.0 nC (nanocoulombs) = 3.0 × 10^-9 C, and the distance from the center of the ring is half of the diameter, which is 5.0 cm = 0.05 m. Plugging in the values, we have: V = (8.99 × 10^9 N m²/C²) * (3.0 × 10^-9 C) / 0.05 m. Calculating this expression will give us the electric potential at the center of one of the rings.

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The power series representation for the function f(x) = ln(7 - x) is f(x) = ln(7) - ∑(n=0 to ∞) [(x - 7)^n / (n+1)].

To find the power series representation, we can use the Taylor series expansion of the natural logarithm function ln(1 + x):

ln(1 + x) = x - (x^2 / 2) + (x^3 / 3) - (x^4 / 4) + ...

In this case, we have the function f(x) = ln(7 - x), which can be rewritten as f(x) = ln(1 + (x - 7)).

Using the Taylor series expansion, we substitute (x - 7) in place of x:

f(x) = (x - 7) - [(x - 7)^2 / 2] + [(x - 7)^3 / 3] - [(x - 7)^4 / 4] + ...

Simplifying, we can write this as:

f(x) = -∑(n=0 to ∞) [(x - 7)^n / (n+1)]

Next, we add ln(7) to the series to account for the constant term:

f(x) = ln(7) - ∑(n=0 to ∞) [(x - 7)^n / (n+1)]

The radius of convergence, R, can be determined by using the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms in the series is L, then the series converges absolutely when L < 1 and diverges when L > 1.

In this case, we take the absolute value of the terms in the series and calculate the limit:

lim(n→∞) |(x - 7)^(n+1) / [(n+2)(x - 7)^n]|

Simplifying and taking the limit, we find:

lim(n→∞) |x - 7| / (n + 2)

Since this limit approaches zero for any value of x, the series converges for all values of x. Therefore, the radius of convergence, R, is infinity.

The power series representation for the function f(x) = ln(7 - x) is f(x) = ln(7) - ∑(n=0 to ∞) [(x - 7)^n / (n+1)]. The series converges for all values of x, indicating that the radius of convergence, R, is infinity.

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Explanation:

inversely proportional

Answer:

directly proportional to force applied

Answer:

Energy moves between the particle of the medium.

Explanation:

The complete reaction of 1.21 mol of carbon (C) produces 2.42 mol of hydrogen gas (H2), which is equivalent to approximately 54.5 liters.

The balanced chemical equation for the reaction between carbon and hydrogen gas is:

[tex]C + 2H_2 -- > CH_4[/tex]

From the balanced equation, we can see that for every 1 mole of carbon (C), 2 moles of hydrogen gas (H2) are required to form 1 mole of methane (CH4). Since the reaction is complete, all the carbon will be consumed, and therefore the moles of hydrogen gas produced will be twice the moles of carbon.

Given that we have 1.21 mol of carbon, we can calculate the moles of hydrogen gas produced by multiplying the moles of carbon by 2:

2.42 mol H2 = 1.21 mol C × 2 mol H2 / 1 mol C

To convert the moles of hydrogen gas to liters, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation to solve for V, we have:

V = (nRT) / P

Substituting the given values into the equation and solving for V, we get:

V = (2.42 mol H2 × 0.0821 L·atm/mol·K × 319 K) / 1.0 atm

V ≈ 54.5 liters

Therefore, approximately 54.5 liters of hydrogen gas are formed from the complete reaction of 1.21 mol of carbon.

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Answer:

Efficiency of the inclined plane is 56%

Explanation:

efficiency = (work output / work input) x 100%

efficiency = ([load force x load distance] x [effort force x effort distance]) x 100%

efficiency = (250 kg x 10 m) / (180 kg x 25 m) x 100%

efficiency = (2500kg/m) / (4500kg/m) x 100%

efficiency = 0.555 = 0.56 x 100%

efficiency = 56%

Answer:

32

Explanation:

Answer:

32

Explanation:

Sulfur has 16 protons and 16 neutrons. The atomic number is roughly 32. Therefore, 16 + 16 = 32 nucleons.  

a. The potential at 5.50 cm from the center of the sphere is 9.00 x [tex]10^7[/tex] V.

b. The potential at 30.0 cm from the center of the sphere is 1.65 x [tex]10^8[/tex] V.

c. The potential at 16.0 cm from the center of the sphere is 2.47 x [tex]10^8[/tex] V.

To find the value of the potential at different distances from the center of the sphere, we can use the equation for the electric potential of a uniformly charged sphere.

Given:

Total electric charge (Q) = 5.50 nC

Radius of the sphere (R) = 30.0 cm = 0.30 m

a) To find the potential at 5.50 cm from the center of the sphere:

Distance from the center of the sphere (r) = 5.50 cm = 0.055 m

The equation for the electric potential of a uniformly charged sphere is:

V = k × Q / r

where V is the potential, k is the Coulomb's constant (8.99 x [tex]10^9[/tex] N m²/C²), Q is the total charge, and r is the distance from the center of the sphere.

Substituting the given values into the equation:

V = (8.99 x [tex]10^9[/tex] N m²/C²) × (5.50 x [tex]10^{-9[/tex] C) / 0.055 m

Calculating the value:

V = 9.00 x [tex]10^7[/tex] V

Therefore, the potential at 5.50 cm from the center of the sphere is 9.00 x [tex]10^7[/tex] V.

b) To find the potential at 30.0 cm from the center of the sphere:

Distance from the center of the sphere (r) = 30.0 cm = 0.30 m

Using the same equation as above:

V = (8.99 x [tex]10^9[/tex] N m²/C²) × (5.50 x [tex]10^{-9[/tex] C) / 0.30 m

Calculating the value:

V = 1.65 x [tex]10^8[/tex] V

Therefore, the potential at 30.0 cm from the center of the sphere is 1.65 x [tex]10^8[/tex] V.

c) To find the potential at 16.0 cm from the center of the sphere:

Distance from the center of the sphere (r) = 16.0 cm = 0.16 m

Using the same equation as above:

V = (8.99 x [tex]10^9[/tex] N m²/C²) × (5.50 x [tex]10^{-9[/tex] C) / 0.16 m

Calculating the value:

V = 2.47 x [tex]10^8[/tex] V

Therefore, the potential at 16.0 cm from the center of the sphere is 2.47 x [tex]10^8[/tex] V.

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Answer:

[tex]F_{elect} = \frac{kq_1q_2}{d^2}[/tex]

Explanation:

Consider the given variables:

Felect = Electrostatic Force between charged particles

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of second charge

d = distance between the charges

The relationship among these variables is given by the Coulomb's Law:

[tex]F_{elect} = \frac{kq_1q_2}{d^2}[/tex]

This is the relationship that contains k, q₁, q₂, d on the right-hand side and Felect on the left-hand side.

The height of the cannon where the objects are launched from is 18 feet.

The canons of page construction are historical reconstructions, based on careful measurement of extant books and what is known of the mathematics and engineering methods of the time, of manuscript-framework methods that may have been used in Medieval- or Renaissance-era book design to divide a page into pleasing proportions. Since their popularization in the 20th century, these canons have influenced modern-day book design in the ways that page proportions, margins and type areas (print spaces) of books are constructed.

To determine the height of the cannon where the objects are launched from, we need to find the value of "h" when "t" is equal to zero.

Given the equation: h = (-1/5)×t^2 + 5×(t) + 18

Substituting t = 0 into the equation, we have:

h = (-1/5)×(0)^2 + 5(0) + 18

= 0 + 0 + 18

= 18

Therefore, the height of the cannon where the objects are launched from is 18 feet.

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