A soap bubble is 105 nm thick and illuminated by white light incident perpendicular to its surface. What is the longest wavelength in nm which is constructively reflected, assuming soap has the same index of refraction as water (n = 1.33)? Round your answer in nanometers to the nearest whole nanometer.

Since the rod with a radius of 2r and length l has a shorter length than the rod with a radius of 4r and length 5l, the change in length (Δl) for the first rod will be smaller than the change in length (ΔL) for the second rod.

If the same force F pulls on two cylindrical rods of the same material, we can determine which rod has a greater change in length by considering their dimensions.

Let's compare a rod with a radius of 2r and length l to a rod with a radius of 4r and length 5l.

To analyze this, we need to understand the concept of stress and strain.

Stress is the force acting on a material per unit area, while strain is the change in length per unit length of the material.

The relationship between stress and strain is given by Hooke's Law.

Now, when a force F is applied to the rods, the stress experienced by both rods will be the same, as the force is constant.

However, the strain experienced by the rods will differ due to their dimensions.

To calculate the strain, we use the formula: strain = change in length/original length.

For the rod with radius 2r and length l, the original length is l and the change in length is Δl.

For the rod with radius 4r and length 5l, the original length is 5l and the change in length is ΔL.

Using Hooke's Law, stress is equal to force divided by cross-sectional area.

The cross-sectional area of the rod with radius 2r is π(2r)^2, and the cross-sectional area of the rod with radius 4r is π(4r)^2.

Since stress is constant for both rods, we have:

(F/π(2r)^2) = (F/π(4r)^2)

We can simplify this equation to:

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

4r^2 = 16r^2

r^2 = 4r^2

r^2 - 4r^2 = 0

-3r^2 = 0

From this, we can conclude that the radius of the rods does not affect the strain experienced. Therefore, the change in length is solely determined by the length of the rod.

In conclusion, the rod of radius 4r and length 5l will have a greater change in length when the same force is applied compared to the rod of radius 2r and length l.

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The speed of the coin when dropped from the top of a building, using free fall formula after 2 and 3 seconds are, 19.6 m/s and 29.4 m/s.

The speed of an object in free fall can be determined by multiplying the acceleration due to gravity (which is approximately 9.8 m/s²) by the time elapsed. In this case, after 2 seconds, the coin will have fallen a distance of 19.6 meters and will be traveling at 19.6 m/s. Similarly, after 3 seconds, the coin will have fallen a distance of 44.1 meters and will be traveling at 29.4 m/s.

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Therefore, the voltage of the cell (silver-silver chloride electrode || saturated calomel electrode) is +0.044 V.

(a) The half-reaction for the silver-silver chloride electrode is:

AgCl(s) + e⁻ → Ag(s)

(b) The half-reaction for the saturated calomel electrode is:

Hg₂Cl₂(s) + 2e⁻ → 2Hg(l) + 2Cl⁻(aq)

(c) To determine the voltage of the cell, we can subtract the potential of the anode (silver-silver chloride electrode) from the potential of the cathode (saturated calomel electrode):

Ecell = Ecathode - Eanode

Given that the potential for the Ag|AgCl electrode in a saturated KCl solution is +0.197 V (Eanode = +0.197 V) and the potential for a calomel electrode is +0.241 V (Ecathode = +0.241 V), we can calculate the voltage of the cell:

Ecell = +0.241 V - (+0.197 V)

Ecell = +0.044 V

Therefore, the voltage of the cell (silver-silver chloride electrode || saturated calomel electrode) is +0.044 V.

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Answer:

Agriculturally suitable lands have adequate precipitation and moderate temperatures as well as good soils. Farmers regularly have to contend with wet and dry events to grow crops, even in hospitable climates.

Explanation:

Rita's momentum  is 294 kg.m/s.

The product of a particle's mass and velocity is called momentum. Being a vector quantity, momentum possesses both magnitude and direction. According to Isaac Newton's second equation of motion, the force applied on a particle is equal to the time rate of change of momentum.

Given that Rita is riding a 7.0 kilogram bike at a velocity of 15 meters/second.

Mass of Rita = 42 kg.

Hence, her momentum is = mass × velocity

= 7.0 × 42 kg.m/s

= 294 kg.m/s.

Her momentum  is 294 kg.m/s.

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Answer: C water

Explanation: The explanati is not hard cause the main elements of earth is fire ice water earth

Answer:

A. Gold (Au)

Explanation:

Based on the given informations, the orbital period of the outer edges of the primordial disk was calculated to be approximately 1515 years.

Assuming that the mass of Neptune is 17 times that of the Earth and that the distance of Neptune from the Sun is about 30 astronomical units (AU), we can use Kepler's third law of planetary motion to solve for the period of the outer edges of the primordial disk.

Using the equation P² = (4π²/GM) x a³, where P is the period, G is the gravitational constant, M is the mass of the Sun, and a is the semi-major axis of the orbit, we can rearrange the equation to solve for P:

P = sqrt((4π²/GM) x a³)

Since the matter that made up the planetesimals was spread out evenly on the edges of the primordial disk, we can assume that the semi-major axis of their orbit was about 35.5 AU (the radius of the disk).

Plugging in the values, we get:

P = sqrt((4π²/6.6743 x 10⁻¹¹ x 1.9885 x 10³⁰) x (35.5 x 1.496 x 10¹¹)³)

P = 1515 years (approx.)

Therefore, the orbital period of the outer edges of the primordial disk was approximately 1515 years.

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Answer:

The pressure at the bottom of the river is less than that at the bottom of the lake.

Explanation:

From Bernoulli's equation, the pressure difference is given by

ΔP = ρgΔh + ρ(v₂² - v₁²)/2 where ρ = density of water, g = acceleration due to gravity, Δh = depth, v₁ = velocity at top, v₂ = velocity at bottom

For the lake, v₁ = v₂, since the velocity at the top and bottom are the same. So,

ΔP₁ = ρgΔh + ρ(v₁² - v₁²)/2 = ρgΔh + 0 = ρgΔh

P₂ - P₁ = ρgΔh

P₂ = P₁ + ρgΔh

For the river, v₁ < v₂, since the velocity at the top of the river is greater than at the bottom.

So,

ΔP₂ = ρgΔh + ρ(v₂² - v₁²)/2.

Since  v₁ < v₂, ρ(v₂² - v₁²)/2 will be negative,

So,

ΔP₂ = ρgΔh - ρ(v₂² - v₁²)/2.

Since ρ(v₂² - v₁²)/2 is negative, making ΔP less than that in the lake.

So, ΔP₂ = ΔP₁ -  ρ(v₂² - v₁²)/2.

ΔP₂ = P₃ - P₁

P₃ - P₁ = P₂ - P₁ -  ρ(v₂² - v₁²)/2.

P₃ = P₂ - ρ(v₂² - v₁²)/2.

where P₃ = pressure at bottom of the river and P₂ = pressure at bottom of the lake and P₁ = atmospheric pressure at top of river and lake respectively.

Since the factor ρ(v₂² - v₁²)/2 is removed from the pressure at the bottom of the lake, the pressure at the bottom of the river is therefore less than that at the bottom of the lake.

This question involves the concepts of work done and elastic potential energy.

The work done by the spring when its extension changes from X to X/4 is "0.56 E".

The work done by the spring is equal to the elastic potential energy stored in the spring, which is given as follows:

\(W=E\\W=E=\frac{1}{2}kX^2---- eqn(1)\)

where,

W = work done = ?

E = elastic potential energy

k = spring constant

X = extension

Now, the spring moves to an extension of X/4, so the change in extension will be:

\(\Delta X = X-\frac{X}{4}\\\\\Delta X = \frac{3X}{4}\)

Hence, the work done will become:

\(W=\frac{1}{2}K\Delta X^2\\\\W=\frac{1}{2}K(\frac{3X}{4})^2\\\\W=\frac{1}{2}KX^2(0.56)\\\\using\ eqn(1):\)

W = 0.56 E

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At this speed, the velocity of satellite takes 1.5 hours to orbit the Earth's 24,000-mile perimeter. The astronaut does not point if he wants to land.

G is the gravitational constant, M is the mass of the attracting mass, and r is the distance from the center of that mass. This equation for the escape velocity, denoted as vesc, is written as vesc = Square root of 2GM/r.

The mass of the planet and the radius of the spacecraft's orbit path are the two factors that nature uses to calculate the speed of an orbiting spacecraft.

calculation:

velocity of satellite = 1.5 hrs

G = gravitational constant

distance = 24000 mile

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Explanation:

hope it helps

Answer:

Both have the same horizontal velocity.

Ball A will take longer to reach the ground.

Ball A will travel farther.

Both have the same horizontal velocity like Ball A will take longer to reach the ground and Ball A will travel farther.

velocity can be defined as the rate of change of the object’s position with respect to  reference and which is complicated but velocity is basically speeding in a specific direction.

Velocity is nothing but a vector quantity means both magnitude (speed) and direction are needed to define velocity. The SI unit of velocity is meter per second (ms-1) and if  magnitude or the direction of velocity of a body changes, then it will be said to be accelerating.

Speed and velocity are the two closest term and can be a little confusing but the major difference between speed and velocity is that speed gives us an idea of the rate of faster movement of an object where as  velocity  speed up as well as   tells us the direction the body

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(a) The change in internal energy of the gas is -6250 J.

(b) The change in temperature of the gas is -568.18 K.

(a) To find the change in internal energy, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, the chemical reaction transfers 6250 J of thermal energy into the gas, so the heat added to the system is 6250 J.

Since the system expands, no work is done on the surroundings, so the work done by the system is 0 J. Therefore, the change in internal energy is equal to the heat added, which is -6250 J.

(b) To calculate the change in temperature, we can use the ideal gas law, which states that the pressure times the volume of a gas is equal to the number of moles of the gas times the gas constant times the temperature. We can rearrange this equation to solve for the change in temperature. Given that the pressure is constant, we have:

P1 * V1 = n * R * T1

P2 * V2 = n * R * T2

Dividing the second equation by the first equation, we get:

(P2 * V2) / (P1 * V1) = T2 / T1

Plugging in the given values, we have:

(0.800 atm * 2.00 L) / (1.25 x 10⁵ Pa * 9.00 L) = T2 / T1

Solving for T2, we find:

T2 = (0.800 atm * 2.00 L * T1) / (1.25 x 10⁵ Pa * 9.00 L)

Substituting the given value of T1, we can calculate T2, which is approximately -568.18 K.

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A: Ranking of wire-mass systems based on fundamental wavelength:

2-unit wire with 2 units of mass, 3-unit wire with 1 unit of mass, 1-unit wire with 4 units of mass, 3-unit wire with 4 units of mass, 2-unit wire with 1 unit of mass, 1-unit wire with 1 unit of mass.

The fundamental wavelength of a vibrating wire is inversely proportional to its length and directly proportional to the square root of its tension. Since all wires have the same tension, the ranking is based solely on length. The longer the wire, the longer the fundamental wavelength.

B: Ranking of wire-mass systems based on wave speed:

1-unit wire with 4 units of mass, 3-unit wire with 4 units of mass and 2-unit wire with 2 units of mass, 1-unit wire with 1 unit of mass and 3-unit wire with 1 unit of mass, 2-unit wire with 1 unit of mass.

The wave speed of a vibrating wire is directly proportional to the square root of its tension and inversely proportional to the square root of its mass per unit length. Since all wires have the same mass per unit length, the ranking is based solely on length and mass. The shorter the wire and the heavier the mass, the faster the wave speed.

C: Ranking of wire-mass systems based on fundamental frequency:

1-unit wire with 1 unit of mass, 2-unit wire with 1 unit of mass, 3-unit wire with 1 unit of mass, 1-unit wire with 4 units of mass, 2-unit wire with 2 units of mass, 3-unit wire with 4 units of mass.

The fundamental frequency of a vibrating wire is inversely proportional to its length and directly proportional to the square root of its tension and mass per unit length. Since all wires have the same tension, the ranking is based solely on length and mass. The shorter the wire and the heavier the mass, the higher the fundamental frequency.

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Answer:

A C E

Explanation:

got it right on edge 2021

Answer:It’s A C E

Explanation: I took the assignment and got it right :D

Answer:

Force = mass × acceleration

Acceleration:

\({ \tt{15 = (22 \times a)}} \\ { \tt{a = \frac{15}{22} \: {ms}^{ - 2} }}\)

From first Newton's equation of motion:

\({ \bf{v = u + at}} \)

Change = v - u:

\({ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \: {ms}^{ - 2} }}\)

Answer:

0° FROM or 180° TO

Explanation:

A course deviation indicator or the CDI may be defined as an avionics tool or instrument that is used in the aircraft navigation which is used to determine an aircraft's lateral position with relation to a course.

While using the VOT service, we should tune the VOT frequency of the VOR receiver. And with the CDI leveled centered, the omnibearing selector or the OBS should read as 0° with the TO/FROM indicator showing 'FROM'. Else the OBS should read as 180° with the TO/FROM indication showing 'TO.'

If I move 3 m east and 4 m north, then my displacement would be 5m.

In geometry and mechanics, displacement is defined as a vector whose length is the shortest distance from initial to the final position of a point P undergoing the motion.

If an object moves relative to the reference frame—for example, if professor moves to the right relative to whiteboard or passenger moves toward the rear of airplane—then object's position changes. And this change in position is known as displacement.

Given AB= 3m east and BC= 4m north

AC= √ 3² + 4²= 25

So, AC, displacement = 5m

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Answer:

C=A>B>E>D

Explanation:

The magnitude of the tension in each radial strand is approximately 4.91×\(10^{-3}\) N.

To find the magnitude of the tension, t, we can use the equation:

t = (m * \(v^2\)) / r. Where m is the mass of the spider, v is its velocity, and r is the radius of the circular path it is moving along. However, since we are given that the spider is stationary and hanging from radial strands, we can use a simpler formula:

t = m * g

Where g is the acceleration due to gravity, which is approximately 9.81 \(m/s^2\) on Earth.

Substituting the given mass of the spider, we get:

t = (5.0×\(10^{-4\) kg) * 9.81 \(m/s^2\)

t = 4.91×\(10^{-3}\) N

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Answer:

false

Explanation:

Because it make things curve not hold it near center

The tennis ball goes approximately 19.53 meters high on its 10th bounce.

Assuming that each bounce is perfect and the ball rebounds to ¾ of its previous height, we can use the formula H = (3/4)^n x 150, where n is the number of bounces and H is the height of the ball after the nth bounce.
To find the height of the ball after the 10th bounce, we plug in n = 10:
\(H = (3/4)^10 x 150H = 0.0563 x 150H ≈ 8.45 meters\)
Therefore, the ball reaches a height of approximately 8.45 meters on its 10th bounce.

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According to the data given in the question the net force on the box is of 5 N.

All of the forces that are applied to an object are added up to form the net force. As a consequence of the fact that it (force) is a vector and therefore that two forces with identical magnitudes and opposing directions cancel each other out, the resultant force is the total of the forces, or put another way, the net force is just the total of all the forces.

Given data :

Force on box to the left side (F1) = 15 N

Force on box to the right side (F2) = 20 N

Because both forces are in opposite direction

Hence,

Net force = F2 - F1

Net force = 20 - 15

Net force = 5 N.

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Answer:

Approximately \(7.2\; {\rm m\cdot s^{-1}}\) (rounded up), assuming that this circle is vertical and \(g = 9.81\; {\rm m\cdot s^{-2}}\).

Explanation:

Let \(v\) denote the tangential speed of the ball, and let \(r\) denote the radius of the circle. Since the ball is in a circular motion, the acceleration on this ball would be equal to the centripetal acceleration \(a = (v^{2} / r)\). The net force on this ball would be \(F_{\text{net}} = m\, a = (m\, v^{2} / r)\).

The net force on this ball is also the vector sum of the tension \(T\) in the rope and the weight of the ball \(m\, g\):

\(F_{\text{net}} = (\text{weight}) + T\).

\(T = F_{\text{net}} - (\text{weight})\).

Note that:

\(\| T \| = \|F_{\text{net}} - (\text{weight})\| \le \|F_{\text{net}} \| + \| (\text{weight})\|\).

In other words, the magnitude of tension \(T\) is at most equal to \(\|F_{\text{net}} \| + \| (\text{weight})\| = (m\, v^{2} / r) + (m\, g)\), which happens when weight and net force are in opposite directions.

When the speed of the ball is maximized, the magnitude of tension \(T\) would be at the largest possible value of \(130\; {\rm N}\). Rearrange the equation and solve for speed \(v\):

\(\displaystyle \frac{m\, v^{2}}{r} + m\, g = \|T\|\).

\(\begin{aligned}v^{2} = \frac{r}{m}\, (\|T \| - m\, g) = \frac{r\, \|T\|}{m} - r\, g\end{aligned}\).

\(\begin{aligned}v &= \sqrt{\frac{r\, \|T\|}{m} - r\, g} \\ &= \sqrt{\frac{(1.2)\, (130)}{2.5} - (1.2)\, (9.81)}\; {\rm m\cdot s^{-1}} \\ &\approx 7.2\; {\rm m\cdot s^{-1}}\end{aligned}\).

Answer:

1). Brake pedal or lever

2). A pushrod also called an actuating rod

Answer:

A

Explanation:

I got this answer because (by my estimation again) you are looking at 0.04 m/s which is the sound of a measure of a 2 Hz sound frequency. Therefore, for each section on the table diagrams I multiplied the period by their wavelength measurement until I got the value 0.04 m/s. In which A was the only option that gave me that value, providing me with a 2 Hz sound frequency.

Answer:

A

Explanation:

Idont know really i asked my grandpa bc he was a teacher and he said it was A if its wrong then im sry

Answer:

68.8 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of box = 18 Kg

Coefficient of friction (μ) = 0.39

Force of friction (F) =?

Next, we shall determine the normal force of the box. This is illustrated below:

Mass (m) of object = 18 Kg

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

Normal force (N) =?

N = mg

N = 18 × 9.8

N = 176.4 N

Finally, we shall determine the force of friction experienced by the object. This is illustrated below:

Coefficient of friction (μ) = 0.39

Normal force (N) = 176.4 N

Force of friction (F) =?

F = μN

F = 0.39 × 176.4

F = 68.796 ≈ 68.8 N

Thus, the box experience a frictional force of 68.8 N.

Answer:

Approximately \(0.37\; {\rm m}\) (assuming that \(g = (-9.81)\; {\rm m\cdot s^{-2}}}\).)

Explanation:

Let \(m\) denote the mass of the student. There are two forces on this student:

The net force on this student would be:

\(F_{\text{net}} = m\, g + F_{\text{normal}}\).

It is given that acceleration of the student is \(a = 3.3\; {\rm m\cdot s^{-2}}\). The net force on this student will also be equal to:

\(F_{\text{net}} = m\, a\).

Note that acceleration is positive since the student is accelerating upwards.

Thus:

\(m\, a = F_{\text{net}} = m\, g + F_{\text{normal}}\).

\(m\, a = m\, g + F_{\text{normal}}\).

Rearrange this equation to find \(F_{\text{normal}}\):

\(F_{\text{normal}} = m\, a - m\, g\).

In other words, the spring exerts a restoring force of \(F_{\text{normal}} = m\, a - m\, g\) on this student. To find the displacement \(x\) of the spring from equilibrium, divide the restoring force by the spring constant \(k\):

\(\begin{aligned}x &= \frac{F_{\text{normal}}}{k} \\ &= \frac{m\,a - m\, g}{k} \\ &= \frac{m\, (a - g)}{k}\end{aligned}\).

Note that the gravitational field strength \(g = (-9.81)\; {\rm m\cdot s^{-2}}\) is negative because the gravitational field near the Earth points downward- towards the center of the planet.

Substitute in mass \(m = 57\; {\rm kg}\), acceleration \(a = 3.3\; {\rm m\cdot s^{-2}}\), and spring constant \(k = 2000\; {\rm N\cdot m^{-1}}\):

\(\begin{aligned}x &= \frac{m\, (a - g)}{k} \\ &= \frac{(57) \, (3.3 - (-9.81))}{2000}\; {\rm m} \\ &\approx 0.37\; {\rm m}\end{aligned}\).

In other words, the compression of this spring would be approximately \(0.37\; {\rm m}\).