A highway speed limit is posted as 90 km/hr. Is this average speed or instantaneous speed? Explain your answer.

The article by M. Dafermos and I. Rodnianski provides a detailed analysis of the black hole stability problem for linear scalar perturbations.

The article written by M. Dafermos and I. Rodnianski in 2010 provides an in-depth analysis of the black hole stability problem for linear scalar perturbations. Numerous researchers over the years. The article offers a critical review of the existing literature on the topic and provides a new perspective on the issue.

The authors begin by discussing the evolution of linear scalar fields in the vicinity of a black hole. They show that the solutions to the wave equation can be expressed as a linear combination of ingoing and outgoing modes. The ingoing mode corresponds to the wave function falling into the black hole, while the outgoing mode corresponds to the wave function escaping to infinity.

The authors then examine the behavior of the solutions to the wave equation as the black hole approaches its final state. They show that the solutions remain smooth and well-behaved as the black hole approaches its final state. This indicates that the black hole is stable to linear scalar perturbations.

The article by M. Dafermos and I. Rodnianski provides a detailed analysis of the black hole stability problem for linear scalar perturbations. The authors offer a new perspective on the issue and provide evidence to support the claim that black holes are stable to linear scalar perturbations.

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Answer:

when the radius is halved, F becomes 4 times

Explanation:

The formula is given by

\(F = \frac{Gm1m2}{r^2}\)

When r is halved:

\(F = \frac{Gm1m2}{(1/2r)^2}\)

=> \(F = \frac{Gm1m2}{1/4 r^2}\)

=> \(F = 4(\frac{Gm1m2}{r^2} )\)

This means when the radius is halved, F becomes 4 times

Answer: c) The line shows a change in speed if its angle changes.

I'm not sure

Explanation:

An atom with great electronegativity is able to attract electrons towards itself in a chemical bond. This means that it is able to form strong covalent bonds with other atoms, and can also participate in ionic bonding by attracting electrons away from other atoms.

Additionally, an atom with high electronegativity is able to exert a greater degree of control over the distribution of charge within a molecule, making it an important factor in determining the overall reactivity and behavior of the molecule.

Electronegativity is a measure of an atom's ability to attract electrons towards itself when it is part of a chemical bond. In other words, it is a measure of an atom's ability to pull electrons away from other atoms in a molecule. Electronegativity is an important concept in chemistry, as it helps predict how atoms will behave in chemical reactions.

Electronegativity is typically measured on a scale called the Pauling scale, named after the American chemist Linus Pauling. The scale ranges from 0.7 (for the least electronegative element, francium) to 4.0 (for the most electronegative element, fluorine). Elements towards the right side of the periodic table, such as the halogens and oxygen, are generally more electronegative than elements towards the left side, such as the alkali metals.

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Answer:

his results in the final angle after the collision of 37.2 degrees basically what we did there is turn the vector into a right triangle. We use sohcahtoa to solve for the angle. Being.

Explanation:

Answer:

How does the kinetic energy of the ice cream change once you put it in the hot chocolate?

Explanation: ice cream is cold when the hot chocolate is hot  

Answer:

Explanation:

Heat always flows from HOT to COLD.

KE of hot chocolate decreases.

KE of ice cream increases.

Temp. of hot chocolate decreases (it cools down), temp. of ice cream increases (it warms up).

Temp. of each will stop changing when the 2 substances reach equilibrium (no more flow of heat).

Light bends when it moves from one transparent medium to another due to a change in speed. The correct answer is d.

When light travels from one transparent medium to another, such as from air to water or from air to glass, its speed changes. This change in speed causes the light to bend or refract at the boundary between the two mediums. This phenomenon is known as refraction.

When light passes through a medium with a different refractive index, its speed changes because the medium affects the propagation of light waves. As the speed of light changes, its direction also changes, resulting in the bending of the light rays. The amount of bending depends on the difference in refractive indices frequency between the two mediums and the angle of incidence of the light ray.

It is important to note that while refraction causes the path of light to bend, it does not change the color, frequency, or amplitude of the light waves. These properties of light remain constant during refraction, with only the direction and speed of the light being affected.

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The potential difference across a cell wall can be calculated using the formula:

ΔV = Ed

where ΔV is the potential difference, E is the electric field strength, and d is the distance or thickness of the cell wall.

Plugging in the values given in the problem, we get:

ΔV = Ed = 360 × 10^-9 × 6.5 × 10^-3 = 2.34 × 10^-6 volts

Therefore, the potential difference across the cell wall is 2.34 microvolts (μV).

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The velocity of the cannonball along the x axis after 7.59 s is therefore:vx = (150 m/s)(cos 0°)(7.59 s)vx = 1139.85 m.(d) Dimensionless velocity and dimensionless time are given by:v' = v/vTt' = 2gh/ vT²These expressions simplify the solution to part (c).

(a)  Terminal velocity along y axis is the maximum velocity that the cannonball can reach along y-axis as it falls. When it reaches terminal velocity, the acceleration of the cannonball becomes zero since its weight is balanced by air resistance. According to Stoke’s law, the drag force (Fd) experienced by a sphere moving slowly in a fluid is given by: Fd = 6πηrvwhere:η is the kinematic viscosity of the fluidv is the speed of the sphere, andr is the radius of the sphereHence, the terminal velocity (vT) of the cannonball is given by:  mg = 4/3 πr³ρg   [weight of cannonball = volume of cannonball x density of cannonball x acceleration due to gravity]6πηrvT = mgvT = mg/ (6πηr)The velocity of the cannonball along the y-axis is zero at the start and it reaches terminal velocity after falling through some height h.The velocity of the cannonball at any time t is given by:v = (2gh/ 3πr² ρ)½The velocity of the cannonball along the y axis is 131.3 m/s.(b)At 95% of terminal velocity, the velocity of the cannonball is 124.74 m/sUsing the expression:v = (2gh/ 3πr² ρ)½124.74 = (2gh/ 3πr² ρ)½h = (3/2)(124.74)² πr²ρ/g = 1509.65 mTherefore, it takes 7.59 s to reach 95% of the terminal velocity.(c)At the time determined in part (b), the velocity of the cannonball along the x axis can be calculated using the equation below:vx = vo xcosθtwhere vo is the initial velocity along the x-axis, θ is the angle of projection, and t is the time taken.At launch, the initial velocity along the x axis is 150 m/s and the angle of projection is 0°.

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The efficiency of the heat engine is 0.2, which corresponds to option A.

To calculate the efficiency of a heat engine, we can use the formula:

Efficiency = (Useful output energy / Input energy)

In this case, the useful output energy is the mechanical power produced by the engine, which is 300 W. The input energy is the total energy input to the engine, which is the sum of the useful output energy and the energy discarded into the environment.

Input energy = Useful output energy + Energy discarded

Input energy = 300 W + 1200 W

Input energy = 1500 W

Now, we can calculate the efficiency:

Efficiency = (Useful output energy / Input energy) = (300 W / 1500 W)

Efficiency = 0.2

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Answer:

El resorte se comprime 0.38 m.

Explanation:

La distancia de compresión del resorte se puede calcular por conservación de la energía:

\( \Sigma E_{i} = \Sigma E_{f} \)

\( E_{p} = E_{e} \)

\( mgh = \frac{1}{2}kx^{2} \)   (1)

En donde:

\(E_{i}\) y \(E_{f}\): son las energías inciales y finales

\( E_{p}\): es la energía potencial gravitacional

\(E_{e}\): es la energía potencial elástica

m: es la masa = 1.50 kg

g: es la gravedad = 9.81 m/s²

h: es la altura

k: es la constante de fuerza = 320 N/m    

x: es la distancia de compresión

Dado que el objeto está 1.20 m sobre el resorte, entonces h es:

\( h = 1.20 + x \)  (2)

Entonces, introduciendo la ecuación (2) en (1) y resolviendo para x tenemos:

\( \frac{1}{2}kx^{2} - xmg - 1.20mg = 0 \)  

\( 160x^{2} - 14.72x - 17.66 = 0 \)  

Resolviendo la ecuación cuadrática anterior tenemos:

x₁ = -0.29 y x₂ = 0.38

Tomando el valor positivo entonces, el resorte se comprime 0.38 metros.

Espero que te sea de utilidad!  

Answer:

Plan B.

Because flexibility is best improved by stretching.

Explanation:

Improving and increasing flexibility is done by having stretching sessions daily which maintains and widens the range of motion in the joints and stretches muscles.

The peak emission wavelength for a campfire with this temperature is 2.32 µm.

Wein's Law is given by the formula: Amax = a/T

We have to find the peak emission wavelength of the campfire using Wein's displacement law which is given by the formula Amax = a/T.

Where Amax is the peak emission wavelength in um, a is a constant that equals 2897 um-K, and T is the blackbody temperature.

Substituting the given values in the formula, we get:

Amax = 2897 / 1250 um-K = 2.32 um

Therefore, 2.32 µm is the peak emission wavelength for a campfire with this temperature

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Given,

The initial frequency of the heartbeat, f₁=1.45 Hz

The increased heartbeat, f₂=1.55 Hz

The frequency of the heartbeat can be described as the number of occurence of the heartbeat per second. That is every second, the heart beats 1.45 times.

(a)

Thus for a minute, the number of the heartbeats is,

Thus 87 beats occur for a minute.

(b)The increase in the frequency of the heartbeat implies the increase in the number of the heartbeat for every second. And hence the beats in a minute increase when the frequency of the heartbeat increases.

(c)

The number of the beats per minute after the increase of the frequency is,

Thus after the increase in the frequency, 93 beats occur in a minute.

Answer:

Explanation:

The cars velocity is 14m/s. That means for every second, the car moves 14m. Now, if the car wants to reach 28m, going 14m/s, it will take 2 seconds for the car to reach 28m.

This statement is True. In an eclipsing binary system, we can observe the periodic eclipses of the two stars as they orbit each other. By measuring the changes in the light and duration of the eclipses, we can calculate the radii of the stars as well as their masses.

In an eclipsing binary system, we can indeed measure both the radii and masses of the stars involved. This is achieved by analyzing the light curves and radial velocity data of the system, which provide information on the stars' sizes and orbital motion.

Combining these measurements allows for the accurate determination of both the radii and masses of the stars.

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The main difference between transverse waves and compression waves is the direction of the particle oscillation relative to the direction of wave propagation. The other differences are as follows:

Transverse waves are waves in which the particles of the medium oscillate perpendicular to the direction of wave propagation. These waves exhibit a characteristic up-and-down motion, like the motion of a rope when it is shaken up and down. Examples of transverse waves include light waves, water waves, and seismic S-waves. On the other hand, compression waves, also known as longitudinal waves, are waves in which the particles of the medium oscillate parallel to the direction of wave propagation. These waves exhibit a characteristic back-and-forth motion, like the motion of a slinky when it is compressed and expanded. Examples of compression waves include sound waves and seismic P-waves. In transverse waves, the particle oscillation is perpendicular to the direction of wave propagation, whereas in compression waves, the particle oscillation is parallel to the direction of wave propagation.

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The best way to describe sound levels is through a ratio scale.

The decibel scale, which measures sound levels, is best described as a ratio scale. While the decibel scale is logarithmic, it can be transformed into a linear ratio scale by using a reference point. By selecting a reference point as the softest sound the human ear can hear unaided (0 dB), we establish a meaningful zero point on the scale. Any sound level above this reference point can be expressed as a positive value, indicating its magnitude relative to the softest audible sound. The logarithmic nature of the decibel scale still holds, with each increase of 10 dB representing a tenfold increase in sound intensity. However, by anchoring the scale to a fixed reference point, we can now quantify and compare sound levels in a meaningful way, making it a ratio scale.

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The heat associated with saunas and hot tubs may cause fetal harm or complications.

Exposure to high temperatures, such as those found in saunas and hot tubs, can potentially lead to an increased risk of certain complications during pregnancy. It is generally recommended that pregnant women avoid prolonged exposure to high temperatures and excessive heat. The main concern is that increased body temperature can potentially affect the development of the fetus and increase the risk of birth defects or pregnancy complications. High temperatures can lead to maternal hyperthermia, which may negatively impact the developing fetus, particularly during the first trimester when organ formation is occurring.

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Newton's second law allows us to find that the correct answer is:

   A) The mass of the baseball never changes so the more force in the pitchers arm, the higher the acceleration of the baseball before it leaves the pitcher's hand.

Newton's second law gives a relationship between force and the product of mass and acceleration of the body

          F = m a

Where F is the force, m the mass and the acceleration of the body.  The bold indicate vectors

The baseball ball is a solid body that has a fixed mass, therefore the pitcher uses different forces in the arm, the acceleration must change proportionally, as the force increases, the acceleration must increase (fastball).

Let's review the different claims

A) true. These statements are in accordance with Newton's second law

B) False. If the force changes any of the other two parameters must change

C) false. Force and acceleration are proportional

D) False. When the ball is in the air it is subjected to the acceleration of gravity

With Newton's second law we find that the correct answer is:

   A) The mass of the baseball never changes so the more force in the pitchers arm, the higher the acceleration of the baseball before it leaves the pitcher's hand.

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Answer:

46.55m/s

Explanation:

To find the distance between the m = 49 and m = 50 interference maxima on the screen, we can use the formula for the fringe spacing in the double-slit interference pattern:

d * sin(θ) = m * λ

d * θ = m * λ

d = (m * λ) / θ

Where:

d is the slit separation,

θ is the angle of the fringe with respect to the central maximum,

m is the order of the fringe,

λ is the wavelength of the light.

In this case, we are given that the separation between two adjacent interference maxima (fringes) near the center of the screen is 3.53 cm. Since the screen is very far away compared to the distance between the slits, we can approximate sin(θ) as θ.

Thus, we have:

d * θ = m * λ

We can rearrange this equation to solve for the slit separation d:

d = (m * λ) / θ

Now, we can substitute the given values into the equation:

m = 50 (order of the fringe)

λ = 500 nm (wavelength)

θ = (3.53 cm) / (2.00 m) ≈ 0.0176 rad

d = (50 * 500 nm) / 0.0176 ≈ 1.42 mm

Therefore, the distance on the screen between the m = 49 and m = 50 maxima is approximately 1.42 m

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The frequency of pku carriers is 0.029

2.9%

q^2= 1/4500

q=sqrt(1/4500)=0.015

p=1-0.015=0.985

2pq = 0.029

A genetic disease that raises blood levels of the amino acid phenylalanine. This may result in developmental delays, behavioural issues, seizures, and mental impairment.

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(3/2) ln|81 + t²| + C.

This expression represents the displacement (s) of the robotic welding device in terms of time (t).

To find the expression for displacement (s) in terms of time (t), we need to integrate the velocity function over the given time interval.

The velocity function is given as: v = 3t / (81 + t²).

To find the expression for displacement, we integrate the velocity function with respect to time:

∫ v dt = ∫ (3t / (81 + t²)) dt.

To evaluate this integral, we can use the substitution method. Let u = 81 + t², then du = 2t dt.

The integral becomes:

∫ (3t / (81 + t²)) dt = (3/2) ∫ (1/u) du.

Integrating, we have:

(3/2) ln|u| + C,

where C is the constant of integration.

Substituting back u = 81 + t²:

(3/2) ln|81 + t²| + C.

This expression represents the displacement (s) of the robotic welding device in terms of time (t).

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The nurse determined that the heart beats periodically 60 times in 60 seconds, which means that the heart beats once every second. which in this case is one heartbeat. the period of the heartbeat is 1 second.

Therefore, the period of the heartbeat is 1 second. Option A (1 Hz) is incorrect because 1 Hz refers to the frequency, which is the number of cycles per second, not the period. Option B (60 Hz) is incorrect because it is an extremely high frequency that is not consistent with the human heartbeat. Option D (60 s) is incorrect because it is too long of a period for one heartbeat.
"A nurse takes the pulse of a heart and determines the heart beats periodically 60 times in 60 seconds. The period of her heartbeat is The period of her heartbeat is C: 1 s.

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Answer:

V = a * t      for uniformly accelerated motion starting at zero speed

V = 4 s * 9.8 m/s^2 = 39.2 m/s

S = V0 t + 1/2 a t^2 = 1/2 a t^2

S = 1/2 * 9.8 m/s^2 * 16 s^2 = 78.4 m

Check:

Average speed = 39.2 / 2 = 19.6 m/s

S = 4 * 19.6 = 78.4 m

Answer:

F = 9/5 C + 32     conversion from C to F

F = 9/5 * F/5 + 32

25 F = 9 F + 800

16 F = 800

F = 50  

Check:

C = 5/9 ( F - 32)  = 5/9 (50 - 32) = 10  as requested

Q = c m change in temp

Q = 1 cal/gm-deg C * 500 gm * 45 deg C = 22,500 calories

50 Fahrenheit heat required.

Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy  between physical systems.

F = 9/5 C + 32     conversion from C to F

F = 9/5 * F/5 + 32

25 F = 9 F + 800

16 F = 800

F = 50

The answer is 50 Fahrenheit.

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The repeated flying into the air and landing on the same spot reinforces to the woodcock that this is the center of its territory.

The phrase "Woodcock sky dances" refers to the woodcock's breeding routine. It is a migratory bird that mates in the United States' Eastern and Central regions. Woodcock "sky dances" or courtship flights, which are a part of its breeding ritual, are familiar to ornithologists and bird lovers.The male woodcock prepares for the display by choosing a place and then creating a tiny opening in the trees or shrubs. It flies into the air at sunset or sunrise, calling out with a distinctive chirping sound. It then lands on the same spot, with each flight being more extensive than the last. The woodcock does this repeatedly, with each flight reaching a greater height than the previous one.

The woodcock's territory is at the center of its sky dancing. The repeated flying into the air and landing on the same spot reinforces to the woodcock that this is the center of its territory. During the breeding season, the woodcock's sky dancing activity is essential. It shows off the male's strength and agility to female woodcocks, which aids in their selection of mates.The woodcock's behavior is a result of natural selection. Through sky dancing, the woodcock's species has developed its breeding ritual over thousands of years. The ability to perform sky dances is an evolutionary benefit to the woodcock, as it aids in species reproduction and survival. Therefore, repeatedly flying into the air and landing on the same spot reinforces to the woodcock that this is the center of its territory.

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Answer:

sorry do not speak Spanish so I would not be able to help you probably cannot read this but I would be happy to help you if you follow me and bake English so thank you