List materials from slowest to fastest: steel, air, glass, water

The magnitude of the force is approximately \(-4.03 * 10^{-5} N\) and the negative sign indicates that the force is attractive.

The magnitude of the force between Fluffy's fur and her plastic brush can be determined using Coulomb's Law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charges are \(+8.0 * 10^{-9}\) coulombs (q1) and \(-1.4 * 10^{-8}\)coulombs (q2), and the distance between them is\(5.0 * 10^{-1}\) meters (r). Thus, the formula to calculate the force is:
F =\(k * (q1 * q2) / r^2\)
Where k is Coulomb's constant, which is equal to \(9 * 10^9 Nm^2/C^2\). Substituting the values given in the problem, we get:
F = \(9 * 10^9 * (8.0 * 10^{-9}) * (-1.4 * 10^{-8}) / (5.0 * 10^{-1})^2\)
F ≈ \(-4.03 * 10^{-5} N\)
The negative sign indicates that the force is attractive, meaning that Fluffy's fur and her plastic brush are being pulled towards each other.

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The two-dimensional curl of the vector field F(x, y) = (-2xy, x²) is computed to be 4x - 2. The region R bounded by y = 0 and y = x(2-x) is sketched as a triangular region in the xy-plane. By applying Green's Theorem in the circulation form, the integrals are evaluated and shown to be equal, confirming the consistency of the computations.

(a) To compute the two-dimensional curl of the vector field F(x, y) = (-2xy, x²), we need to find the partial derivatives of the components of the vector field and take their difference. The curl is given by the expression:

\(\[\nabla \times \textbf{F} = \left( \frac{\partial}{\partial x} (x^2) - \frac{\partial}{\partial y} (-2xy) \right) \textbf{i} + \left( \frac{\partial}{\partial y} (-2xy) - \frac{\partial}{\partial x} (x^2) \right) \textbf{j}\]\)

Simplifying this expression yields:

\(\[\nabla \times \textbf{F} = (0 - (-2x)) \textbf{i} + (4x - 0) \textbf{j} = 2x \textbf{i} + 4x \textbf{j} = \boxed{2x \textbf{i} + 4x \textbf{j}}\]\)

(b) The region R is bounded by the y-axis (y = 0) and the curve y = x(2-x). Sketching this region in the xy-plane, we find that it forms a triangular region with vertices at (0, 0), (1, 0), and (2, 0).

(c) Applying Green's Theorem in the circulation form, which states that the line integral of a vector field around a closed curve is equal to the double integral of the curl of the vector field over the region enclosed by the curve, we can evaluate both integrals. Let C be the boundary of the region R.

Using the circulation form of Green's Theorem, the line integral becomes:

\(\[\oint_C \textbf{F} \cdot d\textbf{r} = \iint_R (\nabla \times \textbf{F}) \cdot d\textbf{A}\]\)

The first integral is evaluated over the boundary curve C, and the second integral is evaluated over the region R. Substituting the given vector field and the computed curl, we have:

\(\[\oint_C \textbf{F} \cdot d\textbf{r} = \iint_R (2x \textbf{i} + 4x \textbf{j}) \cdot d\textbf{A}\]\)

Integrating this expression over the triangular region R will yield a specific result. By evaluating both integrals, it can be verified that they are equal, confirming the consistency of the computations.

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Answer:

10^4 microphones

Hope it helps

Explanation:

10^-2 phones

= 10^(6-2)

= 10^4 microphones

Moving away from the source causes the observer to measure a lower frequency and higher wavelength.

The frequency of the detected sound from a stationary source will change as a result of the observer's movement. Moving away from the source causes the observer to measure a lower frequency and higher wavelength.

The Doppler effect is a shift in sound wave frequency that happens when the source of the sound waves is moving in relation to a listener who is stationary.

The wave propagates the sound energy throughout the medium, typically in all directions and with decreasing intensity as it gets further away from the source.

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a) Two other forces that act on thee sled are the weight of the daughter and the drag force

b) The drag force slows down the sled while the weight presses it down.

We know that motion is the result of the action of unbalanced forces. From the Newton's third law, we know that if the forces that are acting on an object are balanced that the object is not going to move from its state of rest.

In this case, we are told that a  father pushes his daughter and son on a sled down a hill. The fact that the led is moving is due to the fact that there are forces that are acting on the sled.

The forces that are acting on the sled apart  from the force that was applied in pushing the sled is the drag force and the weight of the son.

The drag force would determine how fast that the sled would move because it acts in opposition to the forward force and tries to slow down the sled as it is moving. The weight of the son does press the sled downwards.

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The closest frequencies to 450 Hz that an average person can distinguish as being different from 450 Hz would be in the range of 445-455 Hz.

The closest frequencies to 450 Hz that an average person can clearly distinguish as being different in frequency from 450 Hz are typically within a range known as the "Just Noticeable Difference" (JND) or the "Difference Limen." The JND for frequency perception varies from person to person but is generally around 1-5 Hz for a pure tone in the mid-frequency range. This means that an average person can perceive a difference in frequency if it is within the range of approximately 1-5 Hz from the original frequency. Therefore, the closest frequencies to 450 Hz that an average person can clearly distinguish as being different from 450 Hz would be in the range of 445-455 Hz. Frequencies outside this range would be more easily distinguishable as different from the original 450 Hz tone.

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Here's an example of a MATLAB function file that takes in t as a vector, k as a scale, and eq as the equation number (1, 2, or 3) and returns the solution to the corresponding differential equation:

function y = solveDE(t, k, eq)

switch eq

   case 1 % y' = k*y

       y = exp(k*t);

   case 2 % y'' + k*y = 0

       y = cos(sqrt(k)*t);

   case 3 % y' + k*y^2 = 0

       y = 1./(k*t + 1./exp(k*t));

   otherwise % if eq is not 1, 2, or 3, return an error message

       error('Invalid equation number');

end

end

Here, the switch statement allows us to handle the different cases for each equation number. For example, if eq is 1, then the function returns y = exp(k*t) which is the solution to the differential equation y' = k*y. If eq is 2, then the function returns y = cos(sqrt(k)*t) which is the solution to the differential equation y'' + k*y = 0. And if eq is 3, then the function returns y = 1./(k*t + 1./exp(k*t)) which is the solution to the differential equation y' + k*y^2 = 0. If eq is not 1, 2, or 3, then the function returns an error message using the error function.

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Given:

• Distance = 150 km

• Speed = 50 km/hr

Let's find the time it will take to cover the distance in minutes.

To find the distance, apply the formula:

Thus, we have:

The time in hours is 3 hours.

To convert the time from hours to minutes, we have:

60 minutes = 1 hour

3 hours = 3 x 60 = 180 minutes

Therefore, it time in minutes is 180 minutes.

ANSWER:

180 minutes

Answer:

(You find the epicenter of an area earthquake by using triangulation)

But alright moving towards the graph, in the top row will be 'a' for the seismic station, for the P-wave arrival time (for the top row) it'll be 09:20:42 AM, for the s-wave arrival time (of the top row still) will be 09:21:09 AM, for the S-P interval (in sec) will be 27, and for the Distance to epicenter (km) will be 283 :)

now moving onto the second row, the seismic station is now 'b' , the p-wave arrival time is 09:21:27 AM, the s-wave arrival time will be 09:22:27 AM, the S-P interval (in seconds) is 60, the distance to epicenter is 632,

now moving onto the third row, the seismic station is 'c', the p-wave arrival time is 09:21:49 AM, the s-wave arrival time is 09:23:05 AM, the S-P interval is 76, and the distance to epicenter is 806

that's all for the graph :)

okay so now, for the second part to your question "according to the data table, which type of wave reached station A first, the P waves or the S waves? would you expect this fact to hold data from other seismic" okay so looking at what I've already said earlier about the tables it should be pretty easy to figure out but I'm sure you just want an answer so you can finish your schoolwork so i wont go on to long about it lol, the answer would be the p-waves got there first, and yes it can hold this for data on other seismic station, it also asked you to explain your answer so here's that: in 'b' station and 'c' station the p-wave always reached the stations before the s-wave.

Explanation:

I dont know if you do online school or go to virtual school house, but i do  and my teacher there posted a video in the place where we turn in our work showing us the different tables, anyways i hope this helped :) !!

Answer:

\(9.15\times 10^{13}\ \text{N}\)

Explanation:

\(m=\text{Mass of person}=\dfrac{675}{g}=\dfrac{675}{9.8}\ \text{kg}\)

\(g=\text{Acceleration due to gravity on Earth}=9.8\ \text{m/s}^2\)

\(r=\text{Radius the neutron star}=10\ \text{km}\)

\(M=\text{Mass of neutron star}=1.99\times 10^{30}\ \text{kg}\)

\(G=\text{Gravitational constant}=6.674\times10^{-11}\ \text{Nm}^2/\text{kg}^2\)

Weight on neutron star would be

\(W_n=\dfrac{GmM}{r^2}\\\Rightarrow W_n=\dfrac{6.674\times 10^{-11}\times \dfrac{675}{9.8}\times 1.99\times 10^{30}}{10000^2}\\\Rightarrow W_n=9.15\times 10^{13}\ \text{N}\)

Weight of the person on the neutron star would be \(9.15\times 10^{13}\ \text{N}\)

Answer:

A. the time of fire is 15 seconds, and the angle of fire is approximately 23.794° above the horizontal

B. The angle of fire is increased to approximately 35.4126° above the horizontal

Explanation:

A. The height of the bridge, h = 70 m

The speed with of the shell, v₀ = 100 m/s

The location of the speedboat = 3000 m

The direction of the speedboat = 26 m/s

The acceleration of the speedboat, a = 11 m/s²

Let t represent the time of firing the shells, and let x represent the distance of the speedboat from the bridge, and let θ, represent the angle to fire with, we have;

For the speedboat, t = x/(100 × cos(θ))

We note that the time the shell can travel the 3,000 m = 30 seconds

Therefore an adequate time to fire is, t = 15 seconds

The distance the speedboat covers in 15 seconds is given as follows;

s = u·t + 1/2·a·t²

s = 26 × 15 + 1/2 × 11 × 15² = 1627.5 m

At the point the speedboat had traveled 1,627.5 m, the distance of the speedboat, x is then 3000 - 1,627.5 = 1,372.5 m from the bridge, the angle of fire is given from the following formula;

t = x/(100 × cos(θ))

15 = 1,372.5/(100 × cos(θ))

cos(θ) = 1,372.5/(15 × 100) = 0.915

θ = cos⁻¹(0.915) ≈ 23.794°

The angle of fire, θ ≈ 23.794° above the horizontal in the direction of the speedboat

B. Given that a 10 m/s wind is blowing towards East, we have;

The horizontal velocity towards East = 10 + v₀ × cos(θ)

The angle of firing is therefore, given as follows;

15 = 1,3725.5/(10 + 100 × cos(θ))

(10 + 100 × cos(θ)) = 1,3725.5/15

100 × cos(θ) = 1,372.5/15 - 10

cos(θ) = (1,372.5/15 - 10)/100 = 0.815

θ = cos⁻¹(0.815) ≈ 35.4126°

Therefore, the angle of fire, θ, will be increased to approximately 35.4126° above the horizontal in the remote controlled speedboat direction.

The capacitance of a charged conducting sphere of radius 12 cm that produces an electric field of 49 kn/c at a distance of 21 cm from its center is calculated below. What is the capacitance?

The capacitance is calculated using the following formula:  (Q/V) = C,Q is the charge, and V is the potential difference. The potential difference is given by the electric field E multiplied by the distance between the plates d. For a point at a distance r from the center of the sphere, the electric field is given by: E = Q/4πε0 r2For a uniformly charged sphere, the electric field at a point r within the sphere is given by: E = kQR/r3where k is a constant that is equal to 1/(4πε0).

The electric field at a distance of 21 cm from the center of the sphere is given to be 49 kN/C. For a point at a distance of 21 cm from the center of the sphere, the radius of the sphere is given to be 12 cm. The charge on the sphere is given by Q = 4πε0 R2 E where R is the radius of the sphere. Substituting the values given in the equation above, we getQ = 4π(8.85 × 10−12) (0.12)2 (49 × 10^3)= 3.232 x 10^-7C The potential difference between the surface of the sphere and a point at a distance of 21 cm is given by: V = Ed= 49 × 10^3 × 0.21 = 10.29 × 10^3VThe capacitance of the sphere is calculated by the formula:  (Q/V) = C. Substituting the values of Q and V into the equation above, we get: C = Q/V= (3.232 x 10^-7)/ (10.29 × 10^3) = 3.14 × 10^-11F or 31.4 pF Answer: 31.4 pF.

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The final speed of the electron is the same as its initial speed, which is \(1.4 × 10^7 m/s\). The correct answer is A).

The electric force experienced by an electron in an electric field is given by F = qE, where q is the charge of the electron and E is the magnitude of the electric field. Since the electron has a negative charge, the direction of the force is opposite to the direction of the electric field.

Using the force equation F = ma, where m is the mass of the electron, we can write the acceleration of the electron as:

a = F/m = qE/m

We can use the kinematic equation \(v^2 = u^2 + 2as\), where u is the initial velocity, v is the final velocity, a is the acceleration, and s is the distance traveled.

Substituting the given values, we get:

\(a = (1.6 × 10^-19 C)(120 N/C)/(9.11 × 10^-31 kg) = 2.11 × 10^14 m/s^2\)

s = 3.5 m

\(u = 1.4 × 10^7 m/s\)

Plugging these values into the kinematic equation, we get:

\(v^2 = (1.4 × 10^7 m/s)^2 + 2(2.11 × 10^14 m/s^2)(3.5 m) = 1.96 × 10^15 m^2/s^2\)

Taking the square root of both sides, we get:

\(v = 1.4 × 10^7 m/s\)

Therefore, the final speed of the electron is the same as its initial speed, which is \(1.4 × 10^7 m/s\). The correct answer is A).

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Answer:

The first option- Sun shining through clouds

Explanation:

Answer:

Hold the balloon (negatively charged) just above your head so your hair (positively charged) will be attracted to it and stand up on end. ... The can will start to roll towards the balloon without touching it. The negatively charged balloon repels the electrons of the can so that a positive charge is near the balloon.

Explanation:

I gave my explanation along with the answer.

The investigation presented did offer proof in favor of the existence of imperceptible electric force fields.

The force between two charged objects is the electric force, sometimes referred to as the Coulomb force. The interaction of charged particles produces this fundamental force of nature. The magnitudes of the charges on the two objects and the separation between them define the strength of the electric force.

The tape was observed to interact without touching when it was charged by rubbing the tape against one another. The formation of an electric field surrounding the charged tape, which pulls on other nearby charged items, can be used to explain this interaction.

Even though the electric field surrounding the charged tape was not explicitly measured or quantified, the observed interaction between the tape offers a weak indirection for its presence. Hence, the investigation did offer sufficient proof to back up the existence of undetectable electric force fields.

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Answer:

B. Conductivity.

Explanation:

Conductivity is the quantity of heat passing per second through a slab of unit cross-sectional area when the temperature gradient between the two faces is unity when put in heat.

If a measurable resistance is obtained when testing across a normally closed set of contacts on a relay using an ohmmeter, it indicates a problem known as a "stuck closed" or "welded" contact in the relay.

A relay consists of electromechanical switches that open and close to control the flow of current in a circuit. Normally closed (NC) contacts are designed to be closed when the relay is not energized. However, if a measurable resistance is detected across the normally closed contacts when using an ohmmeter, it suggests that the contacts are not opening as they should.

This can occur due to several reasons, such as welding of the contacts, excessive current or voltage, mechanical damage, or contamination. The stuck closed contacts disrupt the proper functioning of the relay and can lead to unintended circuit operation or failure.

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The temperature for last Sunday is 1.08T.

Expression simply refers to the mathematical statements which have at least two terms which are related by an operator and contain either numbers, variables, or both. Addition, subtraction, multiplication, and division are all possible mathematical operations.

Since thee average temperature was 8 percent higher than the average temperature two Sundays ago and the average temperature two Sundays ago was T degrees Celsius.

The expression for the temperature for last Sunday will be:

= T + (8% × T)

= T + 0.08T

= 1.08T

The expression is 1.08T.

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Complete question

Last Sunday, the average temperature was 8 percent higher than the average temperature two Sundays ago. The average temperature two Sundays ago was T degrees Celsius. What is.l the Temperature for last Sunday?

10 w night light left on for 18 hr is consumes more energy.

a 1 kw hair dryer used for 10 min.

energy consumes = 1kw * 10* 60 = 600kJ.

a 10 w night light left on for 18 hr

18Hr = 18 * 60 * 60 = 64800 sec

Energy consumes = 10 * 64800 =648000 J. = 648 kJ.

Since, 10 w night light left on for 18 hr consumes more energy.

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Answer:

140 m/s

Explanation:

v = u + at

v = 8 + 3.5(40) = 140 + 8 = 140

The surface air around a strengthening low-pressure area normally converges, while above the system, the air normally diverges.

When a low-pressure system strengthens, it means that the pressure at the center of the system is decreasing. As a result, the surrounding air at the surface tends to converge and move towards the low-pressure center.

This convergence of air at the surface creates a cyclonic circulation pattern, where air spirals inward towards the center of the low-pressure system.

At higher altitudes, above the low-pressure system, the air tends to diverge. This means that the air moves away from the center of the system.

The divergence of air at higher altitudes is a result of the vertical motion associated with the low-pressure system.

As air converges at the surface and moves towards the center of the low-pressure system, it rises vertically. This upward motion leads to the divergence of air at higher altitudes.

The combination of surface air convergence and upper-level air divergence is characteristic of a strengthening low-pressure system and contributes to the intensification and development of weather associated with such systems, such as storms and cyclones.

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Answer:

B. It can be answered using measurements.

Explanation:

Scientific inquiry can be defined as the various ways, techniques and approach used by scientists to study the world and provide a detailed description (information) based on the empirical evidence derived from the investigation.

Measurements can be defined as a technique which typically involves the process of identifying and determining the dimensions, quality and quantity of a physical object.

Hence, the statement which best describes a characteristic of a question that can be answered through scientific inquiry is that, it can be answered using measurements. Using scientific inquiry to answer questions entails using measurements to propose specific and evidential explanations, as well as back up theories associated with the question.

Answer:

2500 Newtons

Explanation:

If force equals the mass × acceleration then all should be quite simple

25kg is the mass of your car

100 m/s/s is the acceleration of you car

25kg • 100m/s/s = 2500 Newtons

Answer:

\(\boxed {\boxed {\sf 2500 \ Newtons }}\)

Explanation:

The formula for force is the product of mass and acceleration.

\(F=m*a\)

The mass of the car is 25 kilograms and the acceleration is 100 meters per square second. So,

\(m= 25 \ kg \\a= 100 \ m/s^2\)

Substitute the values into the formula.

\(F= 25 \ kg * 100 \ m/s^2\)

Multiply.

\(F= 2500 \ kg*m/s^2\)

\(F= 2500 \ N\)

The force of the car is 2500 Newtons.

The force that acts when one object moves against another object is friction.

option C is the correct answer.

Friction force or force of friction is the type of force that act between two surfaces that are in contact. This is the type of force that tends to oppose the motion of two surfaces that are contact.

This type of force is of two types and they include the following:

Static friction force is the type of friction force that exist between two stationary surfaces that are in contact.

The kinetic friction force is the type friction force that exists or act between the two surfaces that are in motion.

Thus, we can conclude that friction force is a type of force that act between two surfaces that are either in motion ( kinetic friction force ) or stationary ( static friction force ).

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Volume is -5 m3/s, surface area is -6 m2/s, and a diagonal's length is -1.01 m/sec.

Each object in three dimensions takes up some room. The volume of this area is what is being evaluated. The area occupied within an object's boundaries in three dimensions is referred to as its volume. It is also referred to as the object's potential.

Given l=4m w=h=5 m

dl/dt = 3m/s  dw/dt=3 m/s dh/dt= -7m/s

a) volume

as we know v=lwh

so using chain rule

dv/dt= dl/dt×wh+ lh dw/dt+ lw dh/dt

put the values

dv/dt = (3 × 5 × 5)+ ( 4 × 5 × 3)+ (4 × 5 × (-7))

= 135-140

dv/dt = -5 m^3/s

b) surface area

SA = 2(lw+lh+wh)

dSA/dt= 2 [ dl/dtw + l dw/dt + dl/dt*h + l*dh/dt + dw/dt*h + h*dh/dt]

=2[(3× 5) + (4×3) + (3×5) + (4× -7) + (3×5) + (5× -7)

=2[15 + 12 + 15 - 28 + 15 - 35]

=2[57-63] = - 6

dSA/dt = -6 m^2/s

c) diagonal

D= \(\sqrt{l^2+w^2+h^2}\)

dD/dt= 2l*dl/dt+2w * dw/dt + 2h* dh/dt

dD/dt = 1/    \(\sqrt{l^2+w^2+h^2}\)[l.dl/dt+ w.dw/dt +h.dh/dt ]

put values

dD/dt= 1/ \(\sqrt{16+ 25 + 25}\)[(4*3) + (5*3) + (5*-7)]

= 1/\(\sqrt{66}\)[27-35] = -8/\(\sqrt{66}\)

=dD/dt = -8/\(\sqrt{66}\) m/sec

=dD/dt = -1.01 m/sec

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Answer: Ferromagnetism is a phenomenon that occurs in some metals, most notably iron, cobalt and nickel, that causes the metal to become magnetic. The atoms in these metals have an unpaired electron, and when the metal is exposed to a sufficiently strong magnetic field, these electrons' spins line up parallel to each other.

a)

The distance travelled in the first 3 seconds is 6 meters.

b.)

the total distance travelled is 13 meters.

c.)

the acceleration in the time interval from 4 seconds to 7 seconds is 0.67 m/s^2.

distance = speed × time

distance = 2 m/s × 3 s = 6 meters

b)

The distance travelled in the first 3 seconds=  6 meters.

The distance travelled from 3 seconds to 7 seconds, which can be found by finding the area under the trapezium formed by the line joining (3, 2) and (7, 4), the x-axis and the vertical lines at x=3 and x=7.

The height of the trapezium is (4 - 2) m/s = 2 m/s, and the two bases are 4 s - 3 s = 1 s and 7 s - 3 s = 4 s, respectively.

the area of the trapezium is:

area = (1/2) × (1 + 4) s × 2 m/s = 7 meters

The total distance travelled is therefore:

total distance = distance travelled in the first 3 seconds + distance travelled from 3 seconds to 7 seconds

total distance = 6 meters + 7 meters = 13 meters

c)

change in speed = final speed - initial speed

At t=4 s, t speed is 2 m/s,

and at t=7 s,  speed is 4 m/s.

change in speed = 4 m/s - 2 m/s = 2 m/s

The time interval is:

time interval = 7 s - 4 s = 3 s

acceleration = change in speed / time interval

acceleration = 2 m/s / 3 s = 0.67 m/s^2

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Explanation:

If no force is exerted on a moving object, it will continue to move at a constant velocity. This is described by Newton's first law of motion, which states that an object will remain at rest or continue to move at a constant velocity in a straight line unless acted upon by an external force. In other words, if no net force acts on an object, the object will maintain its velocity, neither speeding up nor slowing down.