What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75

The given mass of water is,

The value of the raise in the temperature is,

The value of the specific heat of the water is,

Thus, the energy needed to increase the temperature of water is,

Hence, the amount of energy required to raise the temperature of the water to 5 degree C is 21 kJ.

Rock can melt at a depth of about below Earth's surface  100 km

On the other hand, interface physics offers a wide variety of spectroscopic & microscopic techniques to characterize that deposition & structure creation process upon that sub-nanometer size and, therefore, to pave the way for effective manufacturing techniques.

It is the study of both the chemical processes that take place at the meeting point of two surfaces, such as solid-liquid, solid-gas, sturdy, liquid-gas, etc. Surface engineering refers to a few surface chemistry applications.

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Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

\((m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}\)

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

\((1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]\)

Answer: C. Soft drink with both ice and carbonation

Explanation: i know its C. because i have had this problem before

Answer:c

Explanation:

The coasts of Africa and South America were touched  125 million years ago due to continental drift.

The time can be computed from the ratio of distance and speed.

Given:

Distance = 5, 000,000

Speed = 4 cm/year = 0.04 m/year

The time is computed as:

Time = Distance / Rate

Time = 5,000,000 meters / 0.04 meters per year

Time = 125,000,000 years

Hence, the coasts of Africa and South America were touched  125 million years ago due to continental drift.

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The work done, in joules, by the net force on the box is 0 J.

The work done by the net force on the box is calculated from the product of net force and distance through which the object travelled and it is given as;

W = F(net) x d

where;

F(net) = F - Ff

where;

ma = F - Ff

where;

when the box moves at a constant velocity, the acceleration = 0, hence net force = 0

W = F(net) x d

W = 0 x d

W = 0 J

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5.73 seconds pass before the centripetal acceleration equals the tangential acceleration in strength.

This enables you to calculate the change in velocity in metres per second squared (m/s²). The change in velocity (v) over the change in time (t) is known as acceleration (a). It can be determined using the formula a = v/t.

a = v²/r

where a is the centripetal acceleration, v is the speed of the car, and r is the radius of the circular track (half the diameter).

a = (2.97 m/s²)

r = (195 m)/2 = 97.5 m

v² = ar = (2.97 m/s²)(97.5 m) = 289.58 m²/s²

v = √(289.58 m²/s²) = 17.01 m/s

at = dv/dt

where at is the tangential acceleration, and v is the speed of the car. Since the car is starting from rest, its initial speed is zero, so we can simplify the formula to:

at = v/t

where t is the elapsed time.

We want to find the time at which the magnitude of the tangential acceleration is equal to the magnitude of the centripetal acceleration, so:

at = ac

v/t = ac

t = v/ac = 17.01 m/s / 2.97 m/s² = 5.73 s

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Answer:

B.  

Explanation:

Correlation is defined as a statistic that determine the relation between two variables.

Both the variables are important while finding the significance of correlation and no variable can alone claim the test results are significant. A test result is significant when both the variables are related to each other because it shows a significant probability that the relationship between the two variables exists.

Hence, the correct answer is "B. "

In correlation, variables are related to each other.

Option B is correct.

When two or more sets of data are linked together, they have a high correlation.

Data sets have a positive correlation when they increase together, and a negative correlation when one set increases as the other decreases.

Correlation refers to the degree of correspondence or relationship between two variables. Correlated variables tend to change together.

Therefore, option B  is correct.

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Answer:

2as=v2-u2

2000=v2

V=44

V=u+at

44/10=t

T=4.4seconds

The best way of presenting the information on density graphically would be to use a D, bar graph.

A bar graph is a type of chart that uses rectangular bars to represent data. The bars are typically arranged in columns, with the independent variable (in this case, the type of wood) on the x-axis and the dependent variable (in this case, the density) on the y-axis.

A bar graph is the best choice for this data because it allows for easy comparison of density of different types of wood. We can see at a glance which type of wood is the densest and which type of wood is the least dense.

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The pressure of the oil with the given properties flowing in the horizontal pipe at the upper portion of the pipe is 157 kPa.

The pressure in the upper portion of the pipe is calculated by applying Bernoulli's equation,

P₁ + ¹/₂ρv₁² + ρgh₁ = P₂ + ¹/₂ρv₂² + ρgh₂

Given;

Susbtsitute the given parameters and solve the for the pressure in the upper portion of the pipe.

237,000 + ¹/₂(892)(1.84)² + (892)(9.8)(0) = P₂ + ¹/₂(892)(3.61)² + (892)(9.8)(8.63)

238,509.9776 = P₂ + 81,252.325

P₂ = 238,509.9776 - 81,252.325

P₂ = 157,257.65 Pa

P₂ ≅ 157 kPa

Thus, the pressure of the oil with the given properties flowing in the horizontal pipe at the upper portion of the pipe is 157 kPa.

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Answer:

40Kg

Explanation:

F=m×a

24N÷0.6m/s^2=40

They slide up to 0.0756 meters to the opposite direction.

What is Velocity ?

Velocity is a vector quantity that describes the rate of change of an object's position. It is equal to the displacement of an object divided by the time interval over which the displacement occurred. Velocity has both magnitude and direction, and its units are typically meters per second (m/s).

The velocity of an object can change over time due to acceleration, and if an object is moving in a circular path, it also has a component of velocity perpendicular to its direction of motion, known as centripetal velocity.

Conservation of momentum:

        60x6 -120x4 =180.v

              v = 120/180 = 2/3 m/s

After falling,

            Kinetic energy = work done against friction

             K=  0.5x180x(2/3)2 = 40 J

Work done against friction,

                       W = f.S = μ.(M +m)g.S = 0.3 x 180x9.8 x S

                            = 529.2 x S

Equating W = K and solving,

                      S = 40/529 = 0.0756 m= 7.56 cm

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Answer:

If the frequency is doubled, the wavelength is only half as long. 3. ... As the frequency slows down, the wavelength increases.

The input to the power station is 5x10^9 W. This means that the power station requires an input of 5x10^9 watts to produce an electrical power output of 1.9x10^9 watts with an efficiency of 38%.

The efficiency of a power station is defined as the ratio of its output power to input power. Therefore, we can use the efficiency and the electrical power output of the power station to calculate its input power as follows:

Efficiency = Output power / Input power

Solving for input power, we get:

Input power = Output power / Efficiency

Substituting the given values, we get:

Input power = 1.9x10^9 W / 0.38

Input power = 5x10^9 W (to two significant figures)

The rest of the input power is lost as heat due to inefficiencies in the power generation process.

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

2.5m/s

Explanation:

Using the law of conservation of energy;

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute the given values in the formula

2(10)-2(5) = (2+2)v

20-10 = 4v

10 = 4v

v = 10/4

v = 2.5m/s

Hence the velocity of the combined object after collision is 2.5m/s

Answer:

Lost Mechanical Power = 7.7565 KW

Head Loss = 26.35 m

Explanation:

First, we will find the useful mechanical power used to transport water to the higher reservoir:

\(P_{useful} = \rho ghV\)

where,

P_useful = Useful mechanical Power = ?

ρ = density of water = 1000 kg/m³

g = acceleration due to gravity = 9.81 m/s²

h = height = 45 m

V = Volume flow rate = 0.03 m³/s

Therefore,

\(P_{useful} = (1000\ kg/m^3)(9.81\ m/s^2)(45\ m)(0.03\ m^3/s)\\P_{useful} = 13243.5\ W = 13.2435\ KW\)

Now, the lost mechanical power will be:

\(Lost\ Mechanical\ Power = Total\ Mechanical\ Power - Useful\ power\\Lost\ Mechanical\ Power = 21\ KW - 13.2435\ KW\\\)

Lost Mechanical Power = 7.7565 KW

Now, for the head loss:

\(Lost\ Mechanical\ Power = \rho g(Head\ Loss)V\\Head\ Loss = \frac{Lost\ Mechanical\ Power}{\rho gV} \\\\Head\ Loss = \frac{7756.5\ W}{(1000\ kg/m^3)(9.81\ m/s^2)(0.03\ m^3/s)} \\\)

Head Loss = 26.35 m

Answer:

\(100\ cm\)

Explanation:

Given

\(Height = 120\ cm\)

\(Distance = 50\ cm\)

Required

Determine the image's distance from the object's distance

Though, it's not stated but I'll assume that we're dealing with a plane mirror.

When an image is reflected in a plane mirror, the distance of the object is the twice the distance between the mirror and the object itself.

Mathematically,

Object Distance from Image = 2 * Image Distance from Mirror

So, we have:

Object Distance from Image = 2 * 50 cm

Object Distance from Image = 100cm

Answer:

atmosphere, gravitational pull, magnetic field

Explanation:

the astroids burn up in the atmosphere, gravitational pull alters the path of the astroids, and the magnanetic field deflects them of path.

Answer:

I would agree with your selection.

Explanation:

The amount the needle on the meter is deflected A. increases

This phenomenon can be explained by Faraday's law of electromagnetic induction. According to this law, when a magnetic field (created by the bar magnet) passes through a coil of wire, it induces an electric current in the wire. This induced current generates its own magnetic field, which interacts with the magnetic field of the bar magnet.

The deflection of the meter needle is a result of this induced current. When the number of coils of wire is increased, there is a greater number of wire loops for the magnetic field to pass through. This leads to a stronger induction of electric current, resulting in a larger deflection of the meter needle.

By increasing the number of coils, more magnetic flux is linked with the wire, resulting in a higher induced electromotive force (emf) and a greater current. This increased current produces a stronger magnetic field around the wire, leading to a larger deflection on the meter. Therefore, increasing the number of coils of wire enhances the magnetic field interaction, resulting in an increased deflection of the meter needle. Therefore, Option A is correct.

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When three particles q1, q2, and q3 are in a straight line, the force experienced by q2 depends on the distance between the particles, the magnitude of the charges of the particles, and the relative position of the particles. The electric force on a charge q1 from a charge q3, which is at a distance r3 from the first charge, is governed by Coulomb’s Law.

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Answer:

100N.

Its 100n. :)

Answer:

The correct answer is D. solids and liquids because their particles are arranged more closely together.

The candidate's speed (m/s), given that the candidate experiences a centripetal acceleration that is 2.8 times the acceleration due to gravity is 17.4 m/s

We understood that the centripetal acceleration is related to speed and radius according to the following formula:

a = v² / r

Cross multiply

v² = ar

Take the square root of both sides

v = √ar

Where

Withe the above formula, we can determin the speed of the candidate. Details below:

v = √ar

v = √(27.44 × 11.05)

v = 17.4 m/s

Thus, the speed of the candidate is 17.4 m/s

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Answer: Hope this helps ;) don't forget to rate this answer !

Explanation:

There are two correct answers:

A) A corporation leader makes direct payment to the candidate.

D) A corporation contributes to a Super PAC that a PAC accepts contributions for a candidate.

Option A describes a scenario where a corporation directly donates money to a presidential candidate, which is allowed as long as it is done within the limits set by campaign finance laws.

Option D describes a scenario where a corporation donates money to a Super PAC, which is a type of political action committee that can accept unlimited donations from individuals, corporations, and other organizations. The Super PAC can then use the money to support or oppose a particular candidate, but it is not allowed to coordinate directly with the candidate or the candidate's campaign.

I hope this helps! Let me know if you have any other questions.

Answer:

ANSWERS ARE 100 % correct

Explanation:

trust me