5. a hockey puck struck by a hockey stick is given an initial speed of 10 m/s. if the coefficient of kinetic friction between ice and puck is 0.10, how far will the puck slide before stopping?

Answer:

Every summer we go and work at a summer camp. My whole family loves it, and it's something that always brings us closer.

Explanation:

Christmastime is without a doubt my favorite time of year, not because of Christmas itself, but because of the atmosphere leading up to it. I've always felt this way, and I believe that is because of my grandmother. My Nanny was the same way. She loved Christmas, and went all out every year. Her house was put in the local newspaper several times because of all the lights and decorations she put up. If only they could have gone inside the house!

Anyway, there are a lot of things I did with my grandma around Christmas, and if I had been asked my favorite tradition when I was a young child, I might have said “decorating,” or “making Christmas treats,” but looking back, it is what we would do every Christmas Eve.

Christmas Eve is last-minute shopping day for the adults of my family, and the only shopping day for the kids. I would head out with my dad in the morning and shop, then go to lunch with my grandpa, and then finally, I would go with my grandma. We would shop a little, and then she would always buy some fake poinsettias. We would then travel to a tiny cemetery that seemed like an hour away back then, and I would walk with her to her father's grave.

Even as a young child, I felt the emotions in the air when we visited his grave. She never cried or showed any signs of sadness, but I could just sense how special the moment was to her. When her mother passed away in 2013, we went to visit both of them, and the atmosphere only intensified.

My grandma died in March of 2015 after a long battle with cancer. Every Christmas Eve, I buy the most beautiful poinsettias I can find, and visit her grave. It's always freezing outside, being December, but I stand for a few minutes, remembering, and then I get back in my car and head home.

My fondest tradition isn't the most fun, or the most festive. It's the one that means the most to me. One day, I hope I can take my children or grandchildren with me, and I hope they'll feel everything that I felt.

To calculate the density of a substance, its mass and volume has to be determined. So that:

i. For the milk in the carton, its mass and volume should be known.

ii. For the two objects given in the question, their individual masses and volumes have to be determined.

Density of a substance (ρ) is the rate of the mass (m) of the substance to its volume (V). It is measure in kg/\(m^{3}\).

i.e ρ = \(\frac{m}{V}\)

The density of a substance can be determined once the mass and its volume have been accurately measured.

1. How to measure the density of the contents of a milk carton.

a. Determine the mass of the milk carton with its content.

b. Determine the mass of the milk carton when empty.

c. Subtract the mass of the milk carton when empty from the mass of the carton with its content. This will give the mass of the milk in the carton.

d. Using a beaker, measure the volume of the milk.

e. Divide the mass of the milk by its volume to determine its density.

2. How to measure the densities of two objects of equal mass.

a. Fill an Eureka's can or a beaker with sprout with water to its brim.

b. Slowly drop the mass into the beaker with water so as to collect the volume of the liquid displaced into another beaker.

c. The volume of the liquid displaced is the volume of the mass.

d. Divide the mass of the object by its volume to determine its density.

NB: These procedures are observed to determine the densities of the two object individually.

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Answer:

The equivalent resistance of the combination is R/100

Explanation:

Parallel Connection of Resistances

If resistances R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

\(\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}\)

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

It's known the 10 parts or resistance R/10 were connected in parallel, thus the electric resistance is:

\(\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}+...+\frac{1}{R/10}\)

Note the sum consists of 10 equal terms. Operating on each term:

\(\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}\)

The sum of 10 identical fractions yields 10 times each fraction:

\(\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}\)

Solving for Re needs to take the reciprocal of both sides of the equation:

\(R_e=R/100\)

The equivalent resistance of the combination is R/100

2.6×10\(^-8\) Ωm,A long, flat wire is used as a rear window defroster and is bonded to the window's interior surface.

Electrical resistivity is a measurement of a material's degree of resistance to current flow. The SI unit for electrical resistivity is the ohm meter (m). It is frequently represented by the Greek letter rho. Materials that easily transmit current and have a low resistance are called conductors.

Length of the wire, l = 12.2 m

width of the wire, w = 1.8 mm

Thickness of the wire, T = 0.11 mm

Potential difference of the battery, v = 12 V

Current in battery, I = 7.5 A

Ohms law says, V = IR.

So that, R = V/I

R = 12/7.5 = 1.6 Ω

The Greek letter rho is frequently used to represent resistance, which is numerically equivalent to the resistance R of a wire-like specimen, multiplied by its cross-sectional area A, and divided by its length.

Resistivity if a material is:

ρ = RA/l

ρ = [1.6 × 1.8 × 10\(^-3\) × 0.11×10\(^-3\)] / 12.2

ρ = (3.168×10\(^-7\)) / 12.2

ρ = 2.596×10\(^-8\)

Therefore, the resistivity of the wire is 2.60×10\(^-8\) Ωm

Resistivity of the wire is 2.60*10^-8 Ωm

Given:

Length of the wire, I = 12.2 m

width of the wire, w = 1.8 mm

thickness of the wire, T = 0.11 mm

Potential difference of the battery, v = 12

Current in the battery, I = 7.5 A

To Find:

Resistivity

Solution: A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents.

Remember, Ohms law says, V = IR

So that, RV/I

R = 12/7.5= 1.6 Q

Resistivity if a material is

p = RA/I

p = [1.6 * 1.8*10^-3* 0.11*10^-3] / 12.2

p = (3.168*10^-7)/12.2

p = 2.596*10^-8

Therefore, the resistivity of the wire is 2.60*10^-8 Ωm

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The capacitance of a parallel plate capacitor is proportional to the area of the plates. Hence option C is correct.

A capacitor is a device that stores electrical energy in an electric field by collecting electric charges on two isolated surfaces. It is a passive electronic component with two terminals.

Capacitance is the effect of a capacitor. While any two electrical conductors in close proximity in a circuit have some capacitance, a capacitor is a component designed to provide capacitance to a circuit. The capacitor was initially referred to as a condenser, a word that is still used in a few compound names, such as the condenser microphone.

Capacitance of the capacitor is directly proprtional to the area of the plates and inversly proprtional to the distance between them.

C = ∈A/d

Hence option C is correct.

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The frequency that the bad guy hear is 12000 hz when the police car is moving with speed of 80m/s.

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Answer:

945 joules

Explanation:

the formula for kenetic energy is:

0.5 × m × v^2

where m is in (kg) ane (v) is in meter per second.

so we can just use the formula;

0.5 × 2.1 × 30^2

so the answer is 945 joules

The main cause of melting along subduction zones is the d. release of water from the subducting plate.

Subduction zones are areas where one tectonic plate moves beneath another, causing the denser plate to sink into the mantle. This process generates a significant amount of heat, which contributes to the melting of rocks in the mantle lithosphere.
As the subducting plate moves deeper into the mantle, it experiences increasing pressure and temperature. The minerals within the subducting plate contain water, which is released as the plate is subjected to these extreme conditions. This released water reduces the melting point of the surrounding mantle rocks, causing them to partially melt.

This partial melting creates magma, which can rise through the mantle lithosphere and eventually reach the Earth's surface, resulting in volcanic activity. The release of water from the subducting plate, therefore, plays a crucial role in generating the magma that leads to volcanic eruptions along subduction zones.
In summary, the main cause of melting along subduction zones is the d. release of water from the subducting plate, which lowers the melting point of surrounding mantle rocks and generates magma. This magma can rise through the mantle lithosphere, causing volcanic activity in these regions.

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The main cause of melting in subduction zones is the release of water from the subducting plate, which lowers the melting temperature of the surrounding rocks and causes them to melt.

The main cause of melting along subduction zones is primarily the release of water from the subducting plate (option d). When the oceanic lithosphere subducts, it carries with it water that has been trapped in the minerals of the crust and upper mantle. This water lowers the melting temperature of the surrounding rocks, causing them to melt and form magma. This is termed 'flux melting'. For example, the subduction of the Pacific Plate beneath the North American Plate in the Cascadia subduction zone causes intense volcanic activity in the Pacific Northwest.

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Answer:

Gravity

Explanation:

After it reaches it's highest point, it falls back down because of gravity.

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

\(a = \frac{f}{m} \\ \)

f is the force

m is the mass

From the question we have

\(a = \frac{250}{70} = 3.5714 \\ \)

We have the final answer as

Hope this helps you

the ball and beam system is a useful tool for studying nonlinear systems and developing control algorithms for various applications.

The ball and beam system is a classic example of a nonlinear equations system that is commonly used in control engineering. This system consists of a ball rolling along a beam, where the position of the ball is controlled by adjusting the angle of the beam. The nonlinear equations that describe this system involve the force of gravity, the angle of the beam, the position and velocity of the ball, and other factors. The content loaded on the ball and beam set can include sensors, actuators, and other components that are used to control the behavior of the system.

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Answer:

40000 N/m²

Explanation:

Applying,

P =  F/A................... Equation 1

Where P = Pressure, F = Force, A = Area.

From the question,

The force(F) exerted by the person's foot is thesame as it's weight.

F = W = mg............ Equation 2

Where m = mass of the person, g = acceleration due to gravity.

Substitute equation 2 into equation 1

P = mg/A................ Equation 3

Given: m = 60 kg, g = 10 m/s², A = 150 cm² = (150/10000) m² = 0.015 m²

Substitute these values into equation 3

P = (60×10)/0.015

P = 600/0.015

P = 40000 N/m²

The magnitude of the resultant force on the elliptical back will be equal to the difference between gravity and normal force.

The weight of the water and the force of gravity pressing on it determine the size of the resulting force on the elliptical tank truck. The water in the tank truck serves as a weight, creating a force of gravity proportionate to its mass. This force acts vertically downwards and is equal to the weight of the water, which can be estimated by multiplying the mass of the water by the gravity acceleration (g = 9.8 m/s^2).

When the tank truck is half full, the weight of the water may be estimated by multiplying the volume of water by its density and gravity acceleration. The volume of water may be calculated by taking the volume of the elliptical tank and dividing it by half. Water's density is constant and equal to 1000 kg/m3.

The total of the vertical components of gravity and normal force equals the size of the resulting force on the elliptical tank truck. The normal force is equal to the weight of the water and acts vertically upwards, against the force of gravity.

The amount of the resulting force on the elliptical back may be calculated by calculating the weight of the water, determining the force of gravity operating on it, and taking into account the normal force acting in the opposite direction.

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a) On a wet basis, the molar composition of the gas is approximately 0.813 mol propane, 0.055 mol n-butane, and 0.132 mol water. The ratio of mol H₂O to mol dry gas is 0.162 mol H₂O/mol dry gas.

b) The required air feed rate is approximately 65.9 kmol/h. If the combustion were only 65% complete, the required air feed rate would increase to approximately 101.4 kmol/h.

a) To calculate the molar composition on a wet basis, we convert the weight percentages to mole fractions using the molar masses of propane, n-butane, and water. The molar composition is determined by dividing the weight percentage by the respective molar mass and normalizing the values to sum up to 1. The ratio of mol H₂O to mol dry gas is determined by dividing the mol water by the sum of mols of propane and n-butane.

b) To calculate the required air feed rate, we use the stoichiometry of the combustion reaction between butane and air. The balanced equation shows that 1 mol of butane reacts with 13.5 mol of air. Considering the 25% excess air requirement, we multiply the stoichiometric air requirement by 1.25. If the combustion is only 65% complete, the remaining butane requires additional air to achieve complete combustion. Therefore, the required air feed rate increases to account for the unreacted butane.

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If we were to make copies of musical content up to the 6th time, Mp3 players would have better sound quality because they have digital signals to produce sound, which do not deteriorate over time.

Cassette tapes work on analog magnetic tape recording format for recording audio and playback whereas mp3 is a type of digital technology that has digital signals to produce sound. This is the reason why cassettes do not sound better than digital mp3 players.

Cassette tapes being magnetic type media suffer from print which causes to hear a bit of sound before the sound is even started and result in deterioration over time whereas digital signals do no deteriorate over time.

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Answer:

Newtons law of gravity I think

Answer:

As per Provided Information

we have been asked to determine the force applied to stop the car. First we will calculate the acceleration of the car.

\( \boxed{\bf \: a \: = \dfrac{(v - u)}{t}}\)

Substituting the value and let's solve it

\( \longrightarrow\sf \: a \: = \dfrac{0 - 13}{5} \\ \\ \\ \longrightarrow\sf \: a \: = \cfrac{ - 13}{5} \\ \\ \\ \longrightarrow\sf \: a \: = - 2.6 \: {ms}^{ - 2} \)

Now, let's calculate the force applied to stop the car .

\( \pink{\boxed{\bf \: F = ma}}\)

Substituting the value we get

\( \longrightarrow \sf \: F = 1400 \times ( - 2.6) \\ \\ \\ \longrightarrow \sf \: F = - 3640 \: N\)

Here , negative sign show that the " Force is acting in opposite direction of the motion"

Therefore,

Answer:

d. 325 kg • m/s

Explanation:

The average speed during the acceleration is 23.95 m/s.

The average speed of the car during the time it takes to accelerate can be calculated as the average of the initial and final speeds.

Speed is equal to the total distance covered by the car divided by the time required to cover the distance.

As per the given information,

The initial speed of the car is 13.5 m/s, and

the final speed is 34.4 m/s.

Therefore, the average speed can be calculated as:

Average speed = (initial speed + final speed) / 2

Average speed = (13.5 m/s + 34.4 m/s) / 2

Average speed = 23.95 m/s

Rounding to two decimal places, the average speed during the acceleration is 23.95 m/s.

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The efficiency of the engine is 20%, and the total work per hour of a four-cylinder engine operating at 1800 rpm cannot be determined without additional details about the engine's specifications.

Engine efficiency is a measure of how effectively an engine converts input energy (in this case, heat) into useful work. In this scenario, the engine requires 5500 J of input heat to perform 1100 J of work. To calculate the efficiency, we divide the work output by the input heat and multiply by 100 to express it as a percentage. Therefore, the efficiency of the engine is (1100 J / 5500 J) * 100 = 20%.

Determining the total work per hour of a four-cylinder engine operating at 1800 rpm requires additional information. The total work per hour depends on various factors, including the displacement of the cylinders, the stroke length, and the efficiency of the engine. Without knowing these specifics, it is not possible to provide an accurate calculation. The engine's specifications, such as the displacement (in liters or cubic centimeters) and the stroke length (in millimeters or inches), are essential to estimate the total work output. Additionally, the engine's efficiency plays a significant role in determining the overall work per hour.

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"Planting more trees brings rain because plants transpire" is the piece of evidence from the page that support the idea that forest promote rainfall.

Water is moved through a plant during transpiration, and it is evaporated from aerial parts of the plant, such as leaves, stems, and flowers. Even though plants need water to survive, only a small portion of the water absorbed by their roots is actually utilised for cellular growth and metabolism.

Transpiration and guttation cause the remaining 97-99.5% to be lost. In most plants, the undersides of the foliage have more of the tiny pores known as stomata that decorate the surface of leaves.

The guard cells that surround the stomata are responsible for opening and closing the pore, as are their stomatal accessory cells. Transpiration occurs through the stomatal apertures and is regarded as a "cost" of the stomata opening, which allows the diffusion of carbon dioxide gas from the atmosphere for photosynthesis.

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Answer:

All of the above is the answer

Answer:

and the power to which the number should be raised:

1.50E2 = 1500

4.167E4 = 41,760

4.167E-4 = .0004176

Mass of the earth = 5.98 * 10^24 kg = 5.98E24 kg

Some examples of scientific notation

Answer:

hey Army

I'm B.T.S Army

안녕하세요

Answer:

Electric current creates a magnetic field that charges the rod.

Explanation:

Answer:

a) 6498.84 kW

b) 0.51

c) 0.379

Explanation:

See the attached picture below for the solution

Given-

Temperature at state 1 \(T_{1}\) is 40 degree cel. This is equal to the 313 K.

Temperature at state 1 \(T_{2}\) is 650 degree cel. This is equal to the 923 K.

To know the mass flow rate use the idol gas equation. Idol gas equation for a substance can be given as,

\(PV=nRT\)

Rewrite this equation for \(n\),

\(n=\dfrac{PV}{RT}\)

Put the value of known variable from the question,

\(n=\dfrac{100\times\dfrac{850}{60} }{0.287\times 313}\)

\(n=15.77\)

Thus the value of n is 15.77 kg/sec.

Temperature at state 2 is,

In Bryon cycle we know that the processes 1-2 and 3-4 are isotropic. Hence,

\(T_{2} =T_{1}\left (1+ \dfrac{r_{p}^{\dfrac{\gamma-1}{\gamma}-1} }{c_{v} } \right )\)

\(T_{2} =313\left (1+ \dfrac{16^{\dfrac{1.35-1}{1.35}-1} }{0.821 } \right )\)

\(T_{2} =700\)

Temperature at state 3 is,

\(T_{3} =\dfrac{T_4}{1+c_p(r_p^{\dfrac{\gamma-1}{\gamma}-1}-1)}\)

Put the value in above equation we get,

\(T_3=1682\)

\(W=mc_p(T_i{n}-T_{out})\)

\(W=mc_p(T_3-T_2-T_4+T_{1})\)

\(W=15.77\times1.109(1682-700-923+313)\)

\(W=65\)

Thus the net power output is 70 watt.

\(r_b=\dfrac{T_2-T_1}{T_3-T_4}\)

\(r_b=\dfrac{700-313}{1682-923}\)

\(r_b=0.51\)

Hence the back work ratio is 0.51.

\(\eta=\dfrac{W}{Q_{in}}\)

\(\eta=\dfrac{65}{mc_p(T_3-T_2)}\)

\(\eta=0.379\)

Hence, for the given problem the net power output is 70 watt,the back work ration is 0.51, and the thermal efficiency is 0.379.

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The acceleration that the same net force would give to an 18-kg tool is 13.3 m/s² (meters per second squared).

According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass.

Mathematically, this can be expressed as F = ma, where F is the net force, m is the mass, and a is the acceleration.

In the given scenario, the net force is accelerating a 4.8-kg tool at 40 m/s². To find the acceleration for an 18-kg tool, we can rearrange the equation to solve for a:

a = F/m

Substituting the values, we have:

a = (F)/(18 kg) = (40 m/s²)/(18 kg) = 2.22 m/s²

Therefore, the net force would give an acceleration of 2.22 m/s² to an 18-kg tool.

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When the Earth is at point C, Equinox happens.

In Equinox, the Sun is directly above the equitor making day and night equal in both hemispheres.

Answer:

A constant velocity of an object ensures that the rate of change of velocity with time is null, and hence, the acceleration of the object is zero. A constant acceleration of an object ensures that the velocity of the object is changing continuously with time, and the velocity will not be constant.

Explanation:

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True. The Drake equation is a mathematical formula developed by astronomer Frank Drake in 1961 to estimate the number of intelligent, communicating civilizations that may exist in our Milky Way galaxy. The equation takes into account several factors such as the number of stars that could potentially host habitable planets, the likelihood of those planets developing life, and the probability of intelligent life emerging on those planets and developing the ability to communicate with other civilizations.

While the exact values of these factors are uncertain, the Drake equation remains an important tool in the search for extraterrestrial life and our understanding of the potential prevalence of intelligent civilizations in our galaxy.

True, the Drake Equation is developed to estimate how many intelligent, communicating civilizations there are in our galaxy. The equation was proposed by Dr. Frank Drake in 1961, and it takes into account various factors such as the rate of star formation, the fraction of stars with planetary systems, the number of planets that can potentially support life, and the likelihood of intelligent life developing and being able to communicate.

By combining these factors, the Drake Equation provides an estimate of the number of civilizations in our galaxy with which we might be able to communicate. However, it is important to note that the equation is based on many assumptions and uncertainties, and therefore the results can vary widely depending on the input values chosen.

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