what is the speed of sound in air if it takes 2.95 seconds to hear an echo from a canyon wall that is 569.71 m away? (be careful!!!)

Answer:

when a car transforms...

Explanation:

Explanation:

Vivimos en un puente en lo que el diseño de coches se refiere, consecuencia de la gran disrupción de la automación que ocurriría en los próximos años.

The recommended minimum length of time that water should remain motionless in service lines prior to first-draw residential lead sampling is 6 hours.

Lead can leach into drinking water from the service lines and plumbing fixtures, particularly in older homes that may have lead pipes or lead-based solder. When water sits stagnant in these pipes for a period of time, such as overnight or during the day when no one is home, the lead particles that have accumulated in the plumbing system can dissolve into the water. This is why it's important to collect first-draw samples after a period of stagnation.

The EPA recommends a 6-hour stagnation period for collecting first-draw samples from residential plumbing systems because this is typically the longest period of time that water remains stagnant in home plumbing systems. This means that the water has been sitting in the pipes long enough to allow any lead particles to leach into the water, but not so long that the water quality may be affected by other factors, such as microbial growth or chemical reactions.

It's important to note that first-draw samples are used to identify the presence of lead in the plumbing system, but they may not be representative of the actual exposure to lead that a person may experience. This is because the lead concentration in the water can vary depending on factors such as the age and condition of the plumbing system, the water chemistry, and the length of time that water has been sitting in the pipes.

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They are useful in providing larger frictional force.

The force produced by two surfaces coming into contact and sliding against one another is referred to as frictional force.

The following variables impact the frictional force: Surface roughness and the amount of force pressing them together have the biggest impact on these forces.

Football shoes have spikes or studs because they offer more frictional force when running on grass than regular shoes do.

The studs make it easier for players to move faster and change directions rapidly without slipping while also preventing them from slipping on the grass.

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The probe is moving at a speed of 0.925c in relation to the earth.

Relative speed We frequently stand on the Earth (or an object attached to the Earth) to measure speeds. A physicist might say that the speed is measured in relation to the Earth or in the Earth frame of reference.

In school, we are taught that the earth orbits the sun in a very nearly round path. The pace at which it travels along this trip is close to 30 kph, or 67,000 mph.

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They have a limited supply in nature, therefore if they are used excessively, they will become exhausted.

Today, we recognise that using fossil fuels has a negative impact on the environment. Fossil fuels produce and utilise local pollutants, and their continued use permanently alters the temperature of our entire world.

Wastes from combustion sources are those that result from carbon pollution (i.e., coal, oil, natural gas). Included in this are all ash and particles taken out of the flue gas.

The fossile fuel is limited in nature. So, it should not waste energy from fossil fuels.

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Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

\(x = \frac{-F}{K}\)

Now  

\(x = \frac{-3500}{14000} \\\\\)

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm

Answer:

because it is due the Intrecation between sub atomic particles

Answer:

Explanation:

We don't need the mass of the car in the equation to solve for final velocity, since the values given for the acceleration and the time it took to accelerate to that velocity are given. The equation we need is the one for acceleration, which is

\(a=\frac{v_f-v_0}{t}\) We are solving for final velocity, we know the initial velocity is 0 (starting from rest), and the time to complete this acceleration (10 m/s/s) is 3 seconds:

\(10=\frac{v_f-0}{3}\) which is the same thing as saying

\(10=\frac{v}{3}\) so

v = 30 m/s

The vast majority of pulsars are known only from their pulses in radio waves. While pulsars emit radiation across the electromagnetic spectrum, it is primarily the radio waves that have been widely used for their detection.

Pulsars are highly magnetized, rotating neutron stars that emit beams of electromagnetic radiation, including radio waves, X-rays, and gamma rays. These beams of radiation are detected on Earth as regular, periodic pulses, giving pulsars their name.

While pulsars emit radiation across the electromagnetic spectrum, it is primarily the radio waves that have been widely used for their detection. Radio telescopes are sensitive to the long-wavelength radio emissions from pulsars, allowing astronomers to observe and study their pulsating signals.

The detection and identification of pulsars are primarily based on the analysis of their radio wave pulses. Astronomers search for periodic signals in the radio data, looking for distinctive patterns that indicate the presence of a pulsar. By analyzing the timing and characteristics of the pulses, astronomers can determine various properties of the pulsar, such as its rotation period, magnetic field strength, and sometimes even its motion and distance.

Therefore, the vast majority of pulsars are known only from their pulses in radio waves, as these have been the most commonly observed and studied emissions from pulsars.

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Answer:

A: 2.8 * 10^(-14) J

Explanation:

Formula to find the energy is;

E = hc/λ

Where;

E is energy of a photon (E)

λ is wavelength of the light

c is speed of light

Thus;

E = (1.99 *10^(-25))/(7 * 10^(-12))

E = 2.8 * 10^(-14) J

The category of selected atoms belonging to a certain specified elements will initiate taking , giving, or sharing of electrons with other atoms for  completing a set of electrons in their outer energy level are considered to be reactive or chemically active element.

The elements present in the lower left corner of the periodic table are the considered the most active metals in terms of reactivity. Lithium,potassium all will effectively react with the water.
Elements which can react quickly or strongly with other elements because they perform simple distribution of electrons are known as active elements.
For instance , fluorine is known as the most chemically active element and effectively react with noble or inert gases like  argon, neon, krypton and helium.
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Part A:

Radius of flywheel = 0.600 m

Angular acceleration of flywheel (αₐ) = 0.200 rad/s²

The tangential acceleration of a point at the start (αₓ) =

= αₓ = αₐ × r

= αₓ = 0.200 × 0.600

= αₓ = 0.12 m/s²

Part B:

Radius of flywheel = 0.600 m

Angular acceleration of flywheel (αₐ) = 0.200 rad/s²

Angular speed = ω = 0 m/s²

Magnitude of radial acceleration of a point on rim at the start (αₙ)=

= (angular speed)² × r

= 0 × 0.600

= 0 m/s²

Part C:

Radius of flywheel = 0.600 m

Angular acceleration of flywheel (αₐ) = 0.200 rad/s²

Resultant acceleration of a point on the rim at the start =

= α =√(αₙ² + αₓ²)

= α = √ (0² + 0.12²)

= α = 0.12 m/s²

Part D:

Radius of flywheel = 0.600 m

Angular acceleration of flywheel (αₐ) = 0.200 rad/s²

Angular speed = ω = 0 m/s²

The tangential acceleration of a point after 60° turn (αₓ₁) = The tangential acceleration of a point at the start (αₓ)

= αₓ₁ = αₐ × r

= αₓ₁ = 0.200 × 0.600

= αₓ₁ =  0.12 m/s²

Part E:

Radius of flywheel = 0.600 m

Angular acceleration of flywheel (αₐ) = 0.200 rad/s²

Angular speed = ω = 0 m/s²

Angular speed after 60° turn = ω₁ = √(ω² + (2×α×θ))

To find θ,

= θ = 60Π / 180

= θ = Π/30

= θ = 1.04 rad

Thus, ω₁ = √(0 + 2 × 1.04 × 0.2)

= ω₁ = 0.644 rad/s

The radial acceleration of a point after 60° turn (αₓ₂) =

= αₓ₂ = r × ω₁²

= αₓ₂ = 0.600 × 0.644²

= αₓ₂ = 0.248 m/s²

Part F:

Radius of flywheel = 0.600 m

The tangential acceleration of a point after 60° turn (αₓ₁) = 0.12 m/s²

The radial acceleration of a point after 60° turn (αₓ₂) = 0.248 m/s²

The magnitude of resultant acceleration of a point on the rim after 60° turn (α₃) =

= α₃ = √ (αₓ₂² + αₓ₁²)

= α₃ = √ (0.12² + 0.248²)

= α₃ = 0.275 m/s²

Part G:

Radius of flywheel = 0.600 m

Angular acceleration of flywheel (αₐ) = 0.200 rad/s²

Angular speed = ω = 0 m/s²

The tangential acceleration of a point after 120° turn (αₓ₁) = The tangential acceleration of a point at the start (αₓ)

= αₓ₁ = αₐ × r

= αₓ₁ = 0.200 × 0.600

= αₓ₁ =  0.12 m/s²

Part H:

Radius of flywheel = 0.600 m

Angular acceleration of flywheel (αₐ) = 0.200 rad/s²

Angular speed = ω = 0 m/s²

Angular speed after 120° turn = ω₁ = √(ω² + (2×α×θ))

To find θ,

= θ = 120Π / 180

= θ = 2Π/3

= θ = 2.09 rad

Thus, ω₁ = √(0 + 2 × 2.09 × 0.2)

= ω₁ = 0.836 rad/s

The radial acceleration of a point after 120° turn (αₓ₂) =

= αₓ₂ = r × ω₁²

= αₓ₂ = 0.600 × 0.836²

= αₓ₂ = 0.502 m/s²

Part I:

Radius of flywheel = 0.600 m

The tangential acceleration of a point after 120° turn (αₓ₁) = 0.12 m/s²

The radial acceleration of a point after 120° turn (αₓ₂) = 0.502 m/s²

The magnitude of resultant acceleration of a point on the rim after 120° turn (α₃) =

= α₃ = √ (αₓ₂² + αₓ₁²)

= α₃ = √ (0.12² + 0.502²)

= α₃ = 0.515 m/s²

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The horizontal acceleration of the softball after she lets go of the ball is  0 m/s².

option B is the correct answer

During a projectile motion, the vertical acceleration of a projectile decreases as the projectile moves upwards and eventually become zero as the object reaches the maximum height.

As the object begins move downwards, the vertical velocity of the projectile increases and eventually becomes maximum before the object hits the ground.

During the horizontal motion of a projectile, the horizontal velocity of the projectile does not change. That is the initial horizontal velocity of the projectile equals the final horizontal velocity of the projectile.

a = v - u/t

where;

Since, final horizontal velocity = initial horizontal velocity

v = u

a = 0

In summary, the vertical velocity of the a projectile changes while the horizontal velocity does not change. This results in a zero horizontal acceleration of the projectile.

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(a) The intensity of the helium-neon laser beam projected onto a circular spot with a diameter of 1.00 mm is approximately 796.84 W/m².

To calculate the intensity, we need to convert the power output from milliwatts to watts and then divide it by the area of the circular spot.

Intensity = Power / Area

Area = π * (radius)²

= π * (diameter/2)²

= π * (0.5 mm)²

Intensity = 0.250 mW / (π * (0.5 mm)²)

Using appropriate unit conversions, we find:

Intensity ≈ 796.84 W/m².

(b) The peak magnetic field strength of the laser beam can be calculated using the relationship between intensity and magnetic field strength. However, without the specific information regarding the wavelength of the laser beam, we cannot provide a numerical value for the peak magnetic field strength.

(c) Similarly, without the specific information about the wavelength, we cannot provide a numerical value for the peak electric field strength of the laser beam.

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Answer:

I think it's 3/4 or 25% i am not sure

Answer:

Approximately \(2.1\; \rm km\), assuming that \(g = -9.8\; \rm m \cdot s^{-2}\).

Explanation:

Let \(t\) denote the time required for the package to reach the ground. Let \(h(\text{initial})\) and \(h(\text{final})\) denote the initial and final height of this package.

\(\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})\).

For this package:

Solve for \(t\), the time required for the package to reach the ground after being released.

\(\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}\).

\(\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}\).

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at \(95\; \rm m \cdot s^{-1}\).) From calculations above, the package would travel forward at that speed for about \(22.588\; \rm s\). That corresponds to approximately:\(95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km\).

Hence, the package would land approximately \(2.1\; \rm km\) in front of where the plane released the package.

A BTU is the amount of energy required to raise the temperature of one pound of water by one degree Celsius: False

The energy needed to raise the temperature of one pound of water by one degree Fahrenheit is known as a BTU.

British thermal units are used to assess the capacity of the majority of air conditioners (BTU). One pound (0.45 kilos) of water needs one BTU of heat to increase its temperature by one degree Fahrenheit (0.56 degrees Celsius). One tonne is 12,000 BTU in terms of heating and cooling.

Heat is measured in British thermal units. One BTU is about 1055 joules (J), the joule being the current SI unit for heat energy.  This represents the fact that the temperature change of a mass of water caused by the injection of a particular quantity of heat (measured in energy units, often joules) varies somewhat depending upon the water's beginning temperature.

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Answer:

Explanation:

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

force = mass × acceleration

From the question we have

force = 2 × 5

We have the final answer as

Hope this helps you

Answer:

B

Explanation:

The fault will eventually build mountians given time.

Please find attached photograph for your answer. Hope it helps. Please do comment

The total vapor pressure of the solution is (c) 99.3 mmHg.

A substance's or an object's temperature is a measurement of how hot or cold it is. It is a characteristic of matter that has to do with the typical kinetic energy of the atoms or molecules that make up the thing or material. Normally, temperatures are expressed in either degrees Celsius (°C) or degrees Fahrenheit (°F).

Raoult's law, because substance A's mole fraction is 0.35, it follows that substance B's mole fraction must be 0.65. (since the two mole fractions must add up to 1). The partial pressures of A and B in the solution can be determined using Raoult's law:

A's partial pressure is calculated as 0.35 x 87 mmHg, or 30.45 mmHg.

B's partial pressure is equal to 79.3 mmHg (0.65 x 122 mmHg).

The partial pressures of A and B are added to determine the solution's overall vapor pressure:

Total vapor pressure equals partial pressures of A and B, or 30.45 mmHg, 79.3 mmHg, and 109.75 mmHg respectively.

Rounding to the nearest tenth, we obtain 109.8 mmHg, which is the value that is most similar to option (c) 99.3 mmHg. Hence, the correct response is (c) 99.3 mmHg.

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Answer: 472.556m⋅s−1

Explanation: λ=7.83cm=7.83⋅(1cm)=7.83⋅(1cm×0.01m1cm)=0.0783m

Answer:

50 cm3 of 1M hydrochloric acid is a six-fold excess of acid. In this reaction, the magnesium and acid are gradually used up. However the acid is in excess, so it is mainly the loss of magnesium (surface area becomes smaller) that causes the change in the rate.

Explanation:

The equation for the reaction is: magnesium + hydrochloric acid → magnesium chloride + hydrogen

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Students follow the rate of reaction between magnesium and the acid, by measuring the amount of gas produced at 10 second intervals.

3 cm of magnesium ribbon typically has a mass of 0.04 g and yields 40 cm3 of hydrogen when reacted with excess acid. 50 cm3 of 1M hydrochloric acid is a six-fold excess of acid.

In this reaction, the magnesium and acid are gradually used up. However the acid is in excess, so it is mainly the loss of magnesium (surface area becomes smaller) that causes the change in the rate.

If a graph of volume (y-axis) against time (x-axis) is drawn, the slope of the graph is steepest at the beginning. This shows that the reaction is fastest at the start. As the magnesium is used up, the rate falls. This can be seen on the graph, as the slope becomes less steep and then levels out when the reaction has stopped (when no more gas is produced).

The reaction is exothermic, but the dilute acid is in excess and the rise in temperature is only of the order of 3.5˚C. There is some acceleration of the reaction rate due to the rise in temperature. Some students might notice the flask becoming slightly warm and they could be asked how this would affect the rate of reaction, and how they might adapt the experiment to make it a ‘fair test’.

Additional information

This is a resource from the Practical Chemistry project, developed by the Nuffield Foundation and the Royal Society of Chemistry. This collection of over 200 practical activities demonstrates a wide range of chemical concepts and processes. Each activity contains comprehensive information for teachers and technicians, including full technical notes and step-by-step procedures. Practical Chemistry activities accompany Practical Physics and Practical Biology.

To solve this we must be knowing each and every concept related to Le Chatelier′s Principle. Therefore, when concentration of hydrochloric acid was changed. the concentration of product will also change.

When a stress is given to a chemical system in equilibrium, the equilibrium shifts to alleviate the tension, according to Le Chatelier′s Principle. In other words, it can anticipate the outcome of a chemical reaction with response to changes in temperature, concentration, quantity, or pressure.

While Le Chatelier's concept can be used to anticipate the reaction to a change from equilibrium, it doesn't explain why the system behaves as it does (at the molecular level).

Mg(s) + 2 HCl(aq) \(\rightarrow\) MgCl\(_2\)(aq) + H\(_2\)(g)

According to  Le Chatelier′s Principle, when concentration of hydrochloric acid was changed. the concentration of product will also change.

Therefore, when concentration of hydrochloric acid was changed. the concentration of product will also change.

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The capacitance of the parallel-plate capacitor is 7.44 pF when connected to a 3.0-V battery.

The capacitance of a parallel-plate capacitor is given by:

C = ε₀A/d

where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

In this problem, we have:

A = \(2.10 cm^2 = 2.10 × 10^-4 m^2\)

d = 2.50 mm = \(2.50 × 10^-3 m\)

The permittivity of free space is:

ε₀ = \(8.85 × 10^-12 F/m\)

We can now substitute these values into the equation to find the capacitance:

C = ε₀A/d = \((8.85 × 10^-12 F/m) × (2.10 × 10^-4 m^2) / (2.50 × 10^-3 m) = 7.44 × 10^-12 F\)

To convert this value to picofarads (pF), we can use the fact that 1 pF = \(10^-12\) F:

C = \(7.44 × 10^-12 F\)= 7.44 pF (to two significant figures)

Therefore, the capacitance of the parallel-plate capacitor is 7.44 pF when connected to a 3.0-V battery.

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The kinetic energy of the proton when it reaches point B is +1.92x10^-7 J (option D) based on the electric potential difference between A and B in the given electric field.

When the proton moves against the electric field from point A to point B, its potential energy decreases and is converted into kinetic energy. The electric potential difference (ΔV) between A and B can be calculated as ΔV = -E * d, where E is the electric field magnitude and d is the distance between A and B. Plugging in the values, ΔV = -95 N/C * 0.0075 m = -0.7125 V. As the proton starts from rest, its initial potential energy is zero. Therefore, the final kinetic energy is equal to the magnitude of the electric potential difference, which is 0.7125 J (option D).

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Diameter = 3 cm

Length = 40 cm

Density = 2170 kg/m³

B = 3T

r = 12 cm

Net magnetic field on the wire 1 is

B₁ = B + (μ₀I / 2πr)

B₁ = 3.0000125 T

Net magnetic field on the wire 2 is\

B₂ = B + (μ₀I / 2πr)

B₂ = 3.0000125 T

By newton's second law,

ilB = T sin θ

(πd²l/4)lg = T cos θ

So, tension force on the thread is

T = (ilB₁)₂ + [ (πd²l/4) lg ]²

T = 11.7261398 N

So, sin θ = (ilB₁/T)

θ = 50.13°

Now, the angle created by two threads is

θ = 2θ = 2 x 50.13 = 100.26°

Therefore, the magnetic field wire conducts electricity at a particular angle.

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Answer:

R=60Ω

Explanation:

You are looking for the total resistance of a parallel circuit so the formula is

R=R1+R2+R3, where R1=10Ω, R2=20Ωand R3=30Ω

R=10Ω+20Ω+30Ω

R=60Ω

Answer:

False.

Explanation:

First off, the Earth orbits the sun. The sun does not orbit the Earth.

Also, it takes 365 days (1 year) for the Earth to orbit the sun.

I hope this helped.