At what part of the cardiac conduction system does the electrical impulse travel most rapidly?

Answers

Answer 1

Answer:

Purkinje Fiber

Explanation:

In normal circumstances, the SA(sinoatrial) node the heart natural  pacemaker, produces electrical activity automatically. This electrical impulse is transmitted across the whole right atrium and into the bundle of Bachmann to the left atrium, arousing the atria's myocardium to contract. And the part through which  the electrical impulse travel most rapidly is called Purkrnje fibers, which is a specialized conducting fiber consisting of electrically excitable cells and these conduct cardiac action potentials more quickly and efficiently than any other cells in the heart.


Related Questions

In a popular classroom demonstration, a cotton ball is placed in the bottom of a strong test tube. A plunger fits inside the tube and it makes an air-tight seal. It is then pushed down very rapidly, and the cotton flashes and burns. This happens because ______.

Answers

Answer:

An increase in air temperature because of its compression.

Explanation:

The Gay-Lussac's Law states that a gas pressure is directly proportional to its temperature in an enclosed system to constant volume.  

[tex] P = kT [/tex]  

where P: is the gas pressure, T: is the gas temperature and k: is a constant.

Therefore, due to Gay-Lussac's Law, when the plunger is pushed down very rapidly, the pressure of the air increase, which leads to its temperature increase. That is why cotton flashes and burns.      

I hope it helps you!

The __________ behind a vehicle is one of the most common causes of collisions while ____________.

Answers

The main cause of accidents related to vehicles is rash driving, inexperienced drivers, and bad driving manners. Lack of awareness regarding the driving methods.

Most of the commonly made mistakes are due to drinking and driving. Most people feel drossy at night and some even don't horn when appropriate.  Hence the blind spot after the vehicles is the most common cause of collisions  

Learn more about the vehicle is one of the most common causes of collisions.

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Final answer:

The blind spot behind a vehicle is a common cause of collisions while driving. Safety measures like properly adjusted mirrors and checking over the shoulder before changing lanes can help. Cars with lighter plastic components crumple upon collision, increasing collision time and reducing the force exerted on occupants for increased safety.

Explanation:

The blind spot behind a vehicle is one of the most common causes of collisions while driving. This pertains to the area of the road that cannot be directly seen by the driver while looking forward or through either the rear-view or side mirrors. The risk of such collisions can be minimized by proper adjustment of mirrors, use of blind spot mirrors, and always checking over one's shoulder before making a lane change.

Upon collision, cars with lighter weight are able to offer better protection to their occupants due to longer collision time, because of their plastic components. The crumpling of the car in an event of a collision or the deployment of dashboard padding and airbags increases the collision time and hence, the force exerted on the occupants to bring them to a stop is significantly less. Modern automobiles use this physics principle to enhance safety during collisions.

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When does the most significant drop in physical activity typically occur in people's lives?

Answers

Answer:

Later high school years and freshman year of college

Explanation:

The transition from high school to college is an important developmental milestone that holds the potential for personal growth and behavioral change.  A cohort of 2,025 students was recruited during the summer before they matriculated into college and completed Internet-based surveys about their participation in a variety of behavioral risks during the last three months of high school and throughout the first year of college. Alcohol use, marijuana use, and sex with multiple partners increased during the transition from high school to college, whereas driving after drinking, aggression, and property crimes decreased. Those from rural high schools and those who elected to live in private dormitories in college were at highest risk for heavy drinking and driving after drinking.

The images output from your new color laser printer seem to be a little too blue. What can you do to fix this?

Answers

Answer:

The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.

Explanation:

This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.

If another driver does not follow the right-of-way rules at a YIELD sign, what should you do?

Answers

Answer:Stop

Explanation:

To avoid accident even thought you in the right

A golfer starts with the club over her head and swings it to reach maximum speed as it contacts the ball. Halfway through her swing, when the golf club is parallel to the ground, does the acceleration vector of the club head point straight down, parallel to the ground, approximately toward the golfer's shoulders, approximately toward the golfer's feet, or toward a point above the golfer's head? Explain.

Answers

Answer:

approximately toward the golfer's feet

Explanation:

The golf club experiences two accelerations. when the club is horizontal, centripetal acceleration acts on it toward the center parallel to the ground. The tangential acceleration acts towards the ground in downward direction. So when the golf club is horizontal, the acceleration vector is given by the combination of tangential and centripetal acceleration. the acceleration vector thus points  in between horizontal and vertical direction . hence towards the golfer's feet.

Answer:

approximately toward the golfer's feet

Explanation:

The club head undergoes circular motion so it has a centripetal acceleration. As the speed of the club head is increasing, it also has an acceleration component pointed vertically down. Thus, the acceleration vector is pointed as described in (c)

A 35-kg girl is standing near and to the left of a 43-kg boy on the frictionless surface of a frozen pond. The boy throws a 0.75-kg ice ball to the girl with a horizontal speed of 6.2 m/s. What are the velocities of the boy and the girl immediately after the girl catches the ice ball?

Answers

Answer:

The velocity of the boy is 0.11 m/s to the right.

The velocity of the girl is 0.13 m/s to the left.

Explanation:

Hi there!

The momentum of the system is conserved. That means that the initial momentum is equal to the final momentum (after throwing the ice ball for the boy and after catching the ball for the girl).

Let´s calculate the momentum of the system boy-ball:

The equation of the momentum is the following:

p = m · v

Where:

p = momentum.

m = mass of the object.

v = velocity.

Initially, the ball and the boy are at rest, so that the momentum of the system boy-ball is zero:

initial momentum = mib · vib + mb · vb

Where:

mib = mass of the ice ball.

vib = velocity of the ice ball.

mb =mass of the boy.

vb = velocity of the boy.

The initial momentum is zero because both velocities are zero.

After throwing the ball, the momentum of the system will be:

final momentum = mib · vib + mb · vb

final momentum = 0.75 kg · (- 6.2 m/s) + 43 kg · vb (Let´s consider the left as negative).

Since final momentum = initial momentum

0 kg · m/s =- 4.65 kg · m/s + 43 kg · vb

Solving for vb:

4.65 kg · m/s/43 kg = vb

vb = 0.11 m/s

The velocity of the boy after releasing the ball is 0.11 m/s. Since there is no friction, this velocity is constant. When the girl catches the ball, the velocity of the boy will be 0.11 m/s (to the right).

Now let´s do the same for the system girl-ball.

initial momentum = mg · vg + mib · vib

initial momentum = 35 kg · 0 m/s + 0.75 kg · (-6.2 m/s)

initial momentum = -4.65 kg · m/s

The final momentum is the momentum of the girl plus the ball:

final momentum = (mg + mb) · vgb

final momentum = (35 kg + 0.75 kg) · vgb

Since initial momentum = final momentum

-4.65 kg · m/s = (35 kg + 0.75 kg) · vgb

-4.65 kg · m/s / (35 kg + 0.75 kg) = vgb

vgb = -0.13 m/s

The velocity of the girl after catching the ball is -0.13 m/s (0.13 m/s to the left).

The velocity of the boy after catching the ice is 0.108m/s right and the velocity of the girl after catching the ice is 0.132m/s left.

Data;

mass of girl = 35kgmass of the boy = 43kgmass of the ice ball - 0.75kgspeed of the ball = 6.2m/s

Conservation of linear momentum

The conservation of linear momentum expects that the momentum of the boy or girl must be equal to the momentum of the ball.

for the boy

[tex]m_1v_1 = m_2v_2\\43*v = 0.75*6.2\\v = 0.108m/s[/tex]

The velocity of the boy after catching the ice is 0.108m/s right

for the girl

[tex]m_1v_1 = m_2 v_2\\0.75*6.2 = 35 * v\\v = 0.132 m/s[/tex]

The velocity of the girl after catching the ice is 0.132m/s left.

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A particle on a string is whirled in a circle at a constant speed, in the absence of a gravitational field. a top view of the circular motion at one instant is

Answers

Answer:

This is an incomplete question. The diagram and a part of the question is missing.

Explanation:

In the context, it is given that the particle tight to a string is whirled in a circular direction in a constant speed. And there is no gravitational field.

Now since the particle's speed is constant, hence the tangential acceleration is zero. The particle exhibits radial acceleration only which will act along the radius towards the center of the circle (towards the left).

Therefore the correct representation of the component of the acceleration is given below --  

Final answer:

A particle being whirled in a circle at a constant speed demonstrates uniform circular motion. The changing direction of velocity results in a centripetal acceleration, determined by an external centripetal force. Similar motion is observed in centrifuges and orbiting satellites.

Explanation:

A top view of a particle on a string whirled in a circle at a constant speed would show uniform circular motion. In this simplest form of curved motion, the particle moves along a fixed radius at a constant speed. Although the magnitude of its velocity doesn't change, the direction of the velocity is constantly changing, leading to what's referred to as centripetal acceleration. This is caused by a net external force - known as the centripetal force.

This type of motion is similar to other examples of circular motion, like a satellite orbiting Earth, a race car moving around a racetrack, or a ball being swung in a circle. Whenever there's no centripetal force to cause circular motion, inertia carries the object along a line tangent to the circle observed as outward motion in a centrifugal frame of reference. This inertial effect is seen in various physical phenomena.

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With each bounce off the floor, a tennis ball loses 18% of its mechanical energy due to friction. When the ball is released from a height of 2.9 m above the floor, what height will it reach after the fourth bounce?

170 cm

130 mm

130 cm

13 cm

Answers

Answer:

130 cm

Explanation:

After each bounce, the mechanical energy decreases by 18%, which means the maximum height decreases by 18%.  Modeling this as exponential decay:

h = 290 (1 − 0.18)ⁿ

h = 290 (0.82)ⁿ

h = 290 (0.82)⁴

h = 130

The ball reaches a height of 130 cm.

Answer:

130 cm

Explanation:

Gradpoint

Early LCD backlights use _______________ technology, popular for its low power use, even brightness, and long life.

Answers

Answer:

cold cathode fluorescent lamp

Explanation:

A cold cathode fluorescent lamp (CCFL) is a lighting system that uses two phenomena: electron discharge and fluorescence.

CCFLs are mainly used as light sources for backlights, since they are compact and durable than ordinary fluorescent lamps. They exhibit a wide range of brightness and color (color temperature and chromaticity) that can be achieved by varying the pressure and type of the material injected into the glass tube. The thickness and type of the phosphor used to coat the inner wall of the tube also plays a role in altering the color and brightness.

Final answer:

Early LCD backlights used CCFL technology for its efficiency and longevity. LCDs rely on a backlight to illuminate the liquid crystals in pixels. Each pixel can control red, blue, or green filters to create the full color spectrum on the screen.

Explanation:

Early LCD backlights use CCFL (Cold Cathode Fluorescent Lamp) technology, popular for its low power use, even brightness, and long life. LCD, which stands for liquid crystal display, relies on the light modulating properties of liquid crystals. These crystals themselves do not emit light. In flat screen LCD televisions, the backlight illuminates the screen through millions of tiny units called pixels. Each pixel has three cells with red, blue, or green filters, each controlled independently to vary the picture contrast and color representation by adjusting the voltage applied to the liquid crystal.

"Q 12.10: A 98 N object is placed 1/3 of the way up a 120 N ladder, which makes an angle of 53 degrees with the horizontal. The lower end is also fastened to the wall by a rope, and both the wall and floor are frictionless. What is the tension in the rope?"

Answers

Answer:

68.829 N

Explanation:

The given parameters are:

Weight of ladder, [tex]W_l[/tex] = 120 N

Weight of object, [tex]W_o[/tex] = 98 N

Angle, [tex]\theta[/tex] = 53°

And we also know that, while

Length of ladder = L

Distance the object is placed = L/3

If we apply translational equilibrium horizontally, then [tex]T-N=0[/tex], so [tex]T = N[/tex]

If we apply rotational equilibrium about the distance the object is placed, then

[tex]W_o\frac{L}{3}cos(\theta) + W_l\frac{L}{2}cos(\theta) - NLsin(\theta) = 0\\2NLsin(\theta) = 2W_o\frac{L}{3}cos(\theta)+W_lLcos(\theta)\\2Tsin(\theta) = \frac{2}{3}W_ocos(\theta)+W_lLcos(\theta)\\T = \frac{1}{3}W_ocot(\theta) + \frac{1}{2}W_lcot(\theta)\\T = \frac{1}{3} * 98 * cot(53) + \frac{1}{2} * 120 * cot(53) = 69.829[/tex]

A fire hose is turned on the door of a burning building in order to knock the door down. This requires a force of 1000 N. If the hose delivers 40 kg per second, what is the minimum velocity of the stream needed, assuming the water doesn't bounce back?

Answers

Answer:

Minimum velocity is 25m/s

Explanation:

Given force F =1000N

Mass flow rate m/t= 40kg/s

Velocity=v

Recall Newton second law of motion F =MV/t

1000=40v

V=1000/40

25m/s

An object has a position given by r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s2)t2] j^, where all quantities are in SI units. What is the magnitude of the acceleration of the object at time t = 2.00 s?

Answers

Final answer:

The magnitude of the acceleration of the object at time t = 2.00 s is 6.40 m/s².

Explanation:

The magnitude of acceleration can be found by taking the second derivative of the position function with respect to time.

Given the position function r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s²)t²] j^, we can differentiate it twice to get the acceleration function a(t) = 5.00 m/s i^ - 4.00 m/s² j^.

Plugging in t = 2.00 s gives us a(2.00 s) = 5.00 m/s i^ - 4.00 m/s² j^, and calculating the magnitude of this vector gives us 6.40 m/s² as the magnitude of acceleration at t = 2.00 s.

The magnitude of the acceleration of the object at [tex]\( t = 2.00 \, \text{s} \) is \( 4.00 \, \text{m/s}^2 \).[/tex]

To find the magnitude of the acceleration of the object at time [tex]\( t = 2.00 \, \text{s} \)[/tex], we first need to determine the acceleration vector from the given position vector. The position vector[tex]\( \mathbf{r}(t) \)[/tex]is given by:

[tex]\[\mathbf{r}(t) = [2.0 \, \text{m} + (5.00 \, \text{m/s})t] \hat{i} + [3.0 \, \text{m} - (2.00 \, \text{m/s}^2)t^2] \hat{j}\][/tex]

Find the Velocity Vector

The velocity vector [tex]\( \mathbf{v}(t) \)[/tex] is the time derivative of the position vector [tex]\( \mathbf{r}(t) \)[/tex]:

[tex]\[\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}\][/tex]

Differentiate each component with respect to time  t :

For the [tex]\( \hat{i} \)[/tex] component:

[tex]\[v_x(t) = \frac{d}{dt}[2.0 + (5.00 \, \text{m/s})t] = 5.00 \, \text{m/s}\][/tex]

For the[tex]\( \hat{j} \)[/tex] component:

[tex]\[v_y(t) = \frac{d}{dt}[3.0 - (2.00 \, \text{m/s}^2)t^2] = -2.00 \times 2t = -4.00 \, t \, \text{m/s}\][/tex]

So the velocity vector is:

[tex]\[\mathbf{v}(t) = (5.00 \, \text{m/s}) \hat{i} + (-4.00 \, t) \hat{j}\][/tex]

Find the Acceleration Vector

The acceleration vector[tex]\( \mathbf{a}(t) \)[/tex] is the time derivative of the velocity vector [tex]\( \mathbf{v}(t) \)[/tex]:

[tex]\[\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}\][/tex]

Differentiate each component with respect to time t:

For the [tex]\( \hat{i} \)[/tex]component:

[tex]\[a_x(t) = \frac{d}{dt}[5.00] = 0 \, \text{m/s}^2\][/tex]

For the [tex]\( \hat{j} \)[/tex] component:

[tex]\[a_y(t) = \frac{d}{dt}[-4.00 \, t] = -4.00 \, \text{m/s}^2\][/tex]

So the acceleration vector is:

[tex]\[\mathbf{a}(t) = 0 \hat{i} - 4.00 \hat{j}\][/tex]

Calculate the Magnitude of the Acceleration

The magnitude of the acceleration [tex]\( |\mathbf{a}(t)| \)[/tex] is:

[tex]\[|\mathbf{a}(t)| = \sqrt{a_x^2 + a_y^2}\][/tex]

Substitute[tex]\( a_x = 0 \) and \( a_y = -4.00 \)[/tex]:

[tex]\[|\mathbf{a}(t)| = \sqrt{0^2 + (-4.00)^2} = \sqrt{16.00} = 4.00 \, \text{m/s}^2\][/tex]

The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Answers

Answer:20 cal/s

Explanation:

Given

Heat transfer rate is [tex]\dot{Q}=10 cal/s[/tex]

Also heat rate is given by

[tex]\dot{Q}=kA\frac{dT}{dx}[/tex]

where [tex]k=thermal\ conductivity[/tex]

[tex]A=area\ of\ cross-section[/tex]

[tex]dT=change\ in\ temperature[/tex]

[tex]dx=change\ in\ length[/tex]

[tex]10=k\frac{\pi }{4}d^2\frac{dT}{L}----1[/tex]

For [tex]d'=2d, Length L'=2L[/tex]

[tex]\dot{Q}=k\frac{\pi }{4}(2d)^2\frac{dT}{2L}---2[/tex]

dividing 1 and 2 we get

[tex]\frac{10}{\dot{Q}}=\frac{2d^2}{4d^2}[/tex]

[tex]\dot{Q}=20 cal/s[/tex]                            

The power rate of the steel is mathematically given as

P=20cal/s

What rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Question Parameter(s):

The rod conducts heat from one end to the other at a rate of 10 cal/s

Generally, the equation for the Power   is mathematically given as

Power = (kA∆T)/l

Therefore

Power = [k × 4(πd^2)/4 × ∆T]/2l

Therefore

Power  = 2(kA∆T)/l

Hence, Initial Power

iP = (kA∆T)/l

iP= 10cal/s

Where

dT is the same

2[(kA∆T)/l] = 2 × 10

P=20cal/s

In conclusion, The power rate is

P=20cal/s

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Ian infinite nonconducting sheet has surface charge density on one side. How far apart are equi-potential surfacews whose potentials sdiffer by 50 V?

Answers

Answer:

8.85 × 10 ⁻³ m

Explanation:

The electric field produced by infinite sheet of charge relates by

E = σ / 2ε₀

where  σ = surface charge density = 0.1 μC/m² = 0.1 × 10⁻⁶C/m², ε₀ = 8.85 × 10⁻¹² C⁻²/Nm²

also

V = E Δs where Δs is the distance apart in meters, V is the voltage of the

make E subject of the formula

V/Δs = E

equate the two equation

V / Δs = σ / 2ε₀

make Δs subject of the formula

V2ε₀ / σ = Δs

substitute the values into the expression

Δs = (50 × 2 × 8.85 × 10⁻¹²)  / (0.1 × 10⁻⁶) = 8.85 × 10 ⁻³ m

A particle with a positive charge q is moving in the positive x direction with velocity of magnitude v0. If there were an electric field in the positive y direction with magnitude E0, and the velocity of the particle is unchanged, find the magnitude and direction of magnetic field.

Answers

Final answer:

The magnitude of the magnetic field in the given scenario is zero.

Explanation:

The magnitude and direction of the magnetic field can be determined by analyzing the force experienced by the positively charged particle q as it moves in the positive x direction and encounters an electric field in the positive y direction. The force experienced by the particle, due to the Lorentz force law, is given by:

F = qvB sin

Where F is the force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Since the velocity of the particle remains unchanged, the force experienced by the particle in the y direction must be zero. This means that the angle θ between the velocity and the magnetic field must be 90 degrees. Therefore, sin θ = 1. Substituting these values into the equation, we get

0 = qv0

Solving for B, we find that

B = 0

Therefore, the magnitude of the magnetic field in this scenario is zero.

2020 NV Cargo’s electronically controlled Multi-Control Valve on the V8 engine helps enhance ________.

Answers

Answer:

Power when towing

Explanation:

The V8 engine has Direct Injection Gas (DIG) technology, which provides better wide-open throttle performance and improved fuel consumption and emissions performance (compared to a non-direct-injection system) by reducing engine knock, improving combustion stability and offering precise injection control.

The multi-control valve on the V8 engine helps to enhance power when towing by reducing the amount of torque the engine must expend to move the vehicle, allowing it to commit more to moving the load.

What did antoine lavoisier contribute to the atomic theory

Answers

Explanation:

Antoine lavoisier is most famous for his role in  discovering of oxygen. He recognized and name two important element oxygen(1778) and hydrogen(1779).

In a chemical reaction, Lavoisier observed the mass is retained. A chemical reaction's total mass of the products will always be the same as the total mass of the reactant materials used in the reaction. His results led to one of the basic laws of chemical behavior: the law of material preservation, which states that matter is preserved in a chemical reaction.

Final answer:

Antoine Lavoisier's major contributions to atomic theory include the discovery of the law of conservation of mass and transforming chemistry into a quantitative science by disproving the phlogiston theory and identifying the role of oxygen in combustion. These discoveries paved the way for the future development of the atomic theory.

Explanation:

Contributions of Antoine Lavoisier to Atomic Theory

Antoine Lavoisier, often regarded as the 'father of modern chemistry', made several groundbreaking contributions to atomic theory. He is particularly famous for discovering the law of conservation of mass. This principle, established through careful experimentation, states that mass in an isolated system is conserved over time, regardless of the type of chemical reactions happening within. This fundamental law formed a cornerstone for John Dalton to later develop the modern atomic theory.

Additionally, Lavoisier conducted experiments that disproved the phlogiston theory and instead suggested that oxygen was a key element in combustion. His innovative approach to chemical reactions included the use of precise weighing measures and conducting reactions in sealed environments to accurately track the mass of gases involved. Lavoisier's meticulous quantitative methods transformed chemistry from a qualitative to a quantitative science.

Through these contributions, Lavoisier laid down the basis for atomic theory as well as the concept of chemical elements, thereby fundamentally changing our understanding of chemical processes. He also impacted other areas such as explaining the role of oxygen in respiration and the composition of air.

BOTH INTERNATIONAL & INLAND A vessel is "engaged in fishing" when ________.
Select one:
a. she is using any type of gear, other than lines
b. she is using fishing apparatus which restricts her maneuverability
c. her gear extends more than 100 meters from the vessel
d. she has any fishing gear on board

Answers

Answer:

(b) she is using fishing apparatus which restricts her manoeuvrability

Explanation:

The explanation is spelled out in Rule 3(d) of the inland and water ways  laws that state that: A vessel engage in fishing is one whose fishing apparatus restricts manoeuvrability. I f the fishing gear does not hamper her movements, then she is engage in fishing. The rules state categorically that a troller  with hook-spangled lines following in her wake is not restricted by her gear. Rule 3(d) might appear to be splitting hairs, but it has a point to make. Although a vessel might not be hampered as to be unable to keep clear of others, her fishing gear could still make following a script difficulty

How can a surface be smooth/polished for some waves and not for others? (This is an important question, refer to the homework on WebAssign and question regarding whether a surface is smooth/polished for some electromagnetic waves verses others)

Answers

Answer:

Depending upon the size of the irregularities on the surface and the wavelength of the surface a surface can be smooth for the waves of smaller wavelengths than the size of the irregularities.

Explanation:

Depending upon the size of the irregularities on the surface and the wavelength of the surface a surface can be smooth for some of the waves but not for the others.

More specifically, when a surface has irregularities of the size greater than or equal to the wavelength of a wave then the surface is rough for that particular wave and causes irregular reflection in the incident light.

Contrarily, when the surface has the irregularities less than the size of the wavelength then it reflects most of the light in a uniform manner.

A person is in a closed room (a racquetball court) with ????=453m3 hitting a ball (m=42.0g) around at random without any pauses. The average kinetic energy of the ball is 2.30 J. (a) What is the average value of ????2x? Does it matter which direction you take to be x? (b) Applying the methods of this chapter, find the average pressure on the walls? (c) Aside from the presence of only one "molecule" in this problem, what is the main assumption in Pressure, Temperature, and RMS Speed that does not apply here?

Answers

Answer:

(a) [tex]v^{2} _{x} = 36.508 m^2/s^2[/tex]

(b)  0.1235 Pa

(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of Van der Waal is greater than zero.

Explanation:

In the question, we are given the following variables:

V = 453 m^3

[tex]m_{ball} = m = 42.0 g[/tex]

Kinetic energy (K.E) = 2.30 J

(a) The equation for kinetic energy is:

[tex]K.E = \frac{1}{2}mv^{2}[/tex]

Since the average velocity components of the ball in the three dimensions are equal. Thus, we have:

[tex]v_{x} ^{2} = v^{2} _{y} =v^{2} _{z}[/tex]

In addition:

[tex]v^{2} =[/tex][tex]v_{x} ^{2} + v^{2} _{y} +v^{2} _{z}[/tex] = [tex]3v^{2} _{x}[/tex]

Therefore:

[tex]K.E = \frac{1}{2}m[3v^{2} _{x}][/tex] = [tex]\frac{3}{2}mv^{2} _{x}[/tex]

Thus:  

[tex]v^{2} _{x} = \frac{KE}{1.5*m} = \frac{2.3}{1.5*0.042} = 36.508 m^2/s^2[/tex]

It does not matter which direction we take as x-direction because the average velocity component in every direction is the same.

(b) The average pressure on the walls can be calculated using the equation below:

[tex]P =\frac{mv^{2} _{x} }{V}[/tex]

where V = 453 m^3

Thus: P = (0.042*36.508^2)/453 = 0.1235 Pa

(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of van der Waal is greater than zero.

Final answer:

The average value of u² is 54.8 m²/s². It does not matter which direction is taken as x. The average pressure on the walls can be found using the force and area equations.

Explanation:

(a) The average value of u² can be found by dividing the average kinetic energy by the mass of the ball: u² = KE / m. In this case, u² = 2.30 J / 0.042 kg = 54.8 m²/s².

It does not matter which direction you take to be x. The average value of u² will be the same regardless of the direction.

(b) The average pressure on the walls can be found using the equation P = F / A, where P is the pressure, F is the force, and A is the area of the walls. The force exerted on the walls can be found using the equation F = ∆p / ∆t, where ∆p is the change in momentum and ∆t is the change in time. Since there are no pauses in the ball's motion, ∆p = m * ∆v, where m is the mass of the ball and ∆v is the change in velocity. The average pressure will be the force divided by the area of the walls.

(c) The main assumption in Pressure, Temperature, and RMS Speed that does not apply here is that there are a large number of gas molecules moving in random directions. In this problem, there is only one molecule, so the assumption of a large number of molecules does not hold.

A car moves 4 times as fast as another identical car. compared to the slower car, the faster car has:_______
a. 4 times the ke.
b. 8 times the ke.
c.12 times the ke.
d.16 times the ke.

Answers

Answer:

Option d

d. 16 times the ke

Explanation:

The kinetic energy is defined as:

[tex]Ke = \frac{1}{2}mv^{2}[/tex]  (1)

Where m and v are the mass and velocity of the object.

The kinetic energy represents how much energy have an object as a consequence of its movement.

For the case of a car moving 4 times as fast as another identical car.

By means of equation 1 it is gotten:

[tex]Ke = \frac{1}{2}m\cdot(4v)^{2}[/tex]

[tex]Ke = \frac{1}{2}m\cdot16v^{2}[/tex]

Hence, the faster car has 16 times the kinetic energy.

If you push against a wall, the wall pushes back against you with ____ force.
1. equal
2. less
3. more
4. no

Answers

Answer:

1. Equal.

Explanation:

Newton's third law of motion;

   According to this law every action have its reaction with same magnitude but in the opposite direction.

If we push a wall with some amount of force then wall pushes back with same magnitude but in the opposite direction.This law tell us about action and reaction forces which are in same magnitude but in the opposite direction.

Therefore the answer will be 1.

1. Equal.

Final answer:

When pushing a wall, the wall pushes back with equal force as stated by Newton's Third Law of Motion. This law means forces have equal magnitude, but opposite direction.

Explanation:

When you push against a wall, the wall pushes back against you with equal force. This is demonstrated by Newton's Third Law of Motion, which states that 'for every action, there is an equal and opposite reaction'. The force you exert on the wall and the force the wall exerts on you are action-reaction force pairs, meaning the force exerted on the wall is the same magnitude, but in the opposite direction to the force the wall exerts on you.

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When a person strikes and lights a match, potential energy in the match is transformed into which types of energy?
a. electrical and thermal
b. thermal and radiant
c. radiant and nuclear
d. nuclear and kinetic

Answers

It would be electrical and thermal because it s giving heat to light the match
It's would be b
Thermal as it makes fire
Radiant as it give rays out to make the fire lit up

If Air Traffic Control advises that radar service is terminated when the pilot is departing Class C airspace, the transponder should be set to code?

Answers

Explanation:

1200 feet.

In the given scenario we would be operation under VFR.

VFR is visual flight rules are set of regulations under which the pilot operates aircraft in a weather which generally clear enough that facilitates pilot to see the flight where it is going.

What is the volume of water in 150ml of the 35% w/w of sucrose solution with a specific gravity of 1.115?

Answers

Answer:108.71 mL

Explanation:

Given

Volume of sample V=150 mL

concentration of sucrose solution 35 % w/w i.e. In  100 gm of sample 35 gm is sucrose

specific gravity =1.115

Density of solution [tex]\rho _s=1.115\times density\ of\ water[/tex]

Thus

[tex]\rho _s=1.15\times 1 gm/mL=1.115\ gm/mL[/tex]

mass of sample [tex]M=1.115\times 150=167.25\ gm[/tex]

mass of sucrose [tex]m_s=0.35\times 167.25=58.53\ gm[/tex]

mass of Water [tex]m_w=108.71 gm[/tex]

Volume of water [tex]=108.71\times 1=108.781 mL[/tex]

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 27.0°below the horizontal. If it strikes the ground 59.0 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)(a) the time of flight.(b) the inital.(c) the speed and angle of the velocity vector with respect to the horizontal at impact.

Answers

a) The time of flight is 3.78 s

b) The initial speed is 17.0 m/s

c) The speed at impact is 46.4 m/s at [tex]70.3^{\circ}[/tex] below the horizontal

Explanation:

The picture of the previous problem (and some data) is missing: find it in attachment.

a)

The motion of the ball is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:

[tex]y=u_y t+\frac{1}{2}at^2[/tex]

where we have:

y = -45.0 m is the vertical displacement of the ball (the height of the building)

[tex]u_y=u sin \theta[/tex] is the initial vertical velocity, with u being the initial velocity (unknown) and [tex]\theta=-27.0^{\circ}[/tex] the angle of projection

t is the time of the fall

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity

Along the x-direction, the equation of motion is instead

[tex]x=(u cos \theta)t[/tex]

where [tex]ucos \theta[/tex] is the horizontal component of the velocity. Rewriting this equation as

[tex]t=\frac{x}{ucos \theta}[/tex]

And substituting into the previous equation, we get

[tex]y=xtan \theta + \frac{1}{2}gt^2[/tex]

And using the fact that the horizontal range is

x = 59.0 m

And solving for t, we find the time of flight:

[tex]t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s[/tex]

b)

We can now find the initial speed, u, by using the equation of motion along the x-direction

[tex]x=u cos \theta t[/tex]

where we know that:

x = 59.0 m is the horizontal range

[tex]\theta=-23^{\circ}[/tex] is the angle of projection

[tex]t=3.78 s[/tex] is the time of flight

Solving for u, we find the initial speed:

[tex]u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s[/tex]

c)

First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to

[tex]v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s[/tex]

The vertical velocity instead changes according to the equation

[tex]v_y = u sin \theta + gt[/tex]

Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:

[tex]v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s[/tex]

Where the negative sign means it is downward.

Therefore, the speed at impact is

[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.6)^2+(-43.7)^2}=46.4 m/s[/tex]

while the direction is given by

[tex]\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-43.7}{15.6})=-70.3^{\circ}[/tex]

So, [tex]70.3^{\circ}[/tex] below the horizontal.

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What will the temperature of the cosmic microwave background be when the average distances between galaxies are 1.7 times as large as they are today? (Hint: The peak wavelength of photons in the background will then also be 1.7 times as large as it is today.)

Answers

Answer:

[tex]T'=1.58K[/tex]

Explanation:

As the Universe expands, the photons of the cosmic microwave background increase its wavelength, making its temperature inversely proportional to the scale factor of the Universe. That is, as the average distances between galaxies increase, the temperature of the cosmic microwave background decreases by the same factor. Therefore, the temperature when the distances between galaxies are 1.7 times as large as they are today will be:

[tex]T'=\frac{T_{now}}{1.73}\\T'=\frac{2.725K}{1.73}\\T'=1.58K[/tex]

Final answer:

When the average distances between galaxies are 1.7 times their present distances, the temperature of the cosmic microwave background (CMB), as a result of Wien's Law, will be about 1.61 Kelvin, which is 1.7 times lower than its current close to 2.73 Kelvin figure.

Explanation:

The cosmic microwave background radiation (CMBR), an aftermath of the Big Bang event, exhibits a relationship between wavelength and temperature called Wien's Law. At present, the temperature of the CMBR is approximately 2.73 Kelvin (K), a finding made possible by scientists Penzias and Wilson who associated radiation intensity at different wavelengths to a temperature of roughly 3.5 K.

As the galaxies continue to recede from one another, their emitted radiation wavelengths become redshifted, effectively stretching them out. This movement is also reflected in the increase in the peak wavelength of the CMBR. If the average distances between galaxies were to magnify by 1.7 times, the peak wavelength of the CMBR would equally elongate by the same factor.

Applying Wien's Law, this increase will consequently result in a decrease in the CMB's temperature. Thus, when galaxies have expanded such that their average distances are 1.7 times larger than their current distances, the temperature of the cosmic microwave background will be 1.7 times lower than its current temperature of approximately 2.73 K. Hence, the new temperature would approximately be 1.61 K.

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Laser light of wavelength lambda passes through a thin slit of width a and produces its first dark fringes at angles of +/- 30 degree with the original direction of the beam. The slit is then reduced in size to a circle of diameter a. When the same laser light is passed through the circle, the first dark fringe occurs at:________
A) +/- 66.9 degree.
B) +/- 45.0 degree.
C) +/- 37.6 degree.
D) +/- 36.6 degree.
E) +/- 15.0 degree.

Answers

Answer:

D)  θ₂= 36. 6º  

Explanation:

In this diffraction experiment it is described by the equation

               sin θ = m λ

The first dark strip occurs for m = 1 and since the angle is generally small we can approximate sine to the value of the angle

              θ₁ = λ/ a

This equation is valid for linear slits, in the case of a circular slit the problem must be solved in polar coordinates, so the equation changes slightly

             θ₂ = 1.22 λ / a

In the proposed exercise we start with a linear slit of width a, where tes1 = 30º and end with a circular slit of the same diameter

             θ₂ = 1.22 λ / a

 

Let's clear (Lam/a) of equalizing the two equations

             θ₁ = θ₂/ 1.22

             θ₂ = 1.22 θ₁

             θ₂ = 1.22 30

             θ₂= 36. 6º

When reviewing the correct results is D

Suppose you want to heat a 600 mL cup of water in the microwave from 20° to 100°C. Water has a specific heat capacity of 4.184 J/g°C and a density of 1.00 g/mL in this temperature range. How much energy is required to heat the cup of water?

Answers

Answer:

The quantity of heat required to to heat the cup of water through a temperature change of 80°C = 200.83 kJ

Explanation:

Density = mass/Volume.

∴ mass = Density ×  volume

Where Density of water = 1.00 g/mL, volume of water = 600 mL

Mass = 1 × 600 = 600 g.

(Q) = cm(ΔT)...................... equation 1

Where Q = quantity of heat, c = specific heat capacity of water, m = mass of water, ΔT = T₁-T₂ = change in temperature.

Where c = 4.184 J/g°C, m = 600 g, ΔT = T₁-T₂ = 100 - 20 = 80°C

Substituting these values into equation 1,

Q = 4.184 × 600 × 80

Q = 200832 J = 200.83 kJ

Therefore the quantity of heat required to to heat the cup of water through a temperature change of 80°C = 200.83 kJ

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