does air move from areas of high pressure to low pressure

Answers

Answer 1

Explanation:  Gases move from high-pressure areas to low-pressure areas. And the bigger the difference between the pressures, the faster the air will move from the high to the low pressure.


Related Questions

A barber wants to set up a salon in a room measuring length 3m by 3m he has a simple wooden chair,three large mirrors & a bulb. Using the knowledge of shadows & reflection advise the barber on how to arrenge a good saloon using the only items he has

Answers

Here are some ideas for setting up the barber's salon based on the size of the space and the products available: The wooden chair should be positioned in the middle of the space, facing a wall.

The barber's workspace will be this. The room's other three walls should be covered with the three enormous mirrors. This will give the impression that there is more space present and enlarge the room. The mirrors should be angled to reflect both the client in the chair and the barber's work area. Over the chair, suspend the lightbulb from the ceiling. The barber salon will be able to operate in enough lighting thanks to this.The wooden chair should be positioned in the middle of the space,  The barber can set up a white sheet or a reflecting surface to improve illumination even further.

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how far, in centimeters, would you have to compress this spring to store this energy?

Answers

Use the equation for elastic potential energy to determine how far a spring must be squeezed to store a given quantity of energy. Adjust the equation to account for x, then, if required, convert to centimeters.

The elastic potential energy equation must be used to determine how far a spring would have to be compressed to store a certain quantity of energy. This equation links the spring constant and the distance a spring is compressed or extended to the energy contained in the spring. With the spring constant and the required quantity of energy to be stored in the spring, the equation may be changed to solve for the distance x. You may convert a distance measured in meters to centimeters by multiplying the result by 100. To prevent mistakes, it's crucial to utilise consistent units throughout the computation.

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suppose your planet at 1 meter from the basketball represents a distance of 4 x 107 km (-0.3 al) from the star. the next closest star to the sun is 4 x 1013 km away. how far away from the model star/planet would you have to be for the distances in the system to be to scale? express your answer in meters and kilometers.

Answers

Answer: The model star/planet would have to be 1,000 km away from the next closest star.

Explanation:
We need to find out the distance required for the distances in the system to be in scale.

Let's use the proportion to solve the problem:

1 m/4 × 10⁷ km = x/4 × 10¹³ km

Where x is the distance required for the distances in the system to be in scale.

Cross-multiply: 4 × 10¹³ km × 1 m = 4 × 10⁷ km × x

Simplify: 4 × 10¹³ m = 4 × 10⁷ x

Divide both sides by 4 × 10⁷ :1 × 10⁶ = x

Therefore, the distance required for the distances in the system to be in scale is 1 × 10⁶ m or 1,000 km.

So the model star/planet would have to be 1,000 km away from the next closest star.

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when electron density is unevenly distributed around atoms in a covalent bond, it is considered to be ____________.

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When the electron density is unevenly distributed around atoms in a covalent bond, it is considered to be polar.

What is Polar covalent bonds?

Polar covalent bonds are a type of chemical bond formed between atoms that share electrons unequally. In a covalent bond, two atoms share electrons in order to achieve a stable electron configuration.

Polar covalent bonds occur when the electronegativity difference between the two atoms in the bond is significant enough to cause the electrons to be shared unequally. The atom with the higher electronegativity attracts the shared electrons closer to itself, resulting in a partial negative charge, while the other atom has a partial positive charge. This unequal distribution of charges in the bond creates a dipole moment, giving rise to a polar bond.

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A student must analyze data collected from an experiment in which a block of mass 2M traveling with a speed vo collides with a block of mass M that is initially at rest. After the collision, the two blocks stick together. Which of the following applications of the equation for the conservation of momentum represent the initial and final momentum of the system for a completely inelastic collision between the blocks? Justify your selection. Select two answers. A. 2Mo = 3Muf, because the blocks stick together after the collision.
B. 3Mvo = 3MUf, because the blocks stick together after the collision. C. 2MVo = 2MU + Muf, because the blocks stick together after the collision. D. 2MVo = M0o + 3 Muf, because the blocks do not stick together after the collision.

Answers

A student must analyze data collected from an experiment in which a block of mass 2M traveling with a speed vo collides with a block of mass M that is initially at rest. After the collision, the two blocks stick together. Thus, the correct options are A and B.

What is Momentum?

The initial momentum of the system = the momentum of block 1 = (2M)vo. The final momentum of the system = the momentum of the combined blocks = (2M + M)uf = 3Muf. Therefore, the correct applications of the equation for the conservation of momentum that represent the initial and final momentum of the system for a completely inelastic collision between the blocks are:

2Mo = 3Muf, because the blocks stick together after the collision. 3Mvo = 3MUf, because the blocks stick together after the collision.

Therefore, the correct options are A and B.

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A within-subjects experiment with 30 volunteers is used to test the effect of light color on mood. The experiment has a single light bulb type with 3 different ambient color schemes. The experiment has 3 different rooms, each room with a different colored light bulb. Which of these is the factor in this experiment?

Answers

In this experiment, the factor is the ambient color schemes. In other words, the independent variable is the ambient color schemes.

What are ambient color Schemes?

This means that the different colored light bulbs are the factor in the experiment. Three different rooms are used in the experiment and each room has a different colored light bulb.

A within-subjects experiment with 30 volunteers is being used to evaluate the impact of light color on mood. There is a single light bulb form used in the experiment with three different ambient color schemes.

Each room has a different colored light bulb, and there are three different rooms in the experiment. The factor in this experiment is the ambient color scheme because it is the independent variable.

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Water is flowing in a circular pipe varying cross-sectional area, and at all points, the water completely fills the pipe.a) At one point in the pipe the radius is 0.150 m. What is the speed of the water at this point if the water is flowing into this pipe at a steady rate of 1.20 m3/s?b) At a second point in the pipe the water speed is 2.90 m/s. What is the radius of the pipe at this point?

Answers

The speed of water at the point with a radius of 0.150 m is 16.97 m/s while the radius of the pipe at the point where the water speed is 2.90 m/s is 0.0682 m.

a) To find the speed of the water at a point of a circular pipe where the radius is 0.150 m if the water is flowing into this pipe at a steady rate of 1.20 m³/s, we'll use the equation;

Q = A₁V₁ = A₂V₂ Where Q = Flow rate (m³/s)A₁ = Cross-sectional area at one point (m²)V₁ = Velocity of water at one point (m/s)A₂ = Cross-sectional area at a second point (m²)V₂ = Velocity of water at the second point (m/s)At one point in the pipe, the radius is 0.150 m.Therefore, the cross-sectional area, A₁ is given by:

A₁ = πr₁² = π (0.150 m)² = 0.0707 m²Given that the water is flowing into the pipe at a steady rate of 1.20 m³/s, we can write;Q = A₁V₁1.20 m³/s = 0.0707 m² V₁V₁ = 1.20/0.0707V₁ = 16.97 m/s.Therefore, the speed of water at the point with a radius of 0.150 m is 16.97 m/s.

b) To find the radius of the pipe at a point where the water speed is 2.90 m/s, we'll use the same equation as in part (a);Q = A₁V₁ = A₂V₂At a second point in the pipe, the water speed is 2.90 m/s.Given that the water completely fills the pipe, we know that the volume flow rate, Q will remain constant at 1.20 m³/s.So, we have:

Q = A₁V₁ = A₂V₂We know that A₁ = πr₁²So, Q = πr₁²V₁Also, we know that A₂ = πr₂²So, Q = πr₂²V₂Since the volume flow rate is constant, we can equate both equations,πr₁²V₁ = πr₂²V₂Dividing both sides of the equation by π, we have;r₁²V₁ = r₂²V₂But we are interested in finding the radius of the pipe at the second point, r₂.So, we can express r₁ in terms of r₂ using the relationship between the cross-sectional areas;

A₁ = A₂r₁² = (A₂/A₁)²r₂²r₁ = r₂ (A₂/A₁)^(1/2).We know that A₁ = πr₁²We can find A₂ using the fact that the water completely fills the pipe;

A₁V₁ = A₂V₂πr₁²V₁ = A₂V₂π(0.150 m)²(16.97 m/s) = A₂(2.90 m/s)A₂ = π(0.150 m)²(16.97 m/s)/(2.90 m/s)A₂ = 0.0707 m²

So,r₂ = r₁(A₂/A₁)^(1/2)r₂ = 0.150 m × (0.0707 m²/π)/(0.0150 m²)^(1/2)r₂ = 0.0682 m. Therefore, the radius of the pipe at the point where the water speed is 2.90 m/s is 0.0682 m.

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a satellite is in a circular orbit around an unknown planet. the satellite has a speed of 1.89 x 104 m/s, and the radius of the orbit is 2.76 x 106 m. a second satellite also has a circular orbit around this same planet. the orbit of this second satellite has a radius of 6.98 x 106 m. what is the orbital speed of the second satellite?

Answers

The orbital speed of the second satellite is 6.55 × 10³ m/s.

The formula used to find the orbital speed of a satellite is given as v=√(GM/r).

Therefore, the value of the first satellite's speed is given as v₁=1.89×104 m/s, and the radius is r₁=2.76×106 m. Using the above formula, the mass of the planet is given as:

M= v²r/G= (1.89×104 m/s)² (2.76×106 m)/(6.6743 × 10⁻¹¹ Nm²/kg²) = 5.31 × 10²⁴ kg.

Now, the orbital speed of the second satellite, given as v₂, is equal to:

v₂ = √(GM/r₂); where G = gravitational constant = 6.6743 × 10⁻¹¹ Nm²/kg²;

M = mass of the planet = 5.31 × 10²⁴ kg;

r₂ = radius of orbit of the second satellite = 6.98 × 10⁶ m.

Substituting the values given above, we get:

v₂ = √(GM/r₂)= √[(6.6743 × 10⁻¹¹ Nm²/kg²) × (5.31 × 10²⁴ kg) / (6.98 × 10⁶ m)] = 6.55 × 10³ m/s

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4. Once the child in the sample problem reaches the bottom of the hill,
she continues sliding along flat; snow-covered ground until she comes
to a stop. If her acceleration during this time is -0.392 m/s², how long
does it take her to travel from the bottom of the hill to her stopping
point?

Answers

Answer:

8.04 seconds

Explanation:

Assuming that the child starts from rest at the bottom of the hill and travels until she comes to a stop, we can use the following kinematic equation:

v_f^2 = v_i^2 + 2ad

where v_f is the final velocity (which is zero since the child comes to a stop), v_i is the initial velocity (which is the velocity at the bottom of the hill), a is the acceleration (-0.392 m/s²), and d is the distance traveled.

We can solve for d:

d = (v_f^2 - v_i^2) / (2a)

= (0 - v_i^2) / (2-0.392)

= v_i^2 / 0.784

Since the child is sliding along flat snow-covered ground, there is no change in elevation, so we can use the distance traveled from the bottom of the hill to the stopping point as the distance d.

To find the time it takes for the child to travel this distance, we can use the following kinematic equation:

d = v_it + 0.5a*t^2

where t is the time and all other variables are as previously defined.

Substituting the expression for d obtained above, we get:

v_i^2 / 0.784 = v_it + 0.5(-0.392)*t^2

Solving for t, we get:

t = (2 * v_i) / 0.392

We still need to find the value of v_i, the initial velocity of the child at the bottom of the hill. To do so, we can use conservation of energy. The child starts at rest at the top of the hill, so all the initial energy is potential energy. At the bottom of the hill, all the potential energy has been converted to kinetic energy. Assuming no energy is lost to friction, we can equate these two energies:

mgh = 0.5mv_i^2

where m is the mass of the child, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the hill.

Solving for v_i, we get:

v_i = √(2gh)

Substituting this expression for v_i into the expression for t obtained earlier, we get:

t = (2 * √(2gh)) / 0.392

Plugging in the values of g, h, and a, we get:

t = (2 * √(29.820)) / 0.392 = 8.04 seconds

why is a polarized filter helpful to a photographer? A. it transmits all light

Answers

Answer:

It blocks some light, but not all.

Explanation:

The point of polarization is to get the light to travel in a single plane. The light waves occur in a single plane. The direction of the vibration of the waves is the same. With two polarized filters, it is possible to block out nearly all the light.

A typical neutron star has a mass of about 1.5Msun and a radius of 10 kilometers Calculate the average density of a neutron star. Express your answer in kilograms per cubic centimeter to two significant figures.

Answers

The average density of the neutron star that has a mass of about 1.5Msun and a radius of 10 kilometers rounded off to two significant figures is 5.9 × 10¹⁴ kg/cm³

The average density of a neutron star can be calculated using the following formula;`d = (3M)/(4πr³)`where `d` is the average density of the neutron star, `M` is the mass of the neutron star, and `r` is the radius of the neutron star.Using the given values in the formula, we get;`d = (3 × 1.5 × 1.989 × 10³⁰)/(4π × (10 × 10³)³)` = 5.9 × 10¹⁷ kg/m³To convert kg/m³ to kg/cm³, we can use the following conversion factor;1 m³ = 10⁶ cm³Therefore,1 kg/m³ = 10⁻³ kg/cm³So, the average density of the neutron star in kg/cm³ is;`d = (5.9 × 10¹⁷) × (10⁻³)` = 5.9 × 10¹⁴ kg/cm³Therefore, the average density of the neutron star is 5.9 × 10¹⁴ kg/cm³ (rounded to two significant figures).Answer: 5.9 × 10¹⁴ kg/cm³.

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the inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light sensitive paper, is named

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The inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light-sensitive paper is named William Henry Fox Talbot.

What is photography?

Photography is the art, process, and practice of creating photographs, which are images recorded by light or other electromagnetic radiation, either electronically or chemically, onto an image sensor or other light-sensitive material.

Photography has made its way from the ancient Chinese invention of the camera obscura in the fifth century BCE to the worldwide photographic society of the present. The first photographic image was taken by French inventor Joseph Nicéphore Niépce in 1826, but the earliest surviving photograph was taken by French photographer Louis Daguerre in 1837.

William Henry Fox Talbot, an English scientist, produced the first photographic negative, which enabled him to make multiple prints, in 1835. Fox Talbot also developed the calotype method, which replaced the daguerreotype and allowed for images to be developed on paper that was first coated with silver iodide and then developed in gallic acid.

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Rank these hypothetical moons from oldest to youngest based on their cratering. You can assume the moons have never been volcanically active.-a moon with very few craters-a moon completely covered in craters, old and new-a moon partially covered with craters

Answers

We can see the moons should be ranked in the following order from oldest to youngest:

A moon completely covered in craters, old and newA moon partially covered with cratersA moon with very few craters

What is a moon?

A moon is a natural satellite that orbits a planet. Moons are typically much smaller than their parent planets and are held in orbit by the planet's gravity. They come in a variety of sizes and shapes, and can be composed of a wide range of materials, such as rock, ice, or a mixture of both.

Moons play an important role in our solar system. They help stabilize the orbits of planets, contribute to tidal forces, and may even play a role in the formation and evolution of planets themselves.

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The countercurrent mechanism functions primarily in the A. renal corpuscle. B. proximal convoluted tubule. C. distal convoluted tubule D. nephron loop.

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The nephron loop is where the countercurrent mechanism predominantly operates. The Henle loop is the name of the nephron component.

Reabsorbing water and sodium chloride from the filtrate is its principal purpose. By creating urine that is extremely concentrated, this helps the body preserve water. Kidney cross slice showing the nephron (kidney tubule) in detail and its associated blood supply.

Renal tubular fluid is defined as fluid present within the lumen of a nephron loop. Glomerular filtrate is the term used to describe the fluid found inside Bowman's capsule. Urine would be regarded as the fluid contained in the collecting duct.

The fact that loop diuretics work on the Loop of Henle gives them their name. They function by blocking NKCC2 transporters in the thick ascending limb, preventing the reabsorption of sodium, potassium, and chloride.

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If the resulting trajectory of the charged particle is a circle, what is ⍵, the angular frequency of the circular
motion?
Express ⍵ in terms of g, m, and Bo.

Answers

The angular frequency of circular motion is given by the expression:

ω = [tex]\sqrt{qB/m}[/tex]

If the resulting trajectory of the charged particle is a circle, the angular frequency (ω) of the circular motion can be expressed in terms of g, m, and Bo as follows:

ω = [tex]\sqrt{qB/m}[/tex]

where q is the charge of the particle, B is the magnetic field strength, and m is the mass of the particle.

This formula is known as the cyclotron frequency equation.

The circular motion occurs because the magnetic force (F = qvB) on the charged particle is perpendicular to its velocity (v) and results in a centripetal force that keeps the particle in a circular path with a constant speed.

The angular frequency (ω) represents the rate at which the particle completes a full revolution (2π radians) around the center of the circular path per unit of time.

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Question 15 (3. 33 points) Solve: What work is done when 3. 0 C is moved through an electric potential difference of 1. 5 V?

A)
0. 5 J

B)
2. 0 J

C)
4. 0 J

D)
4. 5 J

Answers

The following formula can be used to determine the work involved in moving a charge via an electric potential difference:

W = qΔV

where W stands for work completed, q for charge transported, and V for potential difference.

Inputting the values provided yields:

W = (3.0 C) x (1.5 V) = 4.5 J

As a result, 3.0 C moving across a 1.5 V electric potential differential requires 4.5 J of labour.

Response: D) 4.5 J

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A ford is traveling with a speed of 15m/s and is 200 meters ahead of a Chevy that is traveling in the same direction but at a speed of 20m/s. How far will the chevy travel before catching up to the Ford?

Answers

The Chevy will travel a distance f about 600m before catching up with the ford in the same direction of the motion.

The ford if travelling at 15m/s and it is 200 m ahead of the Chevy that is travelling in the same direction with a speed of 20m/s.

Now, we can use velocity = distance/time here,

Time of ford will be equal to the time of Chevy,

Time of ford = Distance/velocity of ford

Time of ford = S/15

Now, for the Chevy, the distance will be 200 more and the time will be same, so we will write,

Time of Chevy = (S+200)/20

(S+200)/20 = S/15

15S + 3000 = 20S

5S = -2000

S = -400m

Negative sign is showing the direction of the motion only, so we can ignore that.

So, the Chevy will travel 600 m before catching up with the ford.

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A lightbulb is in series with a 2.0 ohm resistor The lightbulb dissipates 10 W when this series circuit is connected to a battery. What is the current through the lightbulb? There are two possible answers; give both of them. Enter your answers in ascending order separated by commas. I = 1.78,2.96 A

Answers

The current through the lightbulb when a lightbulb is in series with a 2.0 ohm resistor is I = 1.78 A, 2.96 A.

The current through the lightbulb when a lightbulb is in series with a 2.0 ohm resistor is given by the Ohm's Law. Ohm's law states that current through a conductor between two points is directly proportional to the voltage across the two points. In other words, V = IR. Where V is the voltage, I is the current and R is the resistance. When the circuit is connected to a battery, the lightbulb dissipates 10W. Power can also be expressed as P = IV, where P is the power, I is the current and V is the voltage. We have V = IR and P = IV.

We can substitute V in terms of I from the first equation to the second equation.P = I²R=> I² = P/R=> I = √(P/R) = √(10W/2Ω) = 2.24 A. There are two possible answers. When we apply Ohm's Law to the circuit, I = V/R Where R = 2.0 Ω and V is the voltage across the resistor.I = V/2 = (10/2)^(1/2) A = 1.78 A.The other possible answer is the current through the light bulb.I = (10/2.0 + 2.0) A = 2.96 A. Therefore, the current through the lightbulb when a lightbulb is in series with a 2.0 ohm resistor is I = 1.78 A, 2.96 A.

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how is the change in momentum of a dynamic cart acted upon by the force of a spring related to the impulse

Answers

The change in momentum of a dynamic cart acted upon by the force of a spring is related to the impulse.Impulse is equal to the change in momentum of an object. The force that acts on an object over a given time period determines the impulse. It is the product of force and time.

Impulse, in fact, is also equal to the total momentum of the object before the force is applied. Impulse is a vector quantity with the same direction as the force, as well as the momentum.

The impulse delivered to the cart by the spring will be equal and opposite to the impulse exerted by the cart on the spring, according to Newton's third law of motion.

As a result, the change in momentum of the dynamic cart due to the force of a spring is related to the impulse.

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what bias has been applied to a pn junction if the cathode is positive with respect to the anode?

Answers

A forward biased has been applied to a P-N junction if the cathode is positive with respect to the anode.

A P-N junction is a type of semiconductor junction formed by joining together two types of semiconductors - P-type and N-type. P-type semiconductors have an excess of positively charged holes, while N-type semiconductors have an excess of negatively charged electrons.

When these two types of semiconductors are brought into contact, the electrons from the N-type material flow into the P-type material, filling some of the holes and creating a region with a net negative charge (the N-side), while the P-type material loses some of its holes and creates a region with a net positive charge (the P-side). This creates a potential difference between the P and N regions, which can be used to generate a current. P-N junctions are essential components in electronic devices such as diodes, transistors, and solar cells.

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Calculate the net force in each scenario below:
1.
2.
3.
4.
5.
20 N
40 N
20 N
8N
10 N
3N
7N
40 N
10 N
10 N
10 N
Net Foros:
Net Force:
Net Force:
Net Force:
Net Force:
Direction of motion:
Place a star inside the boxes that are UNBALANCED

Answers

Answer:

1. Net force: 60N (⭐️)

Direction: West

2. Net force: 60N

Direction: East

3. Net force: 18N (⭐️)

Direction: East

4. Net force: 20N

Direction: No movement

5. Net force: 20N

Direction: No movement

Explanation:

Hope you understand :)

The period of a satellite, the time it takes for a complete revolution, depends on the satellite's a. radial distance. b. mass. c. weight. d. all of these e. none of these

Answers

The period of a satellite, the time it takes for a complete revolution, depends on the satellite's radial distance. Hence, the correct option is a.

What is a satellite?

A satellite is an object in space that revolves around a planet, a moon, or even another satellite. Satellites, particularly those in the field of technology, enable the gathering of information and communication of information between two locations on Earth. Satellites can also be used for weather forecasting and military surveillance.

A revolution is one complete orbit around a central body for a satellite. The amount of time it takes for a satellite to complete one revolution is known as the satellite's period. As a result, it is clear that the period of a satellite depends on its radial distance. The closer a satellite is to the planet, the shorter its period would be, while the farther away it is, the longer its period would be.

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The skater with a mass of 50 kg slides on an ice track with a speed of 5 m/s. How fast will she move if she throws a 2kg stone horizontally, once in front of her and once behind her, at a speed of 2m/s? Friction is not considered.

Answers

Answer:

4.92 m/s for her final velocity.

Explanation:

The momentum of the skater before throwing the stone is:

p1 = m1 * v1 = 50 kg * 5 m/s = 250 kg*m/s

where m1 is the mass of the skater and v1 is her initial velocity.

When the skater throws the stone, the total momentum of the system (skater + stone) is conserved. The momentum of the stone is:

p2 = m2 * v2 = 2 kg * 2 m/s = 4 kg*m/s

where m2 is the mass of the stone and v2 is its velocity.

Let's assume the skater throws the stone in front of her. To conserve momentum, the skater will move in the opposite direction to the stone. Let's call the skater's final velocity v3. Then:

p1 = p2 + p3

where p3 is the momentum of the skater after throwing the stone. Substituting the values we get:

250 kgm/s = 4 kgm/s + 50 kg * v3

Solving for v3, we get:

v3 = (250 kgm/s - 4 kgm/s) / 50 kg = 4.92 m/s

So the skater's speed after throwing the stone in front of her is 4.92 m/s.

If the skater throws the stone behind her, the same conservation of momentum principle applies, and we get the same result of 4.92 m/s for her final velocity.

Sorry if I'm wrong

Q4. Convert these into proper vector notation:

Westward velocity of 42 km/h.

Position 6. 5 measured in m that is North of the reference point.

Downward acceleration measured in m/s2 that has a magnitude of 1. 9.

Answers

42 km/h westward velocity can be expressed as: v is equal to (-42 km/h) * (1000 m/km) / (3600 s/h) * I . Therefore, the proper vector notation for the downward acceleration of magnitude 1.9 m/s^2 is -1.9 m/s^2 in the downward direction (k).

where the unit vector pointing west is called i. If we condense this expression, we get: v = -11.67 m/s * I Hence, -11.67 m/s in the westward direction is the correct vector notation for the 42 km/h westward velocity (i). North of the reference point, position 6.5 measured in metres, can be expressed as: r = 6.5 m * j where j represents the unit vector pointing north. Hence, 6.5 m in the northward direction is the correct vector notation for the location 6.5 m north of the reference point (j). It is possible to express a downward acceleration with a magnitude of 1.9 in m/s2 as follows: a = -1.9m/s^2 * k where k is the unit vector in the downward direction. Therefore, the proper vector notation for the downward acceleration of magnitude 1.9 m/s^2 is -1.9 m/s^2 in the downward direction (k).

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calculate the electrostatic force between 1nc and 1nc charges at a distance of 1 m from each other. do not forget to mention the direction of the force, too.

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The electrostatic force between 1nc and 1nc charges at a distance of 1 m from each other is 9.0 × 10^-9 N. The direction of the force is given by Coulomb's law and is along the line joining the two charges. It is either repulsive or attractive based on the type of the charges.

What is Coulomb's law? Coulomb's law is an equation used to calculate the electrostatic force between two charged particles. According to this law, the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation for Coulomb's law is given by:

F = k * (q1 * q2) / r^2

Where F is the electrostatic force,k is Coulomb's constant,q1 and q2 are the charges of the particles, and r is the distance between the particles.

Given,

Charge of particle 1, q1 = 1 nc

Charge of particle 2, q2 = 1

distance between particles, r = 1

coulomb's constant, k = 9 × 10^9 N m^2/C^2

Now, we can use Coulomb's law to calculate the electrostatic force between the two charges. Substituting the given values in the equation:

F = k * (q1 * q2) / r^2= 9 × 10^9 N m^2/C^2 * (1 × 10^-9 C) * (1 × 10^-9 C) / (1 m)^2= 9.0 × 10^-9 N

Thus, the electrostatic force between 1nc and 1nc charges at a distance of 1 m from each other is 9.0 × 10^-9 N. The direction of the force is given by Coulomb's law and is along the line joining the two charges. It is either repulsive or attractive based on the type of the charges.

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An automatic saw has several forces acting on it. In a Cartesian system, a position-dependent force applied to the saw is =-kxy2j, with k = 2.50 N m³. Let's consider the displacement of the saw from the origin to point C (4.0 m, 4.0 m). Calculate the work done on the saw by if the displacement is along the straight-line y = x that connects these two points.​

Answers

The work done on the saw by the force if the displacement is along the straight-line y = x that connects these two points is -640.0 J.

How to calculate work done?

To calculate the work done on the saw by the force as it moves along the straight-line y = x that connects the two points, we need to first find the displacement vector and then use it to calculate the work done.

The displacement vector from the origin to point C is given by:

r = (4.0 m) i + (4.0 m) j

The force acting on the saw is given by:

F = -kxy² j = -2.50 (N m³) (x) (y²) j

Since it is moving along the straight-line y = x, we can substitute x = y into the expression for F:

F = -2.50 (N m³) (x) (y²) j = -2.50 (N m³) (y³) j

Substituting x = y = 4.0 m:

F = -2.50 (N m³) (4.0 m)³ j = -160.0 j N

The work done by the force is given by the dot product of the force and displacement vectors:

W = F · r = (-160.0 N j) · (4.0 m i + 4.0 m j)

W = (-160.0 N) (4.0 m cos(45°))

W = -640.0 J

Therefore, the work done on the saw by the force as it moves along the straight-line y = x that connects the two points is -640.0 J.

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Use Wien's law and a sunspot temperature of 3800 K to calculate the wavelength of peak thermal emission from a sunspot. Express your answer to three significant figures and include the appropriate units.

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The wavelength of peak thermal emission from a sunspot can be calculated using Wien's law and a sunspot temperature of 3800 K.

Wien's Law states that the wavelength of peak thermal emission is inversely proportional to the temperature of the body emitting radiation. It is given by:

λ_max = b/T

where b is the Wien constant, 2.898 x 10^-3 m K, and T is the temperature of the emitting body. Substituting the given values into the equation,λ_max = b/Tλ_max = (2.898 x 10^-3 m K)/(3800 K)λ_max = 7.63 x 10^-7 m

The answer is expressed to three significant figures as 7.63 x 10^-7 m, with units of meters. Therefore, the wavelength of peak thermal emission from a sunspot is 7.63 x 10^-7 m.

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consider the specific example of a positive charge q moving in the x direction with the local magnetic field in the y direction. in which direction is the magnetic force acting on the particle?

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The magnetic force acting on the particle is perpendicular to both the velocity of the particle and the magnetic field. Therefore, the force is in the z direction.


The magnetic force is acting in the direction of the z-axis. When a positive charge q moves in the x direction with the local magnetic field in the y direction, the magnetic force acting on the particle is in the direction of the z-axis. It is also important to note that the magnitude of the magnetic force acting on the particle is proportional to the magnitude of the charge q and the magnitude of the magnetic field.

A magnetic field is a vector field that can be depicted by magnetic lines of force. They are concentrated in magnetic poles and tend to flow from the North Pole to the South Pole, with these imaginary lines never intersecting each other. Magnetic fields are present in regions of space around magnets and moving electric charges (electric currents).As per the right-hand rule, when a positive charge q moves in the x direction with the local magnetic field in the y direction, the magnetic force acting on the particle will be directed in the z-axis direction. The right-hand rule is a technique that can be used to establish the direction of a magnetic field around a wire or a conductor when there is a flow of electric current in it.

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when one stationary object is replaced by another stationary object, the change between the two objects maybe perceived as the movement of a single object. this creates?

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When one stationary object is replaced by another stationary object, the change between the two objects maybe perceived as the movement of a single object. This creates an optical illusion.

An optical illusion is defined as a visual phenomenon in which the information gathered by the eye is processed in a way that results in a false perception of reality or the visual impression of seeing something that is not present or incorrectly perceiving it. It is a misinterpretation of a visual stimulus caused by the brain's ability to misjudge sensory information.

It can happen when visual information is processed in the brain, and it can create an impression of movement that isn't there. This phenomenon occurs when an object is moving or when the eyes are moving around, but it can also happen when the object being looked at is stationary.

When one stationary object is replaced by another stationary object, the change between the two objects maybe perceived as the movement of a single object. This creates an optical illusion because the visual system is misled into thinking that the object is moving.

The brain continues to process visual information even when the object is stationary, creating the impression that the object is moving. This is why an optical illusion can be used to make a stationary object appear to move or to make a moving object appear to be stationary.

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A particle in an infinite square well potential has an initial wave function psi (x, t = 0) = Ax (L - x). Find the time evolution of the state vector. Find the expectation value of the position as a function of time.

Answers

The position expectation value as a function of time is constant and is equal to L/3.

Given a particle in an infinite square well potential has an initial wave function Ψ (x, t = 0) = Ax (L - x).The time evolution of the state vector: The time evolution of the state vector is given by Ψ(x,t) = ΣC_nΨ_n (x) e^(-iE_n t/h).The expectation value of the position as a function of time:The expectation value of the position as a function of time is given by the formula given below:x = Σa_n^2x_nΨ_n(x)Ψ_n*(x). Where,

a_n is the coefficient for each energy level.

Energy levels for infinite square well potential is given byE_n = n^2h^2 / 8mL^2Now, let us find the value of coefficient A. We know that a particle in a square well is normalized using the following formula:

∫Ψ^2 dx = 1. 0 to L∫Ax(L-x)^2dx = 1A(L^3)/3 = 1, A = √(3/L^3).

Now, the wavefunction for the particle is given by:

Ψ (x, t = 0)

= Ax (L - x)

= √(3/L^3) x (L - x).

Now, we can express this wave function in terms of the energy eigenfunctions as below:

Ψ (x, t = 0)

= Σ a_nΨ_n (x)

= Σa_n sin((nΠx)/L).

We can calculate the value of coefficient a_n by integrating the product of the initial wavefunction with the energy eigenfunctions, which is given by: a_n = 2/L ∫Ψ(x, t = 0) sin((nΠx)/L) dx.

Now, let us calculate the value of coefficient

a_n.a_n = 2/L ∫Ψ(x, t = 0) sin((nΠx)/L) dxa_n

= 2/L ∫√(3/L^3) x (L - x)sin((nΠx)/L) dxa_n = 2√3/L^2 ∫x(L - x)sin((nΠx)/L) dx.

From the previous results of integration,

a_n = (-1)^n+1 24√3/nΠ^3

a_n = (-1)^n+1 24√3/nΠ^3

Ψ(x,t) = ∑ a_nΨ_n(x) exp(-iE_n t/ℏ). Where E_n = n²h²π² / 2mL².

Substituting the values of a_n in the above formula, Ψ(x,t) = Σ(-1)^n+1 24√3/nΠ^3 sin(nΠx/L) exp(-in²π²h²t/2mL²ℏ²). Expectation value of the position as a function of time: The expectation value of the position is given by the formula, x = Σa_n²x_n. Where x_n is the position of nth energy level.

So, x_n = L/nSo,x = L∑a_n²/n From the previous results of coefficient, Σa_n²/n = 1/3. Now, x = L/3. Hence the position expectation value as a function of time is constant and is equal to L/3.

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