A 44-kg ice skater moves across the ice at a constant speed of 2.0 m/s. She is caught by her 76-kg partner, and then the pair continues to glide together. He is at rest when he catches her, and immediately afterward they both coast. Even though the "magnitude of the change in momentum" is requested, you should provide a negative sign if the final momentum is less than (or more negative) than the initial momentum.
Part A: What is their velocity just after the catch? Assume she moves in the positive x-direction before her partner catches her.
Part B: A skater follows the girl at a constant speed of 1.0 m/s. What is the magnitude of the girl's change in momentum in the reference frame of the skater?
Please answer with units

Centripetal acceleration of the cyclist is 8.04 m/s^2, while the moving speed of the cyclist is 15.0m/s.

The centripetal acceleration of an object moving in a circle with a radius "r" and velocity "v" can be calculated using the following formula:

a = v^2 / r

where "a" is the centripetal acceleration.

Given the speed of the cyclist

(v) = 15.0 m/s and

the radius of the bend (r) = 28.0 m,

the centripetal acceleration can be calculated as follows:

= a

= v^2 / r

= (15.0 m/s)^2 / 28.0 m

= 225 m^2/s^2 / 28 m

= 8.04 m/s^2

So the centripetal acceleration of the cyclist is 8.04 m/s^2.

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The fundamental distinction is that RGB is used for electronic displays (cameras and monitors), whereas CMYK is used for printing. Many clients generate or alter their print-ready designs with design apps that employ the RGB colour mode.

It refers to the use of red, blue, and green LEDs in diverse combinations to generate varied light colours. RGB LEDs can intelligently alter colour saturation and hue immediately at the source, ensuring correct colour balance between LED lights, cameras, and existing or ambient light for natural-looking results.

The four ink plates used in certain colour printing are referred to as CMYK: cyan, magenta, yellow, and key (black). The CMYK model masks colours partially or completely.

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Answer:

An electric battery provides a current.. answer is continuous

Answer:

2.6 seconds

Explanation:

You have to divide the distance by the velocity to find the time:

46/18 = 2.55 ≈ 2.6 seconds

The answer you are looking for is approximately 2.6 seconds.

Solution/Explanation:

To get the correct answer, you must divide the distance and velocity.

This means that 46÷18=2.5556.

2.5556≈2.6.

So, therefore, the final answer is approximately 2.6 seconds.

I hope that this has helped you. Enjoy your day, and take care.

Answer:

A form of punctuation that marks the end of a sentence, statement, etc.

Explanation:

. is a period.

It can be used in a sentence, such as this one.

not sure what period you mean, but if you mean in history n stuff: it means a certain era/time something happened. like the "golden age" it's called that because it didn't last forever/ it was just a phase yknow?

Answer:

#2

Explanation:

Answer: B

Explanation: just did the quiz

Vector B shows the direction of the centripetal acceleration at this point.

Centripetal acceleration is the acceleration of a frame traversing a circular path. because the pace is a vector quantity (this is, it has both a magnitude, the rate, and a course), when a frame travels on a circular course, its route continuously adjustments, and accordingly its velocity changes.

Centripetal forces motivate centripetal accelerations. within the special case of the Earth's circular motion around the sun – or any satellite's ground movement around any celestial body – the centripetal force causing the movement is the end result of the gravitational enchantment among them.

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When the extension of the spring becomes 5 cm, the applied force is 66.67 N.

The length produced by a force of 100 N is 7.5 cm.

The spring constant of a spring is also known as force constant, is the ratio of force to the extension of a spring.

The force constant of a spring describes how stiff a spring is or how tough a spring is.

The force constant of the given spring is calculated as follows;

k = F/x

where;

The extension of the spring = 3 cm = 0.03 m

The force is given as 40 N

K = 40 N / 0.03 m

K = 1,333.33 N/m

If the extension of the spring becomes 5 cm, the applied force is calculated as follows;

F = kx

F = 1,333.33 x 0.05

F = 66.67 N

The extension produced by a force of 100 N;

x = F/k

x = 100/1,333.33

x = 0.075 m = 7.5 cm

The assumption made is that the applied force is directly proportional to the extension of the spring.

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Answer:

3F

Explanation:

Gravitational Force is directly proportional to the mass; thus if one mass is tripled then the force is triples as well.

The power developed is 8000 W.

Given data:

Force = 250 N

Distance traveled = 6 m

Velocity = 32 m/s

Let's find out the work done on the object by the applied force.

Work done is given by the product of force and distance covered:

W = F × s

W = 250 × 6 = 1500 J

Thus, the work done on the object by the applied force is 1500 J.

Next, let's determine the power developed.

Power is defined as the rate at which work is done, i.e.,

P = W / t

where P is power, W is work done, and t is time taken to do that work.

We know that velocity = distance / time. Rearranging the above expression, we get:

t = d / v

Substituting the given values, we get:

t = 6 / 32

P = W / t

Substituting the calculated value of W and t, we get:

P = 1500 / (6 / 32)

P = 8000 W

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Answer:

= 24.76m

= 2476 cm

Explanation:

18 meters from point of origin = 18m north 90 degrees?

17 meters from point of origin = 17m west 0 degrees?

                                                                                                        _

If these routes are one after the other then we get a triangle  |

18 side and 17 side

This means we need to do Pythagoras Theorem.

18^2 + 17^2 = c^2

sqrt 324 + sqrt 289 =  c^2

c^2 = sqrt 613 = 24.7588368

= 24.76m

The gravitational force that Saturn exerts on Rhea is 3.546 × 10^17 Newtons.

To calculate the gravitational force that Saturn exerts on Rhea, we can use the formula for gravitational force:

F = G * (m1 * m2) / r^2

Where:

F is the gravitational force

G is the gravitational constant (approximately 6.67430 × 10^-11 N m^2 / kg^2)

m1 is the mass of Saturn

m2 is the mass of Rhea

r is the distance between Saturn and Rhea

Given:

m1 (mass of Saturn) = 5.68 × 10^26 kg

m2 (mass of Rhea) = 2.31 × 10^21 kg

r (distance between Saturn and Rhea) = 527,108 km = 527,108,000 m

a) Calculating the gravitational force:

F = G * (m1 * m2) / r^2

F = (6.67430 × 10^-11 N m^2 / kg^2) * (5.68 × 10^26 kg * 2.31 × 10^21 kg) / (527,108,000 m)^2

Calculating this expression:

F ≈ 3.546 × 10^17 N

Therefore, the gravitational force that Saturn exerts on Rhea is approximately 3.546 × 10^17 Newtons.

b) To find a point between Saturn and Rhea where a spacecraft does not experience any gravitational pull, we need to consider the gravitational force equation.

Since gravitational force depends on the masses of the objects and their distance, there is no point between Saturn and Rhea where a spacecraft would be completely free from gravitational pull.

The gravitational force between two objects decreases with distance, but it never becomes zero unless the distance becomes infinitely large.

So, in the vicinity of Saturn and Rhea, there will always be a gravitational force acting on any object present.

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As a freely falling object picks up downward speed, then the power supplied by gravitational force will increase.

We know that Power = work done /time

Power = W ×t and work = Force × displacement.

W= F(d) = F× speed/time (distance = speed /time)

So, if we increase the speed then work done will also increase, eventually it increases the power.

The longer an object is in free fall, the faster it falls toward the ground due to gravity. means it depend on the mass of object.

A falling object accelerates as it falls downward rapidly. The quantity of air resistance rises in proportion to the speed. The pull of gravity eventually is balanced by the force of air resistance as it grows. The item will cease accelerating since there is no net force at this point in time 0 newton.

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The pressure at the point on the wing is \( P_1 - 18000 \text{ m}^2/\text{s}^2 \).The pressure at a point on the wing of the airplane can be calculated using Bernoulli's principle.

Bernoulli's principle states that as the velocity of a fluid (or air in this case) increases, the pressure decreases, and vice versa.

To calculate the pressure at this point on the wing, we need to use the equation:

\( P_1 + 0.5 \rho v_1^2 = P_2 + 0.5 \rho v_2^2 \)

where \( P_1 \) is the pressure at the standard altitude, \( v_1 \) is the velocity at the standard altitude, \( P_2 \) is the pressure at the point on the wing, and \( v_2 \) is the velocity at the point on the wing.

Given:

\( P_1 = \text{pressure at standard altitude} = ? \)

\( v_1 = \text{velocity at standard altitude} = 270 \text{ m/s} \)

\( v_2 = \text{velocity at the point on the wing} = 330 \text{ m/s} \)

\( \rho = \text{density of air} = \text{constant (we can ignore this for this calculation)} \)

We can rearrange the equation to solve for \( P_2 \):

\( P_2 = P_1 + 0.5 \rho v_1^2 - 0.5 \rho v_2^2 \)

Since we are not given the density of air, we can assume it to be constant and cancel it out from both terms. This simplifies the equation to:

\( P_2 = P_1 + 0.5 v_1^2 - 0.5 v_2^2 \)

Now we can substitute the given values and calculate \( P_2 \):

\( P_2 = P_1 + 0.5 (270 \text{ m/s})^2 - 0.5 (330 \text{ m/s})^2 \)

\( P_2 = P_1 + 0.5 \times 72900 \text{ m}^2/\text{s}^2 - 0.5 \times 108900 \text{ m}^2/\text{s}^2 \)

\( P_2 = P_1 + 36450 \text{ m}^2/\text{s}^2 - 54450 \text{ m}^2/\text{s}^2 \)

\( P_2 = P_1 - 18000 \text{ m}^2/\text{s}^2 \)

Therefore, the pressure at the point on the wing is \( P_1 - 18000 \text{ m}^2/\text{s}^2 \).

Please note that the actual value of \( P_1 \) is not given in the question, so we cannot provide a specific numerical answer. However, you can use this equation to calculate the pressure at any given point on the wing if the standard pressure at the standard altitude is known.

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Answer:

Explanation:

Statement A: Increasing a moving average window will enhance noise dampening.

Statement B: Decreasing a moving average window will enhance impulse response.

Statement C: Increasing the smoothing constant in exponential smoothing will enhance impulse response.

Statement D: Decreasing the smoothing constant in exponential smoothing will enhance noise dampening.

Let's evaluate each statement:

A. Increasing a moving average window will enhance noise dampening.

This statement is true. Increasing the window size of a moving average smooths out the fluctuations in the data and reduces the effect of noise, thus enhancing noise dampening.

B. Decreasing a moving average window will enhance impulse response.

This statement is false. Decreasing the window size of a moving average reduces the number of data points considered, leading to less smoothing and a faster response to changes in the data. It does not enhance impulse response.

C. Increasing the smoothing constant in exponential smoothing will enhance impulse response.

This statement is false. Increasing the smoothing constant in exponential smoothing increases the weight given to more recent data points, which results in smoother and more stable forecasts. It does not enhance impulse response.

D. Decreasing the smoothing constant in exponential smoothing will enhance noise dampening.

This statement is true. Decreasing the smoothing constant in exponential smoothing gives more weight to past observations, which helps in reducing the impact of noise and enhances noise dampening.

Based on the evaluation, 2 out of the 4 statements are correct.

The time it takes the ball to stay in the air is 2.33 seconds.

Time of flight is the total time taken to complete the projectile motion.

To calculate how long it takes the ball to stay in the air, we use the formula of time of flight below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the time of flight is 2.33 seconds.

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Stored elastic potential energy can be utilized in everyday objects such as springs and rubber bands, allowing for the transfer of energy through the conversion of potential energy into kinetic energy.

One common example of stored elastic potential energy being utilized is a compressed spring. When a spring is compressed, work is done on it to store potential energy, which can then be released to do work on other objects. For instance, a spring-loaded toy car will store potential energy in its compressed spring, which is then released when the car is let go, causing it to move forward. This energy transfer is from the spring to the car's kinetic energy.

Another example of stored elastic potential energy is a stretched rubber band. When a rubber band is stretched, energy is stored in its molecular bonds, which can be released when the band is allowed to snap back into its original shape. This potential energy can be utilized in everyday life, for example, in a slingshot. When the rubber band is stretched back in a slingshot, it stores potential energy, which is then released when the projectile is released, converting potential energy into kinetic energy. This energy transfer is from the rubber band to the projectile's kinetic energy.

In both cases, the transfer of energy occurs through the conversion of potential energy into kinetic energy, allowing for work to be done on another object. These examples show how the principle of stored elastic potential energy can be utilized in everyday objects, making them more efficient and useful.

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Answer:

125.66 R/s

Explanation:

First    1200  r / min = 20 r/sec

20 r/s  *  2pi Radians / r = 40 pi Radians / sec = 125.66 R/s

The food substance being used which turned Fehling's solution to brick red is a reducing sugar.

This is an indicator which is used to test for the presence of reducing sugar or aldehydes in a solution.

Brick red precipitate of copper(I) oxide is formed when it tests positive to compounds such as glucose etc.

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Answer:

ytrxrddyoxswsdyxgxghfx

jdjdu3jthh

hhhujusbrnog

hhjfjtinrny

ykrjrhrnirjtjjtt

tkrjthr74uu3jt

hri4urjjrjtjjtjtjy

y

Explanation:

uueuhhhwuejroskanficndui39wn

jebfufkr

Answer:

B. Convection

Explanation:

Convection is the process in which heat is transferred from one point to another by the actual movement of the material particles from a region of high temperature to a region of low temperature.

Answer:

they might adapt or move to a new place

Explanation:

Answer:

variable that doesnt change

Explanation:

contolled variable is that variable that does not change throughout the course of the experiment

Answer:

1.83 meters

Explanation:

1 meter = 100 cm

183cm/100cm= 1.83 meters

Answer:

bada bop boop pow

Explanation:

A condition in which incoming light focuses behind the retina, leading to difficulty focusing on close-up objects, common in older adults, in whom the lens becomes less elastic, is known as presbyopia.

Presbyopia is a vision problem associated with ageing that impairs the capacity to clearly see objects in close proximity. Presbyopia, a disorder where the near point of accommodation moves farther distant, affects many order individuals. It happens as a result of the eye's normal ageing process, notably the lens's loss of flexibility.

Blurred vision is brought on by the eye's inability to focus on close-up things as the lens loses elasticity. Presbyopia is a common eyesight issue that commonly manifests around the age of 40 and worsens as people age. Reading glasses, bifocals, or contacts made specifically for presbyopia can all be used to treat it.

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Answer:

7.35 m

Explanation:

We can solve with kinematics or with energy.

Using kinematics, given:

v₀ = 12 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (12 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 7.35 m

Using energy instead:

PE = KE

mgh = ½ mv²

gh = ½ v²

h = v² / (2g)

h = (12 m/s)² / (2 × 9.8 m/s²)

h = 7.35 m