A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are ăÿÿfrom the vertical and releases her from rest. (a) What is the potential energy for the child just as she is released compared with the potential energy at the bottom of the swing? (b) How fast will she be moving at the bottom of the swing? (c) How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

a) The time after the release of the first stone that the second stone hits the water is 2.0 s.

b) 15.7 m/s is the initial speed of the second stone.

c)  The speed of the first stone as it hits the water is 15.7 m/s.

d) The speed of the second stone as it hits the water is 28.2 m/s.

Velocity is a vector quantity that measures both the speed and direction of an object's motion. It is equal to the rate of change of an object's position with respect to time. Velocity is usually represented by the symbol v and is measured in meters per second (m/s).

a) The time between first and second stone's release is 1.0 s. Since the time of release of first stone and the time of splash of both stones are same, the time between the release of second stone and the splash of both stones is 1.0 s.

Thus, the time after the release of the first stone that the second stone hits the water is 2.0 s.

b) The initial speed of the second stone can be calculated using the equation of motion,

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.

Substituting the values,

v² = (1.7)² + 2(9.8) * 59

v = 15.7 m/s

c) The speed of the first stone as it hits the water can be calculated using the equation of motion,

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.

Substituting the values,

v² = (1.7)² + 2(9.8) * 59

v = 15.7 m/s

d) The speed of the second stone as it hits the water can be calculated using the equation of motion,

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.

Substituting the values,

v² = (15.7)² + 2(9.8) * 59

v = 28.2 m/s

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The electric force between two point charges is given by the equation

[tex]F=k*q_1*q_2/r^2[/tex]

The interaction between two things is measured by the physical quantity known as force. It is a vector quantity, and the sign F is frequently used to denote it. When an object interacts with another object, it feels a push or a pull.

where r is the distance between the charges, q1 and q2 are their magnitudes, and k is the Coulomb constant.

When we enter the problem's specified values, we obtain

[tex]2.2N=8.99*10^9\ N*m^2/C^2*q*-2q/(1.4 m)^2[/tex]

which simplifies to

q = -0.500 N/C.

Thus, the magnitude of charge q is 0.500 N/C.

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According to the question the length of the river span of the bridge in meters is 853.3232 m.

Length is a physical quantity that measures the distance between two points. It is one of the fundamental units in the International System of Units (SI). It is usually measured in meters, although it can also be measured in other units such as centimeters, kilometers, feet, yards, miles, and so on.

The length of the river span of the bridge is 2799.0 ft. To convert this length to meters, we need to use a conversion factor. There are 0.3048 meters in one foot, so the conversion factor we will use is 1 ft
= 0.3048 m.

To convert 2799.0 ft to meters, we multiply by the conversion factor:
2799.0 ft * 0.3048 m/ft
= 853.3232 m

Therefore, the length of the river span of the bridge in meters is 853.3232 m.

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The value of the charge on each particle is [tex]1.05 x 10^-8 C[/tex].

Coulomb's law is a fundamental principle of electrostatics that describes the interaction between electric charges. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. We can use Coulomb's law to solve this problem. Mathematically,

[tex]F = k(q1q2)/r^2[/tex]

where F is the force of attraction or repulsion between the two charged particles,[tex]q1[/tex] and [tex]q2[/tex] are the magnitudes of the charges on the two particles, r is the distance between them, and k is Coulomb's constant, which has a value of [tex]9.0 x 10^9 Nm^2/C^2.[/tex]

In this problem, we know that the charges are equal and the distance between them is 10.0 cm. We also know that the force between them is 0.5 N. Therefore,

[tex]0.5 N = k(q^2)/(0.1 m)^2[/tex]

Solving for q, we get:

[tex]q = \sqrt{[(0.5 N)(0.1 m)^2/k]}[/tex]

[tex]q = \sqrt{(0.5 N)(0.01 m)/(9.0 x 10^9 Nm^2/C^2)}[/tex]

[tex]q = 1.05 x 10^-8 C[/tex]

Therefore, the value of the charge on each particle is [tex]1.05 x 10^-8 C.[/tex]

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Answer:

Explanation:

Because you have velocity along the y axis and time along the x axis, this is a velocity v time graph which is an acceleration graph. The slope of the line in this graph IS the acceleration. We can use 2 points and the slope formula to solve for the acceleration:

(0, 0) and (1, 3):

[tex]m=\frac{3-0}{1-0}=3[/tex] m/s squared, choice D.

Answer:

[tex]80\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

The frequency [tex]f[/tex] of a wave is the number of cycles completed in unit time ([tex]1\; {\rm s}[/tex] in this example.) In this question, [tex]f = 40\; {\rm s^{-1}}[/tex] ([tex]1\; {\rm Hz} = 1\; {\rm s^{-1}}[/tex]) means that the wave would complete [tex]40[/tex] cycles in every [tex]1\; {\rm s}[/tex].

The wavelength [tex]\lambda[/tex] of a wave is the distance the wave travels in each cycle. It is given that [tex]\lambda = 2\; {\rm m}[/tex].

The goal is to find the wave speed, which is the distance that this wave travels in unit time ([tex]1\; {\rm s}[/tex].)

In this question, it is given that [tex]\lambda = 2\; {\rm m}[/tex] and [tex]f = 40\; {\rm s^{-1}}[/tex]. Thus, this wave would travel a total of [tex]40\, (2\; {\rm m}) = 80\; {\rm m}[/tex] for the [tex]40[/tex] cycles completed in each unit time of [tex]1\; {\rm s}[/tex] ([tex]\lambda = 2\; {\rm m}[/tex] for each cycle.) The speed of this wave would be [tex]80\; {\rm m\cdot s^{-1}}[/tex].

Formally, the speed [tex]v[/tex] of this wave can be found by multiplying the wavelength [tex]\lambda[/tex] of this wave by its frequency [tex]f[/tex]:

[tex]\begin{aligned}v &= \lambda\, f \\ &= (2\; {\rm m})\, (40\; {\rm s^{-1}) \\ &= 80\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

A sound that is 7.8x10-4 W/m2 in intensity is equal to (10 dB)log3.2106 W/m21012 W/m2=185 dB.

The decibel, often known as the db or 0.1 bel, is the standard measurement unit. Hence, b = 10 log10 (I/I0) can be used to express the relationship between relative intensities, or b, in decibels. This equation can be used to determine that one decibel equals a 26 percent intensity variations.

The "decibel level" of a sound is a less formal term for relative intensity level. It is not the same as energy; relative intensity level reflects loudness more faithfully by using a logarithmic scale.

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The system's final temperature is roughly 16.7°C.

You may determine your substance's final heat by multiplying the temperature change by the initial temperature. Your water's final temperature would be 24 + 6, or 30 degrees Celsius, for instance, if it started off at 24 degrees Celsius.

The following is the formula for energy conservation:

Q1 + Q2 = 0

Q = mcΔT

Q1 + Q2 = 0

568.8

Simplifying and solving for

6394.4 - 106768 = 0

= 16.7°C

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a. We can actually see here that the girl have kinetic energy which is respect to the escalator.

b. The kinetic energy does not depend on the chosen reference.

Kinetic energy is a form of energy that an object possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its current velocity. Kinetic energy is a scalar quantity, meaning it only has magnitude and no direction. The formula for calculating kinetic energy is:

KE = 1/2 × m × v²

Where KE is the kinetic energy, m is the mass of the object, and v is its velocity.

The concept of kinetic energy was first introduced by the French mathematician Gaspard-Gustave de Coriolis in 1829. It was later developed by other scientists such as James Prescott Joule and Hermann von Helmholtz.

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Answer: d. increase

Explanation:

If the sun were more massive, the gravitational force between the sun and Earth would increase. This means that Earth's gravity with the sun would also increase. Therefore, the correct answer is (D) increase.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. So, if the mass of one of the objects increases, the gravitational force between them will also increase. In this case, if the mass of the sun were to increase, the gravitational force between the sun and Earth would become stronger, and hence, Earth's gravity with the sun would also increase.

A cesium-133 atom's radiation goes through 1.5 million cycles in around 0.1633 microseconds (or 163.3 nanoseconds).

9,192,631,770 hertz (cycles per second) is the frequency of the microwave spectral line that the isotope cesium-133 emits. The basic unit of time is provided by this. Cesium clocks have an accuracy and stability of 1 second in 1.4 million years.

The radiation emitted by cesium-133 has a frequency of 9,192,631,770 cycles per second, or 9.192631770 109 Hz.

The following formula may be used to determine how long 1.5 million radiation cycles take to complete:

Time is equal to the frequency of cycles.

Plugging in the numbers, we get:

time = 1.5 million / 9.192631770 × 10^9 Hz

time = 1.632995101 × 10^-7 seconds

So it takes approximately 0.1633 microseconds (or 163.3 nanoseconds) for radiation from a cesium-133 atom to complete 1.5 million cycles.

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The top surface of the oil on that side of the tube is 0.6 times higher than the surface of the water on the other side of the tube.

The principle of hydrostatics, also known as Pascal's principle, states that when an external pressure is applied to a fluid in a container, that pressure is transmitted uniformly in all directions within the fluid, regardless of the shape or volume of the container. In other words, the pressure applied to a confined fluid will be distributed evenly throughout the fluid and will not change in magnitude at any point within the fluid. This principle is important in a number of applications, such as hydraulic systems, which use fluids to transmit force and pressure from one point to another. It is also used to explain how liquids exert pressure on the walls of their container and how objects can float or sink in fluids.

We can use the principles of hydrostatics to solve this problem. Let's call the height difference between the two water surfaces h. We can assume that the oil completely covers the water on one side of the tube and does not mix with it, so the oil and water form two separate liquid columns with a common interface. Let's call the height difference between the oil and water surfaces on the same side of the tube H.

The pressure at any given point in a fluid depends only on the depth of that point below the surface of the fluid and the density of the fluid. Since the two water columns are at the same height, they experience the same pressure from the atmosphere. Similarly, the two oil columns experience the same pressure from the atmosphere.

Now consider a point on the interface between the oil and water on the same side of the tube. This point is at a depth of h+H below the water surface on the other side of the tube, so the pressure at this point is greater than atmospheric pressure by an amount equal to the product of the density of water, the acceleration due to gravity, and the total depth (h+H):

P = Patm + ρwatergh

where P is the pressure at the interface, Patm is atmospheric pressure, ρwater is the density of water, g is the acceleration due to gravity, and h+H is the total depth.

Similarly, the pressure at this point is less than atmospheric pressure by an amount equal to the product of the density of oil, the acceleration due to gravity, and the depth of the oil column (H):

P = Patm - ρoilgH

Since the interface between the oil and water is at the same pressure, we can equate these two expressions for P:

Patm + ρwatergh = Patm - ρoilgH

Solving for H, we get:

H = h(ρwater/ρoil)

Substituting the given values, we get:

H = 0.6h

Therefore, the top surface of the oil on that side of the tube is 0.6 times higher than the surface of the water on the other side of the tube.

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