What should the height of a slide in a park be so that a child will reach the bottom of the slide with a speed of 15.0 m/s? Use g = 10.0 m/s2. options a) 0.75M B)22.5M C)11.3M D)10.5M E)15.1M

Answer:

See the explanation below

Explanation:

The pressure is defined as the product of the density of the liquid by the gravitational acceleration by the height, and can be easily calculated by means of the following equation.

 \(P=Ro*g*h\)

where:

Ro = density of the fluid [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = elevation [m]

In this way we can understand that the greater pressure is achieved by means of the height of the liquid, that is, as long as the fluid has more height, greater pressure will be achieved at the bottom.

Therefore in order of decreasing will be  

The largest pressure with the largest height of the liquid, container B. The next is obtained with container D, the next with container A and the lowest pressure with container C.  

The pressure decreases as we go from the container B - D - A - C

The power delivered during this time is 358.4 Watts (rounded to the nearest tenth).

The equation is Power (P) = Work (W) / Time (t).

When calculating work, force (F) and distance (d) are multiplied:

Work (W) is equal to force times distance.

Given:

First, let's calculate the work:

Work (W) = Force (F) * Distance (d)

W = 128 N * 1.4 m

W = 179.2 Joules

Now, we can calculate the power:

Power (P) = Work (W) / Time (t)

P = 179.2 J / 0.5 s

P = 358.4 Watts

Therefore, the power delivered during this time is 358.4 Watts .

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Therefore, the electric field (magnitude only) at a distance of 8.0 cm from the wire is approximately 3.24 x 10^4 N/C.

The electric field of a long, thin wire can be determined by Coulomb's law. Coulomb's Law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

For a long, thin wire, the electric field is given by;

E = λ/2πε₀r

Where;

λ = linear charge density = 91 µC/m,

ε₀ = permittivity of free space = 8.85 x 10^-12 C^2/Nm^2

r = distance from the wire = 8.0 cm = 0.08 m.

Substitute the given values into the formula to find the electric field;

E = (91 x 10^-6)/(2 x π x 8.85 x 10^-12 x 0.08)

E≈ 32433.8 N/C

E≈ 3.24 x 10^4 N/C.

Electric field refers to the force per unit charge that one object exerts on another object due to the electric charge present in the objects. It is a vector quantity and is measured in newtons per coulomb (N/C).

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The total fraction of the original unpolarized light transmitted through both polarizers is 44.18% and 55.82% of the original light is absorbed by the analyzer.

To calculate the fraction of unpolarized light transmitted through the analyzer, we can use Malus' Law, which states that the intensity of polarized light passing through a polarizer is proportional to the square of the cosine of the angle between the transmission axis of the polarizer and the polarization direction of the light.

The fraction of unpolarized light transmitted through the first polarizer is 1/2, since unpolarized light contains equal amounts of light waves polarized in all directions. Therefore, the fraction of the polarized light transmitted through the second polarizer is given by:

cos^2(20°) = 0.9397^2 = 0.8836

The total fraction of the original unpolarized light transmitted through both polarizers is:

1/2 × 0.8836 = 0.4418, or approximately 44.18%.

B. The fraction of the original light absorbed by the analyzer is equal to the fraction of the original light that is not transmitted through the analyzer. Therefore, the fraction of the original unpolarized light absorbed by the analyzer is:

1 - 0.4418 = 0.5582, or approximately 55.82%.

Therefore, 44.18% of the original unpolarized light is transmitted through the analyzer, and 55.82% of the original light is absorbed by the analyzer.

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Our Sun is indeed unique compared to other stars in our galaxy due to size, mass, age, composition, proximity and luminosity.

Firstly, the Sun is a relatively average star in terms of size, mass, and luminosity. It falls within the spectral class G2V, known as a yellow dwarf star. This puts it in a category that is neither too hot nor too cool, providing a stable and long-lasting environment for the development and sustainability of life on Earth.

Secondly, the Sun's age is significant. It is estimated to be around 4.6 billion years old, which places it in the middle stage of its stellar evolution. This age has allowed for the gradual development and stabilization of planetary systems, including our solar system.

Furthermore, the Sun's composition is unique. It consists primarily of hydrogen (about 74% by mass) and helium (about 24% by mass), with trace amounts of other elements. This composition is vital for the Sun's nuclear fusion processes, which generate its energy and light.

Lastly, the Sun's proximity to us allows for detailed observations and studies, enabling a deeper understanding of stellar physics and providing a basis for comparisons with other stars in our galaxy and beyond.

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Jupiter and Neptune are both gas giants, meaning they have a thick atmosphere composed mainly of hydrogen and helium.

What is Planet?

A planet is a celestial body that orbits a star, is roughly spherical in shape, and has cleared its orbit of other debris. In our solar system, there are eight planets: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. These planets vary in size, composition, and other characteristics.

They also have strong winds and storms, and both have a system of rings and numerous moons. In terms of size, Jupiter is the largest planet in our solar system, while Neptune is the fourth largest.

Mercury and Venus are rocky planets, similar in composition to Earth. Mercury is the smallest planet in our solar system, while Venus is similar in size to Earth but has a very thick and hot atmosphere.

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Answer:

The net charge on each sphere is -1.5 q

Explanation:

Conductors are materials that allow the electrons which are the carriers of the charges to move between them, and when two conductors come in contact, the available charge is shared by the two conductors and the resultant like charges will spread on the surface of the conductor due to the repellent effect between similar charges such that if the conductors are identical, the resultant charge becomes evenly shared by the conductors when they become separated again

The given parameters of the conducting spheres meant to touch are;

The net charge on the first sphere, Q₁ = +5.0 q

The net charge on the second sphere, Q₂ = -8.0 q

The net charge on each sphere after touching and then separated, 'Q', is given as follows;

\(Q = \dfrac{Q_1 + Q_2}{2}\)

Therefore, by substituting the known values of the variables, we have;

\(Q = \dfrac{5 \ q+ (-8 \ q)}{2} = -\dfrac{3 \ q}{2} = -1.5 \ q\)

The net charge on each sphere, Q = -1.5 q.

The net charge on each sphere after spheres are brought together, allowed to touch is -1.5 q

Conductors are materials that allow the electrons which are the carriers of the charges to move between them, and when two conductors come in contact, the available charge is shared by the two conductors and the resultant like charges will spread on the surface of the conductor due to the repellent effect between similar charges such that if the conductors are identical, the resultant charge becomes evenly shared by the conductors when they become separated again

The given parameters of the conducting spheres meant to touch are;

The net charge on the first sphere, Q₁ = +5.0 q

The net charge on the second sphere, Q₂ = -8.0 q

The net charge on each sphere after touching and then separated, 'Q', is given as follows;

\(Q=\dfrac{Q_1+Q_2}{2}\)

Therefore, by substituting the known values of the variables, we have;

\(\dfrac{5q+(-8q)}{2}=-1.5q\)

Hence the net charge on each sphere after spheres are brought together, allowed to touch is -1.5 q

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Answer:

X rays is wrong on edg

Explanation:

Answer:

The tire fills up just like anything else that holds air when u pump a ball or tire up it fills all the way up cause it is a small confined space and after filling it with air the atoms of the air fill the tire up

Explanation:

Explanation:

what is time in this question

The statement that is true concerning a substance with high specific heat is as follows: the substance cools down slowly after heating. an example substance is aluminum metal (option A)

Specific heat or specific heat capacity refers to the heat capacity per unit mass of a pure substance.

In other words, specific heat is defined as the amount of heat needed to increase the temperature of 1kg of a material by 1K and is expressed in terms of J/kg·K or equivalently J/kg·°C.

The specific heat capacity of a material is a physical property. It is also an example of an extensive property since its value is proportional to size.

Water is an example of a substance that has an extremely high specific heat capacity, which makes it good for temperature regulation.

Therefore, a substance with high specific heat cools down slowly after heating and an example is aluminum metal.

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Answer:

i cant even see that no one can i will answer it if u can make it bigger

Explanation: eyes

Answer:

627.2 N

Explanation:

The Normal Force (N) is the force the ground exerts on the sled, which is equal to the weight (W) of the sled.  These forces act in the y direction.  The other forces, Tension, Applied Force, Force of Friction all act in the x direction.

N = W = mg = (64.0 kg)(9.8 m/s²) = 627.2 N

Answer:

Label A: red

Label B: yellow

Label C: green

Label D: blue

Label E: violet/purple

I think this should be correct. I'm sorry if I'm incorrect

Based on the degree of refraction from least to most refracted, the colors are as follows: Red, Orange, Yellow, Green, Blue, Indigo, Violet.

Refraction of light is the bending of light rays as they cross the boundary between two media of different densities.

White light consists of seven colors of light; Red, Orange, Yellow, Green, Blue, Indigo, Violet.

The red color has the shortest wavelength and is the least refracted while violet color has the longest wavelength and is most refracted.

Therefore, based on the degree of refraction from least to most refracted, the colors are as follows: Red, Orange, Yellow, Green, Blue, Indigo, Violet.

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Answer:

No, it is not necessary for them to have same mass.

Explanation:

Let both bodies have a density d1 and d2 respectively.

Since their volumes are equal V1 = V2

we know that, https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D

Hence, d1 =  and d2 =    

Taking the ratio of densities,we get

This implies that unless the bodies have same densities, the mass of the two bodies will not be same.

Velocity is defined as the rate of change of displacement. It's a vector quantity that specifies both speed and direction. The x-component of the velocity vector is 1.3 m/s², and the y-component of the velocity vector is -1.4 m/s².

To determine the direction of the resultant velocity vector with respect to the + x‐axis, we need to calculate the angle made by the vector with the x-axis.

The tangent of the angle is the ratio of the y-component of the velocity to the x-component of the velocity.

tan θ = (-1.4 m/s²) / (1.3 m/s²)
θ = tan⁻¹ (-1.4/1.3)
θ = -49.78°

Therefore, the direction of the resultant velocity vector with respect to the + x‐axis is -49.78°.

Note: The negative sign in the answer represents that the angle is measured clockwise from the + x-axis.

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The cocountable topology is coarser than the usual topology and is not Hausdorff.

Let X be an infinite set and P (X) the power set of X. We define three topologies on X: the cofinite topology, the left ray topology, and the cocountable topology. We will compare each topology to the usual topology on X. We denote the usual topology by u.  

The Cofinite Topology Let F be the family of subsets of X such that F is either finite or X. That is, F = {A ⊆ X : A is finite or A = X}. The cofinite topology on X is defined by Tcf = {U ⊆ X : X \ U ∈ F} ∪ {Ø}. The open sets in the cofinite topology are the complements of finite sets plus the empty set.

A subset A of X is closed if and only if A is either X or finite. Thus, in the cofinite topology, every infinite subset of X is dense in X. Compared to the usual topology, the cofinite topology has fewer open sets and is coarser. In other words, the cofinite topology is a weaker topology than the usual topology.

The cofinite topology is also Hausdorff since given any two distinct points x, y ∈ X, the complements of the cofinite sets containing x and y are disjoint

. The Left Ray Topology Let F be the family of subsets of X such that F contains the empty set and all sets of the form L(a) = {x ∈ X : x < a}, where a is any element of X. The left ray topology on X is defined by TL = {U ⊆ X : U = ∅ or U contains some set L(a) from F}.

The open sets in the left ray topology are the empty set, all left rays L(a), and all sets that contain a left ray L(a). A subset A of X is closed if and only if A is the empty set, X, or contains the right endpoint of every left ray it meets. The left ray topology is finer than the cofinite topology but coarser than the usual topology.

Thus, the left ray topology is a weaker topology than the usual topology but stronger than the cofinite topology.

The left ray topology is also Hausdorff. The Cocountable Topology Let F be the family of subsets of X such that F is countable or all of X. The cocountable topology on X is defined by Tcc = {U ⊆ X : X \ U ∈ F} ∪ {Ø}. The open sets in the cocountable topology are the complements of countable sets plus the empty set.

A subset A of X is closed if and only if A is either countable or all of X. Thus, in the cocountable topology, every countable subset of X is nowhere dense.

Compared to the usual topology, the cocountable topology is coarser. The cocountable topology is also not Hausdorff since any two nonempty open sets have nonempty intersection. Hence, in the cocountable topology, the closure of a singleton set is the whole space X.

Among the three topologies, the cofinite topology is the weakest topology, and it is also a Hausdorff space. The left ray topology is a topology that is weaker than the usual topology but stronger than the cofinite topology, and it is also a Hausdorff space. Finally, the cocountable topology is coarser than the usual topology and is not Hausdorff.

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Answer:

d.) Newton's first law

Explanation:

This is also called the law of inertia, which means that an object in motion will not stop unless a force is acted upon it, and vice versa. Try this out with a piece of paper and a quarter. Pull the paper from under the quarter slightly quick, and the quarter will stay on the table. Hope i helped you.

Answer:

I agreed because, gravity keeps the moon in orbit and causes a pop fly to smack into your baseball mitt.

Explanation:

yes.

Option C: 0.115 is the correct option.

The correct error between the thread plug gauge pitch diameter and the micrometer measurement is 0.115 mm.

Explanation:

In order to determine the correct error between the thread plug gauge pitch diameter and the micrometer measurement, we first need to calculate the difference between the two.

This will give us the error.

The formula we will use is:

Error = |Pitch Diameter - Micrometer Measurement|

Given that:

             Pitch Diameter = 22.35 mm

             Micrometer Measurement = 22.235 mm

Substituting the values, we get:

              Error = |22.35 - 22.235|

              Error = 0.115 mm

Therefore, the correct error is 0.115 mm.

Option C: 0.115 is the correct option. The correct error between the thread plug gauge pitch diameter and the micrometer measurement is 0.115 mm.

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Answer:

90m

Explanation:

it travels at 30 m/s so 30*3 is 90

A 4kg steel ball moving with a speed of 12m/s in the positive x-direction has a glancing collision with a 6kg ball at rest. After the collision, the 4kg ball moves at an angle of 25 degrees above the x-axis, and the 6kg ball moves at an angle of 40 degrees below the x-axis.The velocities of the two objects after the collision are given by v1f and v2f.

Using the conservation of momentum, we can write: (m1v1i + m2v2i) = (m1v1f + m2v2f) where m1 and v1i are the mass and initial velocity of the 4kg ball, m2 and v2i are the mass and initial velocity of the 6kg ball, and v1f and v2f are the final velocities of the two objects after the collision.

Using the conservation of kinetic energy, we can also write: (1/2)m1v1i² + (1/2)m2v2i² = (1/2)m1v1f² + (1/2)m2v2f² We can use these two equations to solve for v1f and v2f.

First, let's find the initial momentum and kinetic energy of the system:(4kg)(12m/s) + (6kg)(0m/s) = 48kgm/s(1/2)(4kg)(12m/s)² + (1/2)(6kg)(0m/s)² = 288J.

Now, we can use these values to solve for v1f and v2f.

Starting with the momentum equation and substituting in the known values: (4kg)(12m/s) = (4kg)(v1f cos(25°)) + (6kg)(v2f cos(40°)) 48kgm/s = (4kg)(v1f cos(25°)) + (6kg)(v2f cos(40°))

Dividing by 2, we get: 24kgm/s = (2kg)(v1f cos(25°)) + (3kg)(v2f cos(40°))

Solving for v1f cos(25°), we get: v1f cos(25°) = (24kgm/s - 3kgv2f cos(40°))/2kg

Now, let's use the kinetic energy equation to solve for v2f.

Substituting in the known values: 288J = (1/2)(4kg)(v1f sin(25°))² + (1/2)(6kg)(v2f sin(40°))²

Simplifying and solving for v2f, we get: v2f = sqrt((576 - 8v1f² sin²(25°))/(3 sin²(40°)))

Now that we have an expression for v2f, we can substitute it into the momentum equation to solve for v1f.

Substituting in the known values and simplifying, we get: v1f cos(25°) = (24 - 4v1f cos(25°) - 3v2f cos(40°))/2

Solving for v1f, we get: v1f = (24 cos(25°) - 3v2f cos(40°))/6cos(25°) + 2

We can substitute in the value of v2f we found earlier to get:v1f = 2.47m/sv2f = -2.34m/s.

Therefore, the 4kg ball moves at 2.47m/s at an angle of 25 degrees above the x-axis after the collision, and the 6kg ball moves at 2.34m/s at an angle of 40 degrees below the x-axis after the collision.

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Recall the formula,

∆θ = ω₀ t + 1/2 α t ²

where ∆θ = angular displacement, ω₀ = initial angular speed (which is zero because the disk starts at rest), α = angular acceleration, and t = time. Solve for the acceleration with the given information:

50 rad = 1/2 α (5 s)²

α = (100 rad) / (25 s²)

α = 4 rad/s²

Now find the angular speed ω after 3 s using the formula,

ω = ω₀ + α t

ω = (4 rad/s²) (3 s)

ω = 12 rad/s

Answer:

B.

Explanation:

The one that demonstrate the result of waves traveling at different speeds when moving from one medium to another is the bending of light as it moves from air to water. The correct option is B.

Even though waves travel at different speeds through different mediums, light bends or is refracted as it travels from one medium to another.

The degree to which it bends is referred to as the index of refraction. Light bends more as the index of refraction increases.

The change in direction of waves that occurs when they travel from one medium to another is referred to as refraction.

Refraction is always accompanied by a change in wavelength and speed. The bending of waves around obstacles and openings is known as diffraction.

The bending of light as it moves from air to water is an example of the result of waves traveling at different speeds when moving from one medium to another.

Thus, the correct option is B.

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The molar solubility of La(IO₃)₃ in pure water is approximately 1.0 × 10⁻⁴ mol/L.

The molar solubility of a compound refers to the maximum amount of the compound that can dissolve in a given solvent at a specific temperature, typically expressed in moles per liter (mol/L). To determine the molar solubility of La(IO₃)₃ in pure water, we can use the given value of the solubility product constant (Ksp) for La(IO₃)₃, which is 1.0 × 10⁻¹¹.

La(IO₃)₃ dissociates into La³⁺ and IO₃⁻ ions in water. Let's assume x mol/L represents the molar solubility of La(IO₃)₃. According to the balanced chemical equation, one mole of La(IO₃)₃ produces one mole of La³⁺ ions and three moles of IO₃⁻ ions.

Therefore, the solubility product expression for La(IO₃)₃ is:

Ksp = [La³⁺][IO₃⁻]³

Since the concentration of La³⁺ ions is equal to the molar solubility (x) and the concentration of IO₃⁻ ions is three times the molar solubility (3x), we can substitute these values into the Ksp expression:

Ksp = (x)(3x)³

1.0 × 10⁻¹¹ = 27x⁴

Solving for x, we find:

x ≈ 1.0 × 10⁻⁴ mol/L

Therefore, the molar solubility of La(IO₃)₃ in pure water is approximately 1.0 × 10⁻⁴ mol/L.

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The one reason why this graph demonstrates that the component being tested is a diode is because the high current depicted is in only one direction.

This is referred to as an electrical component that allows the flow of current in only one direction and it consists of specialized electronic component with two electrodes called the anode and the cathode.

From the graph, the current is in only one direction which is a major feature or characteristic of a diode and is therefore the reason why it was chosen as the correct choice in this scenario.

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