the moment of inertia of an object of mass m(a)m(a) is an intrinsic property of the object, (b) depends on the choice of axis of rotation, (c)(c) is proportional to mm regardless of the choice of axis, (d)(d) both (b)(b) and (c)(c) are correct.

The total internal energy of the gas is 5104 Joule.

Internal energy of helium gas=

3/2 × 1.3 × 8.31 × 315

= 5104 joule

Thus, we can conclude that the internal energy of the helium gas is 5104 Joule.

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Q3 should be placed equidistant between Q1 and Q2 to make the potential energy of the system equal to zero.

The potential energy of a system of charges depends on their relative positions and magnitudes. In this scenario, we have three charges: Q1, Q2, and Q3. To make the potential energy of the system equal to zero, Q3 needs to be placed in a specific position between Q1 and Q2.

The potential energy of the system is zero when the net force on Q3 is zero. This occurs when Q3 experiences equal and opposite forces from Q1 and Q2. To achieve this, Q3 must be placed equidistant from Q1 and Q2.

By positioning Q3 at the midpoint between Q1 and Q2, the forces exerted on Q3 by Q1 and Q2 will cancel each other out, resulting in a net force of zero. This equilibrium configuration ensures that the potential energy of the system is zero, as desired.

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The speed of the proton immediately after the collision is \(-1.52 *10^7 m/s\), and the speed of the carbon atom immediately after the collision is \(2.70 * 10^6 m/s\).

In a perfectly elastic head-on collision, both momentum and kinetic energy are conserved.

Therefore, we can use the conservation of momentum and energy to find the velocities of the proton and the carbon atom immediately after the collision.

Let m1 be the mass of the proton, v1 be its velocity before the collision, and m2 be the mass of the carbon atom, v2 be its velocity before the collision. After the crash, the speeds are v1' and v2', individually.

Utilizing the preservation of energy, we can compose:

\(m1v1 + m2v2 = m1v1' + m2v2'\)

Since the proton is traveling to the right and collides head-on with the stationary carbon atom, the initial velocity of the carbon atom is zero. Thusly, we can work on the above condition as:

\(m1v1 = m1v1' + m2v2'\)

Utilizing the preservation of motor energy, we can compose:

\((1/2)m1v1^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2\)

Presently, we can tackle these two conditions at the same time to find v1' and v2':

\(m1v1 = m1v1' + m2v2'\)

\((1/2)m1v1^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2\)

Solving for v1', we get:

\(v1' = ((m1 - m2)/(m1 + m2))v1\)

Solving for v2', we get:

\(v2' = ((2m1)/(m1 + m2))v1\)

Subbing the qualities given in the issue, we get:

\(v1 = 2.0 * 10^7 m/s\)

\(m1 = 1.67 * 10^-27 kg\)

\(m2 = 12.0 * 1.67 * 10^-^2^7 kg\)

Therefore,

\(v1' = ((1.67 * 10^-^2^7 - 12.0 * 1.67 * 10^-^2^7)/(1.67 * 10^-^2^7 + 12.0 * 1.67 * 10^-^2^7)) 2.0* 10^7 m/s\)

   \(= -1.52 * 10^7 m/s\)

\(v2' = ((2 * 1.67 * 10^-^2^7)/(1.67 *10^-^2^7 + 12.0 * 1.67 *10^-^2^7)) * 2.0 * 10^7 m/s\)

   \(= 2.70 * 10^6 m/s\)

Therefore, the speed of the proton immediately after the collision is \(-1.52 * 10^7 m/s\), and the speed of the carbon atom immediately after the collision is \(2.70 * 10^6 m/s\).

Note that the negative sign of v1' indicates that the proton is now traveling to the left after the collision.

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The statement that best illustrates that the light behaves like particles is "light bounces off a white cement sidewalk".

Light is electromagnetic radiation which can be seen by Bare eyes. It is the part of the electromagnetic spectrum.

Light is a particulate matter the reflection of light by striking to a surface is the best example to understand its nature.

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Answer:

the fastest would be gas, liquid would be in the middle and solid would be the slowest

Explanation: brainliest pls

Answer:

Solid → Liquid → Gas

Explanation:

The speed of molecules in an object increases as temperature increases.

An example of said molecules at different states of matter:

Radii is a vital feature of the elements, and it can be useful in determining the characteristics of elements in various chemical and physical processes. The radii of atoms and ions of the same element differ due to their various charge and mass characteristics.

Atomic and ionic radii increase as you move down a group on the periodic table, and decrease as you move across a period from left to right due to increased nuclear charge, making the electrons closer to the nucleus. The size of an atom and ion also changes due to the number of electrons charge, and electronic configuration.In order of increasing radius, the arrangement of \(Ne, F-, O2-, Br-, Rb\) is given as follows:

\(Ne < F- < O2- < Br- < Rb+\)

Rb+ has the smallest radius due to its large nuclear charge and fewer electrons in the valence shell.

As a result, they are larger than Rb+. O2- has more electrons than Ne and is the largest among the given ions and atoms. It is important to note that in certain conditions, the trends in radii may not be valid because of hybridization and other factors. Nonetheless, this arrangement is valid for the given ions and atoms.

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The energy of the flywheel can be calculated using the equation E = \(0.5Iw^2\), where I is the moment of inertia and ω is the angular velocity.

For a flywheel with a radius of 1m and mass of 500kg, the moment of inertia is\(500 x (1m)^2 = 500 kg-m2.\) The angular velocity is 500 revolutions per minute, which is equivalent to 8.667 rad/s. Plugging these values into the equation gives us \(E = 0.5 x 500 x (8.667)^2 = 35,584.7 J.\)

The energy of a flywheel is the energy stored in the kinetic form due to its rotation. It is defined as the product of the moment of inertia and the square of the angular velocity of the flywheel.

The energy stored in the flywheel is proportional to the square of the angular velocity and is also proportional to the moment of inertia. The greater the angular velocity, the more energy is stored in the flywheel.

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The planet's climate thermostat, as well as the world's chief greenhouse gas, is primarily water vapor. The correct option is A.

Water vapor is the most abundant greenhouse gas in the atmosphere, accounting for about 60% of the greenhouse effect. It is also the most important feedback mechanism in the climate system. When the Earth's temperature rises, more water evaporates from the oceans and other water bodies. This water vapor then traps more heat in the atmosphere, which further raises the temperature. This feedback loop can lead to runaway climate change.

Carbon dioxide is the second most abundant greenhouse gas in the atmosphere, accounting for about 20% of the greenhouse effect. It is released into the atmosphere by the burning of fossil fuels, deforestation, and other human activities. Carbon dioxide is a very long-lived greenhouse gas, meaning that it can remain in the atmosphere for hundreds of years.

Methane is the third most abundant greenhouse gas in the atmosphere, accounting for about 10% of the greenhouse effect. It is released into the atmosphere by the decomposition of organic matter, such as in landfills and wetlands. Methane is a very potent greenhouse gas, meaning that it has a much stronger warming effect than carbon dioxide.

Ozone is a greenhouse gas that is found in the stratosphere, the layer of the atmosphere that is about 10-50 kilometers above the Earth's surface. Ozone is formed when ultraviolet radiation from the sun splits oxygen molecules into two oxygen atoms. These oxygen atoms then combine with other oxygen molecules to form ozone. Ozone is a very effective absorber of ultraviolet radiation, which helps to protect the Earth from this harmful radiation. However, ozone is also a greenhouse gas, and it contributes to the greenhouse effect.

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Answer: 3 hours

Explanation: 150 km / 50 km = 3 hrs.

Figure 4 is the graph which best represents the relationship between the acceleration of an object falling freely near the surface of Earth and time taken.

This is defined as a pictorial representation of data in an organized and defined manner.

The acceleration of an object falling freely near the surface of Earth is constant which is why option D was chosen.

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Answer: Different types of fuels have varying compositions and release different amounts of pollutants when burned. Here are some common types of fuels and the pollutants associated with them:

Fossil Fuels:

a. Coal: When burned, coal releases pollutants such as carbon dioxide (CO2), sulfur dioxide (SO2), nitrogen oxides (NOx), and particulate matter (PM).

b. Petroleum (Oil): Burning petroleum-based fuels like gasoline and diesel produces CO2, SO2, NOx, volatile organic compounds (VOCs), and PM.

Natural Gas:

Natural gas, which primarily consists of methane (CH4), is considered a cleaner-burning fuel compared to coal and oil. It releases lower amounts of CO2, SO2, NOx, VOCs, and PM.

Biofuels:

Biofuels are derived from renewable sources such as plants and agricultural waste. Their environmental impact depends on the specific type of biofuel. For example:

a. Ethanol: Produced from crops like corn or sugarcane, burning ethanol emits CO2 but generally releases fewer pollutants than fossil fuels.

b. Biodiesel: Made from vegetable oils or animal fats, biodiesel produces lower levels of CO2, SO2, and PM compared to petroleum-based diesel.

Renewable Energy Sources:

Renewable energy sources like solar, wind, and hydropower do not produce pollutants during electricity generation. However, the manufacturing, installation, and maintenance of renewable energy infrastructure can have environmental impacts.

It's important to note that the environmental impact of a fuel also depends on factors such as combustion technology, fuel efficiency, and emission control measures. Additionally, advancements in clean technologies and the use of emission controls can help mitigate the environmental impact of burning fuels.

When a sled is initially pushed up a frictionless incline, it experiences the force of gravity pulling it down and the normal force exerted by the incline. Without the presence of friction, the sled can move up the incline more easily, leading to a smoother and more efficient ascent.

When a sled is pushed up a frictionless incline, there are two primary forces acting on it: the force of gravity pulling it down and the normal force exerted by the incline perpendicular to its surface. Since the incline is frictionless, there is no opposing force due to friction.

In summary, when a sled is given a shove up a frictionless incline, it will experience the force of gravity acting downward and the normal force from the incline. Without the presence of friction, the sled can move up the incline more easily as there is no resistance to its motion. The absence of friction allows for a smoother and more efficient ascent up the incline.

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The complete question is:

A sled is pushed up a frictionless incline at a 22.0-degree angle. It gains enough energy to reach a vertical height that is 1.05 meters higher than its starting point. What was its initial speed?

a) The value of the required velocity is v(t) = t²/2 - 4t + 4e⁻⁵

b) The value of the required distance is S(t) = t³/6 - 2t² + 4e⁻⁵t + 5

c) The value of the velocity when t = 9 is approximately -28.2 m/sec

(a)To find v(t), we need to integrate a(t) first

a(t) = 2∫

a(t) dt = ∫ 2

dt = 2t + C1

Here, C1 is the constant of integration

Next, we need to integrate v(t) by substituting the value of a(t) into it

v(t) = ∫ (2t + C1)

dt= t² + C1t + C2

Here, C2 is another constant of integration

v(0) = - 4⇒ C2 = - 4

Also, given: v(t) = Vt + 4e⁻⁵

Comparing the above two equations:

v(t) = Vt + 4e⁻⁵ = t² + C1t - 4

On differentiating with respect to t, we geta(t) = 2

Therefore, we get the differential equation as:

Vt + 4e⁻⁵ = t² + C1t - 4 - - - - - - (1)

To find v(t), we need to solve this differential equation

Substituting t = 0, in (1), we get

-4 + 4 = 0 + C1(0)⇒ C1 = 0

Therefore, the differential equation becomes:

Vt + 4e⁻⁵ = t² - 4

On differentiating the above equation w.r.t t, we get:

a(t) = 2

Therefore, we get

v(t) = t²/2 - 4t + 4e⁻⁵ - - - - - - (2)

Thus, the required velocity is given by:

v(t) = t²/2 - 4t + 4e⁻⁵

Part (b) To find s(t), we need to integrate v(t) first

v(t) = t²/2 - 4t + 4e⁻⁵∫ v(t)

dt = ∫ (t²/2 - 4t + 4e⁻⁵)

dt= t²/6 - 2t² + 4e⁻⁵t + C3

Here, C3 is the constant of integration

Also, given that S(0) = 5⇒ C3 = 5

Therefore, we get

S(t) = t³/6 - 2t² + 4e⁻⁵t + 5 - - - - - - (3)

Thus, the required distance is given by:

S(t) = t³/6 - 2t² + 4e⁻⁵t + 5

Part (c) To find the velocity when t = 9We can use the value of v(t) obtained in part (a)⇒ v(9) = (9)²/2 - 4(9) + 4e⁻⁵⇒ v(9) = - 28.2 (approx)

Thus, the velocity when t = 9 is -28.2 (approx)m/sec.

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