If the pressure is increased by 2 times of a certain amount of gas at a fixed temperature, then what would be its final volume?.

Answer:

I think it's D

Explanation:

because kinetic basically means moving and potential mean sitting still so ig it's c or D

If the radius of Cosmo is equal to 2Re, the gravitational field strength at Cosmo's surface is equal to that at Earth's surface.

The Earth's radius, as measured by 183 high school students involved in the World Year of Physics project "Measure the Earth with Shadows," is 6563 km, as opposed to the commonly accepted figure of 6371 km for the mean radius.

About ten times as much mass as Mars is the Earth. If an object weighs 200 N on Earth's surface, it will weigh similarly on Mars.

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Calculating ΔG°rxn at 25°C using the provided table, we obtain:

A.) -510.0 kJ/mol B.) -138.92 kJ/mol C.) -1915.46 kJ/mol  

To calculate the \(\Delta G^\circ_\text{rxn}\) for the given reactions at 25°C, you can use the following formula:

\(\Delta G_{\text{rxn}}^{\circ} = \sum \Delta G_{\text{f}}^{\circ}(\text{products}) - \sum \Delta G_{\text{f}}^{\circ}(\text{reactants})\)

Let's calculate the \(\Delta G^\circ_\text{rxn}\) for each reaction:

A.) 2Mg(s) + O₂(g) --> 2MgO(s)

\(\Delta G_{\text{rxn}}^{\circ} = (2 \cdot \Delta G_{\text{f}}^{\circ}(\text{MgO})) - (2 \cdot \Delta G_{\text{f}}^{\circ}(\text{Mg}) + \Delta G_{\text{f}}^{\circ}(\text{O}_2))\)

\(= (2 \times (-255.0 \, \text{kJ/mol})) - (2 \times 0 \, \text{kJ/mol} + 0 \, \text{kJ/mol})\)

= -510.0 kJ/mol

B.) 2SO₂(g) + O₂(g) --> 2SO₃(g)

\(\Delta G^\circ_\text{rxn} = (2 \times \Delta G^\circ_\text{f}(SO_3)) - (2 \times \Delta G^\circ_\text{f}(SO_2) + \Delta G^\circ_\text{f}(O_2))\)

\(= (2 \times (-370.53 \, \text{kJ/mol})) - (2 \times (-300.57 \, \text{kJ/mol}) + 0 \, \text{kJ/mol})\)

= -740.06 kJ/mol + 601.14 kJ/mol

= -138.92 kJ/mol

C.) 2C₂H₂(g) + 5O₂(g) --> 4CO₂(g) + 2H₂O(l)

\(\Delta G^\circ_\text{rxn} = (4 \times \Delta G^\circ_\text{f}(CO_2) + 2 \times \Delta G^\circ_\text{f}(H_2O)) - (2 \times \Delta G^\circ_\text{f}(C_2H_2) + 5 \times \Delta G^\circ_\text{f}(O_2))\)

\(= (4 \times (-394.65 \, \text{kJ/mol}) + 2 \times (-237.35 \, \text{kJ/mol})) - (2 \times 168.43 \, \text{kJ/mol} + 5 \times 0 \, \text{kJ/mol})\)

= -1578.6 kJ/mol - 336.86 kJ/mol

= -1915.46 kJ/mol

Therefore, the \(\Delta G^\circ_\text{rxn}\) values for the given reactions at 25°C are:

A.) -510.0 kJ/mol

B.) -138.92 kJ/mol

C.) -1915.46 kJ/mol

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Answer:

Assume that the tractor and the car are moving in the same direction. If the tractor keeps moving at the same speed, the car need to decelerate at a minimum rate of approximately \(2.8\; {\rm m\cdot s^{-2}}\) (\((25 / 9)\; {\rm m\cdot s^{-2}}\)) to avoid collision.

Explanation:

Relative to the tractor, the car was initially moving at \(80\; {\rm km \cdot h^{-1}} - 20\; {\rm km \cdot h^{-1}} = 60\; {\rm km \cdot h^{-1}}\).

Apply unit conversion; ensure that the unit of velocity is \({\rm m\cdot s^{-1}}\):

\(\begin{aligned} 60\; {\rm km \cdot h^{-1}} &= 60\; {\rm km \cdot h^{-1}} \times \frac{1\; {\rm m\cdot s^{-1}}}{3.6\; {\rm km \cdot h^{-1}}} \\ &\approx \; 16.7\; {\rm m\cdot s^{-1} \end{aligned}\).

In other words, the initial velocity of the car was \(u \approx 16.7\; {\rm m\cdot s^{-1}}\) relative to the tractor.

Since the tractor is moving at constant velocity, the acceleration \(a\) of the car relative to the tractor is the same as the acceleration of the car relative to the ground.

It is given that the initial distance between the tractor and the car was \(50\; {\rm m}\). The final velocity \(v\) of the car should be no more than \(0\; {\rm m\cdot s^{-1}}\). Otherwise, the car would keep moving toward the tractor until the two vehicles collide.

Relative to the tractor, if the deceleration of the car was at the minimum safe value:

Apply the SUVAT equation \(a = (v^{2} - u^{2}) / (2\, x)\) to find the acceleration of the car:

\(\begin{aligned}a &= \frac{v^{2} - u^{2}}{2\, x} \\ &\approx \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (16.7\; {\rm m\cdot s^{-2}})^{2}}{2 \times 50\; {\rm m}} \\ &\approx (-2.8)\; {\rm m\cdot s^{-2} \end{aligned}\).

In other words, the deceleration of the car should be at least \(2.8\; {\rm m\cdot s^{-2}}\) relative to the tractor and to the ground.

Answer: B) 100N

Explanation:usatestprep

the 4.40 μF capacitor in series with the 5.80 kΩ resistor and 150 V emf source will charge up to 63.2% of its maximum voltage after one time constant, and will approach 150 V after several time constants.

When the emf source is connected to the circuit, current will start to flow and charge will begin to accumulate on the capacitor. The rate of charging will be determined by the time constant of the circuit, which is equal to the product of the resistance and capacitance (RC). In this case, the time constant is:

RC = 5.80 kΩ * 4.40 μF = 25.52 ms

After one time constant (25.52 ms), the capacitor will have charged to approximately 63.2% of its maximum voltage. After two time constants, it will have charged to approximately 86.5% of its maximum voltage, and after three time constants it will have charged to approximately 95% of its maximum voltage.

The maximum voltage that the capacitor will reach is equal to the emf of the source (150 V) because there is negligible internal resistance in the source. Therefore, the capacitor will eventually charge to 150 V, but it will take multiple time constants to get close to this value.

the 4.40 μF capacitor in series with the 5.80 kΩ resistor and 150 V emf source will charge up to 63.2% of its maximum voltage after one time constant, and will approach 150 V after several time constants.

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Answer: C. Work.

Explanation: A heat engine is a device that operates on the principles of thermodynamics to convert heat energy into mechanical work.  It does so by taking in heat from a high-temperature source, performing work, and then releasing waste heat to a low-temperature sink.

The process through which a heat engine converts heat into work is known as the thermodynamic cycle.  One common example of a heat engine is the internal combustion engine found in cars. In this engine, the combustion of fuel releases heat, which is then converted into mechanical work to propel the vehicle.

The conversion of heat into work in a heat engine is based on the second law of thermodynamics, which states that heat flows from a high-temperature region to a low-temperature region spontaneously.  This flow of heat can be harnessed to do useful work by controlling the transfer of heat energy and maintaining a temperature difference.

Therefore, when a heat engine operates, it takes in heat from a high-temperature source, uses it to perform work, and then releases waste heat to a low-temperature sink. The work done by the heat engine can be used to power machines, generate electricity, or perform other useful tasks.

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Answer:

Mass of Vehicle = 600 kg

Force = F = 7000 N

Drag force = f = 500 N

Let Fnet is net force on Vehicle then

Fnet = F - f

Fnet = 7000 - 500

Fnet = 6500 N

Let a is acceleration of Vehicle then by Newton's law

Fnet = Ma

a = Fnet / M

a = 6500/ 600

a = 10.83 m/s2

This is acceleration of Vehicle

Explanation:

Answer:

x = 0.78 m = 78 cm

Explanation:

Applying the law of conservation of energy to the spring-mass system, we can write the following equation:

\(Kinetic\ Energy\ at\ Mean\ Position= Potential\ Energy\ at\ extreme\ position\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2\\\\x^2 = \frac{mv^2}{k}\\\\x = \sqrt{\frac{mv^2}{k} }\)

where,

x = maximum displacement = ?

m = mass = 3.6 kg

v = speed at mean position = 5.2 m/s

k = spring constant = 160 N/m

Therefore,

\(x = \sqrt{\frac{(3.6\ kg)(5.2\ m/s)^2}{160\ N/m} }\\\\\)

x = 0.78 m = 78 cm

Answer:

H2O is also known as for water

Answer:

a low protein diet helps you lose weight because lack of protein can make you lose muscle mass

Explanation:

It cuts your strength it makes it harder to keep your balance and slows your metabolism and later you get tired easily it leads to muscle wasting overtime and sheds your fat

Based on the information provided, the direction of the magnetic field produced by the electrons going around a circle in a counterclockwise direction depends on the orientation of the circle .

with respect to the observer's viewpoint. Using the right-hand rule, which states that if you point your right thumb in the direction of the current (or the motion of electrons), the curl of your fingers indicates the direction of the magnetic field, we can determine the direction of the magnetic field in different scenarios:

If the circle is oriented such that the current is flowing counterclockwise and the circle is in a plane perpendicular to the plane of the paper (out of the page), then the magnetic field would be directed to the left.

If the circle is oriented such that the current is flowing counterclockwise and the circle is in a plane parallel to the plane of the paper (in the plane of the page), then the magnetic field would be directed into the page.

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The force on the string with a mass of 1 kg attached it is 250 N.

Force is the product of mass and acceleration. The S.I unit of force is Newton (N). Force is a vector quantity because it can be measured in terms of magnitude and direction.

To calculate the force, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

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An astronomer reports they have measured the parallax angle of 10 arc seconds for a star. which is reasonable as "the smaller a star's parallax angle, the further away it is."

A parallax angle is the distance evaluated from a nearby star between both the Earth at one time year and the Earth six months after that.

Some key features regarding the parallax angle are-

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Answer: Seatbelts relate to Newton's first law of motion because in the case of a car accident, someone in the car would continue forward until they hit somethng else. A seat belt will help prevent this because it will push back on you (action-reaction force).

Explanation:

In this era of economic, technical, and political change, elected leaders, political appointees, government employees, and American citizens are searching for new perspectives on some of the most pressing problems of the day.

What is an example of technology?

All of these items are examples of technology, whether they are useful (like washing machines, dryers, refrigerators, vehicles, flooring, windows, or door knobs) or amusing (like televisions, Blu-ray players, games consoles, recliners, or toys).

Why is technology so crucial?

Technology has a significant impact on how businesses operate. Regardless of the size of your business, technology delivers both concrete and abstract advantages that can help you generate revenue and deliver the outcomes your consumers want.

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Without tipping over, the boy can walk up to 0.69 m from the right end of the beam.

To solve this problem, we need to use the principles of statics, which dictate that the sum of the forces and the sum of the torques acting on a body at rest must be zero.

First, let's find the normal force provided by each of the posts to support the beam:

The weight of the beam is acting downward at its center, which is 2.5 m from each post. Therefore, each post must provide a normal force equal to half the weight of the beam to balance it. The normal force provided by each post is:

N = (1/2)mg = (1/2)(30 kg)(9.81 m/s²) = 147.15 N

Next, let's consider the boy walking along the beam. We can treat the system as two separate parts: the beam with its weight and the normal forces from the posts, and the boy with his weight.

To prevent the beam from tipping over, the sum of the torques acting on the beam-boy system must be zero. We can choose the left post as the pivot point and calculate the torque due to each force:

- The weight of the beam creates a counterclockwise torque of:

τ_beam = (30 kg)(9.81 m/s²)(2.5 m) = 735.75 N·m

- The normal force provided by the left post creates a clockwise torque of:

τ_left = (147.15 N)(2.5 m) = 367.87 N·m

- The normal force provided by the right post creates a clockwise torque of:

τ_right = (147.15 N)(5.0 m - 2.5 m) = 367.87 N·m

- The weight of the boy creates a counterclockwise torque, which depends on his position along the beam. Let's call his distance from the right end of the beam x. Then his torque is:

τ_boy = (40 kg)(9.81 m/s²)(2.5 m + x)

For the system to be in equilibrium, the sum of these torques must be zero:

τ_beam + τ_left + τ_right + τ_boy = 0

Substituting the values we found and solving for x, we get:

(735.75 N·m) - (367.87 N·m) - (367.87 N·m) - (40 kg)(9.81 m/s²)(2.5 m + x) = 0

Simplifying and solving for x, we get:

x = 0.69 m

Therefore, the boy can walk up to 0.69 m from the right end of the beam without it tipping over.

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The new volume corresponding to a new pressure can be calculated using Boyle's law. The volume of the gas when pressure increased to 2.40 atm is 3.01 L.

According to Boyle's law, the volume of a gas is inversely proportional to the pressure of the gas. Thus PV = a constant.

If a system have initial pressure P1 and volume of V1 when changes to final pressure of P2 and volume V2, the relation between them can be written as follows:

P1V1 = P2V2

Thus for the initial pressure of 1.7 atm and 4.25 l volume, the final volume corresponding to 2.4 atm pressure is calculated as follows:

1.7 atm × 4.25 L = 2.4 atm × V2

V2 = (1.7 atm × 4.25 L)/2.4 atm

     = 3.01 L.

Thus,  volume of the gas when pressure increased to 2.40 atm is 3.01 L.

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To derive the algebraic equation for the vertical force, we must consider the forces acting in the vertical direction and apply Newton's second law of motion.

Newton's second law states that the net force acting on an object is equal to the product of its mass (m) and its acceleration (a). In the vertical direction, the forces that typically act on an object are gravity and any other forces such as the normal force or applied forces.

Assume that the vertical force is denoted as Fv. Forces acting in the vertical direction are typically mass (mg) and the normal force (N) exerted by the surface, if present. Therefore, the equation for the vertical force can be expressed as:

Fv = N - mg,

where:

Fv is the vertical force,

N is the normal force,

m is the mass of the object,

g is the acceleration due to gravity (approximately 9.8 m/s² on Earth).

The normal force N is equal to the weight of the object if the object is at rest on a horizontal surface. In this case, N = mg. However, if the object is on an inclined plane or is subject to other forces, the normal force may differ from the weight.

By substituting the appropriate values ​​for the normal force and mass (mg), you can derive an algebraic equation for the vertical force based on the specific scenario you are considering.

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Answer:

Explanation:

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Considering the definition of work,  the work done by the force is 1000 J.

Work is defined as the force that is applied on a body to move it from one point to another. When a force is applied, an energy transfer occurs.

When a net force is applied to the body or a system and this produces displacement, then that force is said to perform mechanical work.

In the International System of Units, work is measured in Joule.

The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves.

In this case, the angle between the force and the displacement is 0 degrees because the force is parallel to a horizontal table. The cosine of 0 degrees will have a value of 1. So, the force can be calculated as:

Work= Force× distance× cosine α

Work= force× distance× 1

Work= force× distance

Being:

you get:

Work= 50 N× 20 kg

Work= 1000 J

Finally, the work is 1000 J.

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0.77 m/s2 directed 35° south of west

net force = (-17,-12)

net force = mass * acceleration

(-17,-12) = 27 * (x-acceleration,y-acceleration)

(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)

angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.

magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)

The energy of the scattered photon increases as the scattering angle in the Compton effect increases. The correct option is A.

The Compton effect refers to the scattering of photons by charged particles, typically electrons. When a photon interacts with an electron, it can transfer some of its energy and momentum to the electron. As a result, the photon changes its direction and its energy.

In the Compton effect, the change in the energy of the scattered photon depends on the scattering angle, which is the angle between the initial and final directions of the photon. According to the Compton formula, the change in energy (ΔE) of the scattered photon is given by:

ΔE = E' - E = \(\frac{{h}}{{m_e c}}(1 - \cos \theta)\),

where E' is the energy of the scattered photon, E is the initial energy of the photon, h is the Planck's constant, me is the mass of the electron, c is the speed of light, and θ is the scattering angle.

From the Compton formula, it can be observed that as the scattering angle increases (θ increases), the term (1 - cosθ) becomes larger, resulting in a larger change in energy (ΔE). Therefore, the energy of the scattered photon increases as the scattering angle increases in the Compton effect. Option A is the correct answer.

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Answer:

Gravity is a non-contact force that was observed famously by Sir Isaac Newton, who watched an apple falling from a tree to the ground. ... Newton was the first to realise that gravity existed in space and it was the force which holds the Moon in orbit around the Earth and planets in orbit around the Sun.

Answer:

This is not an ideal controlled investigation because the answer choice B, "The drafts at each window may be different because the windows aren’t near each other," is correct.

To conduct a controlled investigation, it is crucial to minimize variables that could affect the results.

In this case, the windows on the south and north sides of the house may have different draft levels due to their location and proximity to various environmental factors.

To achieve better control, Lyndon should ideally select windows that are in close proximity to each other, preferably on the same side of the house, to minimize the potential differences in drafts.

This would allow for a more accurate comparison between the two windows.

Additionally, answer choices A, C, D, and E are not directly related to the issue of controlling the investigation.

The size of the windows, the choice of temperature as an indicator, and the type of plastic used are valid factors to consider but do not pertain specifically to the control of the investigation.

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The catch in this one is:  We don't know how much force the student used to push the bike.  

It wasn't necessarily the 100N.  That's just the weight of the bike. But you know that you can push a car, a wagon, or a bicycle hard, you can push it not so hard, you can give it a little push, you can give it a big push, you can push it strong, you can push it weak, you can push it medium.  The harder you push, the more it'll accelerate, but it's completely up to you how hard you want to push.  That's what's so great about wheels !  That's why they were such a great invention ! This is where I made my biggest mistake. This guy came into my store one day and said he's got this great invention, it's definitely going to take off, it'll be a winner for sure, he called it a "wheel".  I looked at it, I turned it over and I looked on all sides. I thought it was too simple.  I didn't know then it was elegant. I threw him out.  I was so dumb.  I could have invested money in that guy, today I would have probably more than a hundred dollars.

Anyway, can we figure out how much force the student used to push with ?  Stay tuned:

-- The bike covered 15 meters in 5 seconds.  Its average speed during the whole push was (15m/5s) = 3 meters/sec.

-- If the bike started out with no speed, and its average speed was 3 m/s, then it must have been moving at 6 m/s at the end of the push.

-- If its speed increased from zero to 6 m/s in 5 seconds, then its acceleration was (6m/s / 5 sec) = 1.2 m/s²

-- The bike's weight is 100N.  

(mass) x (gravity) = 100N

Bikemass = (100N) / (9.8 m/s²)

Bikemass = 10.2 kilograms

-- F = m A

Force = (mass) x (acceleration)

Force = (10.2 kg) x (1.2 m/s²)

Force = 12.24 N

-- Work = (force) x (distance)

Work = (12.24 N) x (15 m)

Work = 183.67 Joules

-- Power = (work done) / (time to do the work)

Power = (183.67 joules) / (5 seconds)

Power = 36.73 watts

Answer:

F_{resultant} = 1 [m/s]

Explanation:

These types of problems can be solved by means of relative velocities, where vectors (forces or velocities) are necessarily handled, these velocities depending on the velocity are added or subtracted.

Forces to the right are taken as positive and negative to the left.

\(F_{velocity}-F_{air}=F_{resultant}\\2-1 = F_{resultant}\\\\F_{resultant} = 1 [m/s]\)