A rock against the frictionless inside wall of a cone with a radius of 2 meters. It moves with the cone with a constant speed of 6m/s. The rock does not slide up or down in the cone as it rotates what is the centripetal acceleration of the rock

The 100-meter dash world record is roughly 10 m/s, The wind is blowing from of the northwest at 30 km/h today.

Velocity and speed describe how quickly or slowly an object moves. We frequently encounter circumstances when we must determine which of two or even more moving objects is going faster.

First, let's review the distinctions between velocity and speed:

- The ratio of an object's distance traveled to its period of travel indicates its speed. Being a scalar only has a magnitude.

The velocity of the object is a vector with a direction determined by the direction of movement and a magnitude equal to the product of the object's displacement and the time it takes to travel in that direction. The quantity and direction of velocity are thus both present.

Let's now examine each choice:

Given that the record holder for the 100-meter dash is approximately 10 m/s, this speed is just of magnitude (10 m/s).

The wind speed currently is 30 km/h from the northwest; as this has both a speed and a direction, it is a velocity.

When a 200,000 kg train derailed, it was moving north at 70 km/h, therefore this has magnitude (70 km/h) plus direction (north), making it a velocity.

A car was given a penalty for exceeding the speed limit on the interstate by 140 km/h, hence this is a speed with only that magnitude (140 km/h).

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Accuracy refers to how close a measurement is to the true or accepted value. Precision refers to how close measurements of the same item are to each other.

Precision and accuracy are two indices of observational error. Precision is how near a collection of measurements is to its true value, whereas accuracy is how close the measurements are to one other. Precision, in other terms, is a description of random errors and a measure of statistical variability.

If you weigh a body weighing 20 kg and obtain 17.4,17,17.3, and 17.1, your weighing scale is exact but not particularly accurate. If your scale returns values of 19.8, 20.5, 21.0, and 19.6, it is more accurate but not particularly precise.

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Answer:

42.6083 mi/h

Explanation:

Given: A car travels 8km in 7 minutes.

To find: Find the speed of the car.

Formula: \(Speed = \frac{Distance}{Time}\)

Solution: Since the formula for the speed of an object (which is the car) is speed = distance ÷ time, divide the distance (8km) by the time (7min)

Speed = 42.6083 miles per hour

Answer:

The graph of the velocity of an object in free fall would look like a straight line sloping downward. As the object falls, its velocity increases at a constant rate, so the graph of its velocity versus time will be a straight line with a negative slope. This is because acceleration due to gravity is a constant -9.8 meters per second squared, so the velocity of a free-falling object will increase by 9.8 meters per second every second.

Therefore, the graph that shows the change in velocity of an object in free fall is a straight line with a negative slope. Here is an example of such a graph:

Free Fall Velocity Graph

The tangential acceleration of the short player's ball is twice the tangential acceleration of the tall player's ball.

The ball of the shorter player experiences twofold of the ball of the taller player's tangential acceleration. The reason behind this is that the magnitude of the tangential acceleration correlates directly with the radius of the circle.

The ball belonging to the smaller player is positioned nearer to the elbow, resulting in a decreased radius. It can be deduced from this statement that the ball of the short player experiences twice the magnitude of tangential acceleration compared to the ball of the tall player.

The answer is: ashort = 2atall

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The Complete Question

A short tennis player hits a ball that is r meters from their elbow with an angular acceleration a. A tall tennis player hits a ball with the same angular acceleration where the ball is 2r from their elbow. How does the tangential acceleration of the short player's ball Ashort compare with the tall player's ball a tall? Choose 1 answer: ashort 2atall ashort Otall ashort 1 atall 2

Answer:

Magnitude of electric field is 1.06 x \(10^5\) V/m along negative X-direction

Explanation:

Given: initial velocity of proton = u = 3.5 x \(10^6\) m/s

final velocity of proton = v = 0 m/s

initial point \(l_i\) = 0.2 m and final point is \(l_f\) = 0.8 m

According to conservation of energy:

change in in kinetic energy = change in potential energy of proton

⇒\(\frac{m}{2}(v^2-u^2 ) = qE(l_i - l_f)\)

where q and m is the charge and mass of proton E is the electric field , \(l_i\) and \(l_f\) is the initial and final position of proton

on substituting the respected values we get,

1.023 x \(10^-^1^4\) = 9.6 x \(10^-^2^0\) x E

⇒ E = 1.06 x \(10^5\) V/m

external force is opposite to the motion as velocity of proton decreases with distance.

Therefore, magnitude of electric field is 1.06 x \(10^5\) V/m along negative X-direction

This question involves the concepts of general gas equation and pressure.

The force exerted by the gas on one of the walls of the container is "74.08 KN".

First, we will use the general gas equation to find out the pressure of the gas:

\(PV = nRT\)

where,

P = Pressure of the gas = ?

V = Volume of cube = (side length)³ = (10 cm)³ = (0.1 m)³ = 0.001 m³

n = no. of moles = 3 (since molecules equal to avogadro's number make up 1 mole)

R = general gas constant = 8.314 J/mol.K

T = Absolute Temperature = 24°C + 273 = 297 K

Therefore,

\(P = \frac{(3)(8.314\ J/mol.k)(297\ K)}{0.001\ m^3}\)

P = 7407.78 KPa

Now, the force on one wall can be given as follows:

\(P =\frac{F}{A}\\\\F=PA\)

where,

A = area of one wall = (side length)² = (0.1 m)² = 0.01 m²

Therefore,

\(F=(7407.78\ KPa)(0.01\ m^2)\\\)

F = 74.08 KN

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Answer:

Explanation:

The Workdone is the product of force and distance, Hence, the Workdone on the box from distance x = 0 meters and x = 16 meters is 0 Joules.

Using the graph given :

The work done from x = 0 to x = 16 metres ;

The distance can be split evenly into :

(x = 0 to x = 8) and (x = 8 to x = 16)

Workdone = Force × distance

Workdone from ; x = 0 to x = 8 ;

Force at a distance of 8 meters = - 40N

Workdone = - 40N × 8 m = -320 Nm

Workdone from ; x = 8 to x = 16 ;

Force at a distance of 16 meters = 40 N

Workdone = 40 N × 8 m = 320 Nm

The total workdone :

(-320 + 320) Nm = 0 J

.

Therefore, the Workdone ls 0 Joules.

Answer:

The higher the energy, the shorter the wavelength.

Explanation:

It takes approximately 28.3 days for the radioactive sample to decay from an initial activity of 1.07 x 105 Bq to an activity of 100 Bq.

The formula for radioactive decay is \(A = A0 * (1/2)^{(t/t1/2)}\) where: A is the final activity A0 is the initial activity is the time elapsed t1/2 is the half-life of the sample

In this problem, the initial activity A0 is given as 1.07 x 105 Bq and the final activity A is given as 100 Bq.

The half-life of the sample is given as 5.7 days.

Substituting these values into the formula, we get: \(100 = 1.07 \times 105 * (1/2)^{(t/5.7)}\)

Dividing both sides by 1.07 x 105, we get:

\(0.000934579 = (1/2)^{(t/5.7)}\)

Taking the natural logarithm of both sides, we get:\(-7.072737 = (t/5.7) * ln(1/2)\)

Dividing both sides by ln(1/2), we get:

\(t = -7.072737 \div ln(1/2) * 5.7t \approx 28.3\) days

Therefore, it takes approximately 28.3 days for the radioactive sample to decay from an initial activity of 1.07 x 105 Bq to an activity of 100 Bq.

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From the histogram given, there were at approximately 7 downloads between 3pm and 4pm . This can be derived by counting the rows in that time period.

A histogram is a graph that depicts the frequency distribution of a few data points from a single variable.

Histograms frequently divide data into "bins" or "range groups" and count the number of data points that belong to each of those bins.

Histograms are frequently used to depict the key properties of data distribution in a handy format. It is especially beneficial when working with huge data sets (more than 100 observations). It can aid in the detection of uncommon observations (outliers) or gaps in the data.2

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Answer:

Step 1: List the mass and velocity of the object. Step 2: Convert any values into SI units (kg, m, s). Step 3: Multiply the mass and velocity of the object together to get the momentum of the object.

Answer:

 \(k_{eq} = \frac{k}{N}\)

Explanation:

For this exercise let's use hooke's law

         F = - k x

where x is the displacement from the equilibrium position.

        x = \(- \frac{F}{k}\)

if we have several springs in series, the total displacement is the sum of the displacement for each spring, F the external force applied to the springs

       x_ {total} = ∑ x_i

we substitute

       x_ {total} =  ∑ -F / ki

       F / k_ {eq} =  -F  \(\sum \frac{1}{k_i}\)

      \(\frac{1}{k_{eq}} = \frac{1}{k_i}\) 1 / k_ {eq} =  ∑ 1 / k_i

if all the springs are the same

     k_i = k

     \(\frac{1}{k_{eq}} = \frac{1}{k} \sum 1 \\\)

     \(\frac{1}{k_{eq} } = \frac{N}{k}\)

     \(k_{eq} = \frac{k}{N}\)

Answer:

8.6602 tons

Explanation:

We first draw the known vector forces.

2fcos30⁰

We have f to be equal to 5tons

Inserting into formula

Σfx = 2(5)cos30⁰

= 8.6602 tons

Σfy is equal to 0, this is because in the y direction, the forces cancel themselves out.

Therefore the resultant force on the ship is equal to 8.6602 tons

I hope this helps!

Please check attachment for diagram.

Answer:

The answer should be C

Explanation:

Answer:

C.49 is yr ans...

The potential and kinetic energy of the system at the following time instances is zero and maximum.

From the given,

The displacement of the system is, x = A cos(3πt/T)

1) At t =0, the displacement of the system is given by, x = cos(3π×0/T)= cos(0) = 1. The displacement is maximum at t=0. Hence, the potential energy is maximum and kinetic energy is zero.

2) At t=T, the displacement, x = cos(3πT/T)= cos(3π) = -1. The displacement is minimum and hence, the potential energy is minimum and kinetic energy is maximum.

3) At t = T/6, the displacement x = cos(3πT/6T)=cos(π/2)=0, the displacement is zero, and hence, both the potential and kinetic energy is zero.

4) At t=T/3, the displacement, x= cos(3πt/T)=cos(3πT/3)= -1. The displacement is minimum and hence, the potential energy is minimum and kinetic energy is maximum.

5) At t=T/2, the displacemetn x = cos(3πt/T) = cos(3πT/2T) = cos (3π/2)=0. Hence, both the potential and kinetic energy is zero.

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(a) The work done by the 150 N force is 877.5 Joules.

(b) The coefficient of kinetic friction between the crate and the surface is approximately 0.437.

To answer this problem, we must take into account the work done by the applied force as well as the work done by friction.

(a) The applied force's work may be estimated using the following formula:

Work = Force * Distance * cos(theta)

where the force is 150 N and the distance is 5.85 m. Since the force is applied horizontally and the displacement is also horizontal, the angle theta between them is 0 degrees, and the cosine of 0 degrees is 1.

As a result, the applied force's work is:

Work = 150 N * 5.85 m * cos(0) = 877.5 J

So, the work done by the 150 N force is 877.5 Joules.

(b) Frictional work is equal to the force of friction multiplied by the distance. The work done by friction is identical in amount but opposite in direction to the work done by the applied force since the crate travels at a constant speed.

The frictional work may be estimated using the following formula:

Work = Force of Friction * Distance * cos(theta)

The net force applied on the crate is zero since it is travelling at a constant pace. As a result, the friction force must be equal to the applied force, which is 150 N.

Thus, the work done by friction is:

Work = 150 N * 5.85 m * cos(180) = -877.5 J

Since the work done by friction is negative, it indicates that the direction of the frictional force is opposite to the direction of motion.

The coefficient of kinetic friction may be calculated using the following equation:

Friction Force = Kinetic Friction Coefficient * Normal Force

The normal force equals the crate's weight, which may be computed as:

Normal Force = mass * gravity

where the mass is 35.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2.

Normal Force = 35.0 kg * 9.8 m/s^2 = 343 N

Now, we can rearrange the equation for the force of friction to solve for the coefficient of kinetic friction:

Force of Friction = coefficient of kinetic friction * Normal Force

150 N = coefficient of kinetic friction * 343 N

coefficient of kinetic friction = 150 N / 343 N ≈ 0.437

As a result, the kinetic friction coefficient between the container and the surface is roughly 0.437.

In summary, the work done by the 150 N force is 877.5 Joules, and the coefficient of kinetic friction between the crate and the surface is approximately 0.437.

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

T=3.12 Sec

Distance d= 9.703m

The distance can refer to a physical length in physics or to an estimate based on other factors in ordinary language (e.g. "two counties over"). It is sometimes used to indicate the distance between two points: display style |AB||AB|. The terms "distance from A to B" and "distance from B to A" are frequently used interchangeably.

It is a generalization of the idea of physical distance. Distance is a non-numerical measurement in psychology and the social sciences; psychological distance refers to how an object feels.

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Answer:The concept of momemtum will be used to solve this question.A moving body's momemtum, which is generally equal to the product of the body's mass and velocity, is a quality that the body has as a result of its mass and motion.

Explanation:

The vector components of the weight force parallel to the plane is 463.4 N, and the component perpendicular to the plane is 166.8 N upward.

A quantity which has both magnitude and direction is termed as a vector quantity. It can be divided into two components in 2-dimensional plane: The vertical component and the horizontal component.

The weight force, which is equal to the force of gravity acting on the yoga instructor, has a magnitude of 490 N, which we can decompose into its vertical and horizontal components using trigonometry:

Vertical component = weight force x sin(20°)

= 490 N x sin(20°)

= 166.8 N (upward)

Horizontal component = weight force x cos(20°)

= 490 N x cos(20°)

= 463.4 N (parallel to the plane)

So the component of the weight force parallel to the plane is 463.4 N, and the component perpendicular to the plane is 166.8 N upward.

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The force that acts on a projectile in the horizontal direction is Gravitational force.

A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.

Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. Hence, The only force acting upon a projectile is gravity.

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The heating element of an electric oven is designed to produce 3.9 kW of heat when connected to a 240-V source. Then the resistance of the element is 14.7Ω

When A 120-V hair dryer has two settings: 950 W and 1400 W then the resistance at 950 W is 15.15Ω and at 1400 W is 10.28Ω.

The pace at which electrical energy is transmitted by an electric circuit is referred to as its power. The watt is the SI unit of power, equal to one joule per second. Watts, like other SI units, have standard prefixes: thousands, millions, and billions of watts are referred to as kilowatts, megawatts, and gigawatts, respectively.

It is a popular misperception that electric power is purchased and sold, whereas electrical energy is. Electricity, for example, is sold to users in kilowatt-hours (kilowatts multiplied by hours), because energy is defined as power multiplied by time.

In this problem,

Given,

Power P = 3.9 KW

Voltage V = 240 V

P = V²/R

3.9 × 10³ = 240²/R

R = 240²/3.9 × 10³

R = 14.7Ω

For hair dryer,

given,

P = 950 W or 1400 W

V = 120 V

P = V²/R

950 = 120²/R

R = 120²/950

R = 15.15Ω

P = V²/R

1400= 120²/R

R = 120²/1400

R = 10.28Ω

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To determine which gas has the higher initial temperature, we can use the formula:

thermal energy = (3/2) nRT

where n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

For gas A, we have:

5000 J = (3/2) (2.5 mol) R T_A

For gas B, we have:

8400 J = (3/2) (2.5 mol) R T_B

Dividing the equations, we get:

5000 J / 8400 J = T_A / T_B

Solving for T_A, we get:

T_A = (5000 J / 8400 J) T_B

T_A / T_B = 0.5952

So gas B has the higher initial temperature.

To find the final thermal energy of gas A, we need to use the conservation of energy:

thermal energy_A + thermal energy_B = total thermal energy

We know that the initial total thermal energy is 5000 J + 8400 J = 13400 J. We also know that the number of moles and the specific heat capacity of each gas do not change. Therefore, we can write:

(3/2) (2.5 mol) R T_A,f + (3/2) (2.5 mol) R T_B,f = 13400 J

Simplifying and substituting R = 8.31 J/mol K, we get:

T_A,f + T_B,f = 4528.6 K

We also know that the two gases are in thermal equilibrium, so they must have the same final temperature:

T_A,f = T_B,f

Substituting into the equation above, we get:

2T_A,f = 4528.6 K

T_A,f = 2264.3 K

Using the formula for thermal energy, we can calculate the final thermal energy of gas A:

thermal energy_A,f = (3/2) (2.5 mol) (8.31 J/mol K) (2264.3 K) = 77646 J

Therefore, the final thermal energy of gas A is 77646 J.

To find the final thermal energy of gas B, we can use the same equation as above, but substitute T_A,f for T_B,f:

(3/2) (2.5 mol) R T_A,f + (3/2) (2.5 mol) R T_A,f = 13400 J

Simplifying and substituting R = 8.31 J/mol K, we get:

2T_A,f = 4528.6 K

T_A,f = 2264.3 K

Using the formula for thermal energy, we can calculate the final thermal energy of gas B:

thermal energy_B,f = (3/2) (2.5 mol) (8.31 J/mol K) (2264.3 K) = 77646 J

Therefore, the final thermal energy of gas B is also 77646 J.

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The general driving forces behind most chemical interactions is the need for atoms to be electrically neutral and to complete their outer electron orbital area (shell).

When both the cases are fulfilled i.e. when the enthalpy change is negative and the entropy change is positive, the reaction is said to be spontaneous and thus enthalpy and entropy are the two thermodynamic driving forces of chemical reactions.

There are several factors that affects the driving force:

1. Release of heat (exothermic reactions are driven)

2. Free energy – the change in free energy should be negative

3. Entropy – An increase in entropy drives a reaction

4. Equilibrium Shift - Removal of products drives the reaction

The general driving forces behind most chemical interactions is the need for atoms to be electrically neutral and to complete their outer electron orbital area (shell).

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We are given two points on the distance-time graph: (1, 40) and (3,130)

This means that:

At time 1 hour, the distance traveled was 40 km

At time 3 hours, the distance traveled was 130 km

We want to find the average speed during this 2 hour interval (from 1 hour to 3 hours).

Average speed is defined as:

Average Speed = Change in Distance / Change in Time

The change in distance is the distance traveled from 1 hour to 3 hours, which is 130 km - 40 km = 90 km

The change in time is 3 hours - 1 hour = 2 hours

Average Speed = 90 km / 2 hours

= 45 km/hr

Therefore, the average speed during this interval of time is 45 km/hr.

Answer:B

Explanation: I just did it on a p e x

273 K gives the object's temperature in the SI unit therefore the correct answer is option B

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

The International System of Units, sometimes known as the SI system of units, is the most frequently used and acknowledged system of units in use nowadays. There are three additional units and 7 SI basic units in this system of SI units.

The three supplemental SI units are radian, steradian, and becquerel, whereas the base SI units are meter, kilogram, second, kelvin, ampere, candela, and mole. These base units can be used to create all other SI units.

Thus,273 K gives the object's temperature in the SI unit therefore the correct answer is option B

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Answer:

30 m

Explanation: