A cylinder with initial volume V contains a sample of gas at pressure p. On one end of the cylinder, a piston is let free to move so that the gas slowly expands in such a way that its pressure is directly proportional to its volume. After the gas reaches the volume 3V and pressure 3p, the piston is pushed in so that the gas is compressed isobarically to its original volume V. The gas is then cooled isochorically until it returns to the original volume and pressure. Find the work W done on the gas during the entire process. Find the amount of work W done on the gas during the entire process. Express your answer in terms of some or all of the variables p and V.

In 2.90 seconds, a train can accelerate from 10.0 m/s to 15.0 m/s, and this train accelerates at a rate of 1.94 ms2/m/s2.

At the start of the trip, the train picks up speed as it accelerates. In the middle of the journey, it remains the same (where there is no acceleration). When the train comes to a stop, it gets smaller as it slows down.

Distance traveled by the train as a percentage of time is its speed. The time it takes for two trains to cross each other is determined by the length of the trains, say lengths a and b, and the speeds of the trains, x and y, respectively.

In 2.90 seconds, a train can accelerate from 10.0 m/s to 15.0 m/s, and this train accelerates at a rate of 1.94 ms2/m/s2.          

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1 km=1000 m

Therefore,

The melting point of the substance in the heating curve is 20°C.

1) The melting point of the substance is the temperature at the point where, the substance is having a phase change from solid state to liquid. So, the temperature at point B, 20°C is its melting point.

2) The boiling point of the substance is the temperature at the point where, the substance is having a phase change from liquid state to gaseous state. So, the temperature at point D, 90°C is its boiling point.

3) The point A is where the substance is at solid state.

4) The substance is at liquid state at the point C.

5) The substance is at gaseous state at the point E.

6) At point B, solid state is changing to liquid state.

7) At point D, liquid state is changing to gaseous state.

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Explanation:

Newton's third law of motion states that every action has an equal and opposite reaction. This means that forces always act in pairs. Action and reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel out.

The time taken by the cars to meet is 5.4 x 10³ s.

Speed of car A, v₁ = 45 km/h = 12.5 m/s

Speed of car B, v₂ = 55 km/h = 15.27 m/s

Distance between the cars, d = 150 km = 15 x 10⁴m

The expression for the time taken by the cars to meet can be given as,

Time = Distance/Average speed

t = d/(v₁ + v₂)

Applying the values of d, v₁ and v₂.

t = 15 x 10⁴/(12.5 + 15.27)

t = 15 x 10⁴/27.77

t = 5.4 x 10³ s

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Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0  

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If,  V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t  

a(0) = 1.5m/s^²  

1.5m/s^²  =  A + 2B ×  0  

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²  

1.5× 1 +B× 1 = 2m/s

B = 2-1.5  

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² ×  1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s  

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

Answer:

（3x - 4）（x - 1）

Explanation:

Your question is wrong

Answer:

is this it?

Explanation:

λ = h/mv, where λ is wavelength, h is Planck's constant, m is the mass of a particle, moving at a velocity v. de Broglie suggested that particles can exhibit properties of waves.

The total mechanical energy at point A is 617,400 J, and the velocity of the coaster at point A is 25 m/s, the velocity of the coaster at point B is 34.4 m/s. and the highest hill the coaster could have gotten over before point A with no additional mechanical energy is 63 m.

To solve this problem, we can use the conservation of mechanical energy, which states that the sum of kinetic energy and potential energy is constant in a closed system where there is no work done by non-conservative forces like friction. We can use this principle to find the answers to the questions.

a) The Total Mechanical Energy of the rollercoaster at Point A

The total mechanical energy of the rollercoaster at point A is the sum of its potential energy and kinetic energy. At point A, the rollercoaster is at its highest point, so its kinetic energy is zero. Therefore, the total mechanical energy at point A is equal to the potential energy, which is given by:

mgh = 1000 kg × 9.8 m/s^2 × 63 m = 617,400 J

b) The velocity of the coaster at point A

To find the velocity of the coaster at point A, we can use the conservation of mechanical energy principle again. At point A, all of the potential energy is converted into kinetic energy. Therefore, we can write:

1/2 mv^2 = mgh

where m is the mass of the rollercoaster, v is the velocity at point A, g is the acceleration due to gravity, and h is the height of point A relative to point E. Solving for v, we get:

v = sqrt(2gh) = sqrt(2 × 9.8 m/s^2 × 63 m) = 25 m/s

Therefore, the velocity of the coaster at point A is 25 m/s.

c) The velocity of the coaster at point B

To find the velocity of the coaster at point B, we can use the conservation of mechanical energy principle again. At point B, the rollercoaster is at a height of 42 m above the ground. Therefore, the potential energy at point B is:

mgh = 1000 kg × 9.8 m/s^2 × 42 m = 411,600 J

At point B, some of the potential energy is converted into kinetic energy, so we can write:

1/2 mv^2 + mgh = mgh_D

where v is the velocity at point B, h is the height of point B relative to point E, and h_D is the height of point D relative to point E. Solving for v, we get:

v = sqrt(2(mgh_D - mgh)/m) = sqrt(2(8780 J)/1000 kg + 2gh) = 34.4 m/s

Therefore, the velocity of the coaster at point B is 34.4 m/s.

d) The highest hill the coaster could have gotten over before point A with no additional mechanical energy

If the rollercoaster had no additional mechanical energy, its total mechanical energy at point A would be equal to its potential energy, which we calculated in part (a) to be 617,400 J. Therefore, the maximum height that the rollercoaster could reach without any additional mechanical energy is given by:

mgh_max = 617,400 J

Solving for h_max, we get:

h_max = 617,400 J / (1000 kg × 9.8 m/s^2) = 63 m

So, the highest hill the coaster could have gotten over before point A with no additional mechanical energy is 63 m.

Therefore, The coaster has a total mechanical energy of 617,400 J at point A, a velocity of 25 m/s, a velocity of 34.4 m/s, and a maximum hill height of 63 m that it could have traversed without using any further mechanical energy.

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The statements that apply to inertia are that it is a force that maintains moving objects in motion, a force that keeps stationary objects at rest, a force that causes stationary objects to naturally resist motion, and a force that causes moving objects to naturally resist changes in velocity.

According to Newton's first law, until pushed to alter its condition by the intervention of an external force, every object will continue to be at rest or in uniform motion along a single direction.

As given all the following the statements that are true about inertia are,

1. Inertia is the force that keeps a moving object in motion.

2. Inertia is the force that keeps a stationary object at rest.

3. Inertia is the natural tendency of a stationary object to resist motion.

4. Inertia is the natural tendency of a moving object to resist a change in its velocity.

Thus, all the given options are correct about inertia.

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Explanation:

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The total electric potential energy is \(\frac{kq_{1} q_{3} }{r_{13} } + \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1} q_{2} }{r_{12} }\).

Electric Potential Energy of a System of Charges :

The system's electric potential energy is equal to the amount of work necessary to create a system of charges by guiding them toward their designated locations from infinity against the electrostatic force without accelerating them. The symbol for it is U.U=W=qV. Electrostatic fields are conservative, therefore the work is independent of the path.

Assume three charges q₁ , q₂ and q₃ bring from infinity to point P.

To bring  q₁ no work is done,

\(V_{p} = \frac{kq_{1} }{r_{1} }\)

where, V = electric potential energy.

            q = point charge.

            r = distance between any point around the charge to the point charge.

           k = Coulomb constant; k = 9.0 × 109 N.

Now bring q₂,

\(V_{2} = \frac{kq_{2} }{r_{2} }\)

Work done by q₁ ;

\(W_{1} = V_{p} q_{2} = \frac{kq_{1}q_{2} }{r_{12} }\)

Now bring  q₃,

\(V_{3} = \frac{kq_{3} }{r_{3} }\)

Work done on q₃ by q₁ and q₂

\(W= q_{3} [ V_{1} + V_{2} ]\)

    \(=\frac{kq_{1} q_{3} }{r_{13} }\)\(+ \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1}q_{3} }{r_{12} }\)

This work done is stored in the form of potential energy.

∴U=W= potential energy of three systems.

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The total electric potential energy is \(\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}\)

Electric Potential Energy of a System of Charges

The labor required to establish a system of charges by guiding them toward their intended positions from infinity against the electrostatic force without accelerating them is equal to the electric potential energy of the system. Its identifier is U.U=W=qV. Due to the conservatism of electrostatic fields, the work is independent of the path.

Consider having three charges. Q1, Q2, and Q3 bring point P from infinity.

No work has been done to bring q1,

\(V_{1} = \frac{kq1}{r1}\)

where, V = electric potential energy.

           q = point charge.

           r = distance between any point around the charge to the point charge.

          k = Coulomb constant; k = 9.0 × 109 N.

Now bring q₂,

\(V_{2} = \frac{kq2}{r2}\)

Work done by q₁ ;

W1 = \(V_{p} q2\) = \(\frac{kq1q2}{r12}\)

Now bring  q₃,

\(V_{3} = \frac{kq3}{r3}\)

Work done on q₃ by q₁ and q₂

W= q3{\(V_{1} + V_{2}\)}

W = \(\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}\)

This work done is stored in the form of potential energy.

∴U=W= potential energy of three systems.

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Answer: the answer is corect of what everone else said

Explanation:

The temperature gradient in the flow of direction is 294525 W.

A temperature gradient is the gradual variance in temperature with distance. The slope of the gradient is consistent within a material. A gradient is established anytime two materials at different temperatures are in physical contact with each other.

Q= T/( L/ KA)

Q= ( 1500 − 450) / 0.15 / 9.35v * 4.35)

   = 294525 W

Units of measure of temperature gradients are degrees per unit distance, such as °F per inch or °C per meter.

Many temperature gradients exist naturally, while others are created. The largest temperature gradient on Earth is the Earth itself. Q= T/Ka.

Therefore, The temperature gradient in the flow of direction is 294525 W.

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The wavelength will remain unchanged.

Explanation:

The velocity \(v\) of a wave in terms of its wavelength \(\lambda\) and frequency \(\nu\) is

\(v = \lambda\nu\) (1)

so if we double both the velocity and the frequency, the equation above becomes

\(2v = \lambda(2\nu)\) (2)

Solving for the wavelength from Eqn(2), we get

\(\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}\)

We would have gotten the same result had we used Eqn(1) instead.

Answer:

the wavelength increases

Explanation:

Sound waves are longitudinal waves. Hence, option (C) is correct.

Longitudinal waves are waves in which the displacement of the medium is in the same (or the opposite) direction as the wave propagation and the medium vibration is parallel ("along") to the direction the wave travels. Because they create compression and rarefaction when passing through a material, mechanical longitudinal waves are also known as compressional or compression waves.

They are also known as pressure waves because they cause pressure changes. Examples from the real world include seismic P-waves (created by earthquakes and explosions)and sound waves (which are the oscillations in pressure, a particle's displacement, and a particle's velocity).

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Approximately 0.123 kg of liquid water at 26 degrees Celsius would be needed to melt 41 grams of ice.

To calculate the minimum amount of liquid water required to melt 41 grams of ice at 0°C, we need to consider the energy required for the phase change from solid to liquid, which is known as the specific heat of fusion of ice.

The energy required to melt 1 kg of ice is 3.33×105 J/kg.

Therefore, the energy required to melt 41 grams of ice is (3.33×105 J/kg) × (41/1000) kg = 13653 J.

To calculate the amount of liquid water required, we use the specific heat capacity of water, which is 4180 J/kg/°C.

Assuming the initial temperature of water is 26°C, the amount of water needed can be calculated as (13653 J) ÷ (4180 J/kg/°C) ÷ (26°C) = 0.123 kg or approximately 123 ml of water.

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Answer:

Explanation:

a )

At a point midway between the wires , magnetic field due to two wires will be equal and opposite so they will cancel each other . Therefore , net magnetic field will be zero .

b )

At a point 25cm to the right of point midway between the wires

distance of this point will be .25 m from one wire and .75 m from other wire .

Magnetic field due to a current carrying wire can be given by the following expression .

B = (μ₀ / 4π ) x 2I / R where I is current and R is distance of the point

= 10⁻⁷ x 2I / R

For the wire at a distance of .25 m

magnetic field B₁ = 10⁻⁷ x 2 x 20 / .25

= 160 x 10⁻⁷ T .

For the wire at a distance of .75 m

magnetic field B₂ = 10⁻⁷ x 2 x 20 / .75

= 53.33 x 10⁻⁷ T .

These fields are opposite to each other , so

net magnetic field = 160 x 10⁻⁷ T -  53.33 x 10⁻⁷ T

= 106.67 x 10⁻⁷ T .

If the wires are horizontal and current is flowing away from the observer , net magnetic field will be in vertically upward direction .

c )

At a point 75cm to the right of point midway between the wires

The point will be .25 m from one wire and 1 .25 m from other wire . And the magnetic field will be aligned in the same direction .

For the wire at a distance of .25 m

magnetic field B₁ = 10⁻⁷ x 2 x 20 / .25

= 160 x 10⁻⁷ T

For the wire at a distance of 1.25 m

magnetic field B₂ = 10⁻⁷ x 2 x 20 / 1.25

= 32  x 10⁻⁷ T

As the fields are aligned in the same direction

Net magnetic field =  160 x 10⁻⁷ T + = 32  x 10⁻⁷ T

= 192 x 10⁻⁷ T .

If the wires are horizontal and current is flowing away from the observer , net magnetic field will be in vertically downward direction.

Knowing the exits are exactly 7 km apart, the cars pass each other at 9:29 and 15 seconds a.m.

The relative velocity of the cars is 30 m/s - 26 m/s = 4 m/s.

The distance between the cars is 7 km = 7000 m.

The time it takes for the cars to pass each other is 7000 m / 4 m/s = 1750 seconds.

1750 seconds is 29 minutes and 15 seconds.

To calculate the time in minutes;

Let:

v_p = the speed of car P (m/s)

v_q = the speed of car Q (m/s)

d = the distance between the cars (m)

t = the time it takes for the cars to pass each other (s)

Given that:

v_p = 30 m/s

v_q = 26 m/s

d = 7000 m

Use the equation for relative velocity to find the velocity of the cars relative to each other:

v_r = v_p - v_q

v_r = 30 m/s - 26 m/s = 4 m/s

Use the equation for distance to find the time it takes for the cars to pass each other:

d = v_r × t

7000 m = 4 m/s × t

t = 7000 m / 4 m/s = 1750 s

Convert 1750 seconds to minutes and seconds:

1750 s = 29 minutes and 15 seconds

Therefore, the cars pass each other at 9:29 and 15 seconds a.m.

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Answer:

the car movves briefly as you ran, however, it stops again after you ran in to the wall

Explanation:

Our net total linear momentum was zero at the time both the train and the boy was at rest. As there is no pressure, the train and boy system's linear momentum can be conserved provided no external forces are working on it that might shift its momentum.

If the boy runs inside of the train with a velocity V1 in the forward direction, the train would have a velocity V2 in the reverse direction to V1 to conserve the system's momentum, resulting in Final momentum.

i.e

\(m \times V_1 + M \times V_2 = 0\)  --- (1)

here;

m = mass of the boy

M = mass of the train

Thus;

\(m \times V_1 =- M \times V_2\) --- (2)

As the boy crashes into the train's wall, a pair of equal and opposing force F intervene on both the train and the boy. This force F causes the boy traveling with momentum m× V1 to come to a halt; its momentum remains zero. As the train moves with about the same momentum as the boy as seen in (2) and is subjected to the same force F, its momentum will be diminished to zero, and it will come to a halt.

Answer: 100km/h

Explanation: This is because the road isn't moving. If you drew the roads velocity and the bus velocity you would find a 100km/h difference. Thus an object inside the bus would experience the same differential.

(a) When you apply a force of 40 N, the magnitude of the horizontal component of the force that the floor exerts on the block is 51 N.

(b) When you apply a force of 59 N, the magnitude of the horizontal component of the force that the floor exerts on the block is 51 N.

The horizontal component of the force that the floor exerts on the block is determined by applying Newton's second law of motion;

\(F = ma\)

\(\Sigma F = 0\\\\F - F_f = 0\\\\F = F_f\)

\(F_f = F\\\\F_f = 51 \ N\)

where;

Since the surface (floor) is constant, the frictional force (horizontal component of the force that the floor exerts on the block ) will be the same on the cases.

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Based on the definition of limiting frictional force;

Force is any push or pull effect which tends to cause a change in the motion or shape of an object.

Based on Newton's third law, every force produces an equal and opposite reaction.

Therefore, the force that the floor exerts on the block is oppositely direction to the force exerted by the horizontal force acting on the block.

Since a horizontal force of 51 N applied to the block is required to keep the block moving across the floor at a constant speed, this is the limiting frictional force.

a) when a force of 40 N is applied to the block, the magnitude of the horizontal component of the force that the floor exerts on the block is equal to 40 N

b) when a force of 59 N is applied to the block, the magnitude of the horizontal component of the force that the floor exerts on the block is equal to 51 N, the limiting frictional force.

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Answer:

I believe the answer is sea floor spreading

Answer:

A. 13 cm

Explanation:

here the total diameter is given 4 cm,

here the radius then will be 2 cm.

use the formula of circumference of circle = 2πr ........where r is the radius

using the formula: 2 * π * 2 = 4π = 12.57 = 13 cm.

(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:

the stone hits the gound with a speed of  20.7 m/s

Explanation:

Becuase gravity is constant we know that the initial upward velocity will be equal to the downward velocity when the stone has returned to its intal location.

We know that:

v² = v₀² - 2ah

Where v is the final velocity whos value at the maximum height is zero.

v₀ is the initial velocity

h is the maximum height

a = g = 10 m/s² the acceleration due gravity on Earth

Solving for v0:

\(v_0=\sqrt{v^2+2ah} \\\\v_0 =\sqrt{ 0+2(10\;m/s^2)(50000\;m)}\\\\v_0 =\sqrt{ 2(10\;m/s^2)(50000\;m)} \\\\v_0 = 1000\;m/s\)

Answer:

1000 m/s ( 990.45 m/s to be exact)

Explanation:

If we use one of the SUVAT equations, i.e.,

2as = (v^2) - (u^2) where;

s = distance, a = acceleration, v = final velocity, u = initial velocity

At max point the rocks are considered stationery, which makes v = 0, the height is 50,000m which makes s = 50,000, the acceleration is -9.81, as the object decelrates to 0

2(-9.81)(50000) = (0)^2 - (u)^2

-981000 = - u^2

u = sqr root( 981000)

u = 990.45 m/s = 1000 m/s

To calculate the volume of a parallelepiped given the sides, we can use the scalar triple product. The formula for the volume of a parallelepiped with sides a, b, and c is:

Volume = |a · (b × c)|

where · represents the dot product and × represents the cross product.

Using the given sides:

a = (7, 2, 4)

b = (4, 7, 6)

c = (3, 4, 7)

First, calculate the cross product of b and c:

b × c = (7*7 - 4*4, 6*3 - 7*7, 4*4 - 2*3)

      = (49 - 16, 18 - 49, 16 - 6)

      = (33, -31, 10)

Next, calculate the dot product of a and the cross product (b × c):

a · (b × c) = 7*33 + 2*(-31) + 4*10

           = 231 - 62 + 40

           = 209

Finally, take the absolute value of the result to obtain the volume:

Volume = |209| = 209 cubic units

Therefore, the correct answer is:

209 cubic units

Answer: current I = 1.875A

Explanation:

If the resistors are connected in series,

Then the equivalent resistance will be

R = 6 + 18 + 15 + 9

R = 48 ohms

Using ohms law

V = IR

Make current I the subject of formula

I = V/R

I = 90/48

I = 1.875A

And if the resistors are connected in parallel, the equivalent resistance will be

1/R = 1/6 + 1/18 + 1/15 + 1/9

1/R = 0.166 + 0.055 + 0.066 + 0.111

R = 1/0.3999

R = 2.5 ohms

Using ohms law

V = IR

I = 90/2.5

Current I = 35.99A