Question 8 a-e plz

Question 8 A-e Plz

Answers

Answer 1

Answer:

(a) t = 0 s

(b) t = 0 s, 30 s, 55 s

(c) t = 40 s to t = 60 s

(d) t = 10 s to t = 15 s

(e) a = 6 m/s^2

Explanation:

(a) The car is at starting position at t = 0 s and v = 0 m/s.

(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.

(c) from t = 40 s to 60 s the car is moving in the negative direction.

(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.

(e) The slope of the velocity time graph gives acceleration.

a = (60 - 0) / (10 - 0) = 6 m/s^2


Related Questions

You walk into a room and you see 4 chickens on a bed 2 cows on the floor and 2 cats in a chair. How many legs are on the ground? (I know this answer just a riddle to see who knows it) (:

Answers

Answer:

18

Explanation:

I'm pretty sure I got it right

Distillation is the separation of multiple Choose... components based on their different Choose... . As the mixture is heated and the first component Choose... , its Choose... form travels through the distillation set-up and Choose... into a different container.

Answers

Answer:

Explanation:

Distillation is the separation of multiple LIQUID components based on their different BOILING POINT. As the mixture is heated and the first component SEPARATES, its PURE form travels through the distillation set-up and GOES into a different container

A system gains 1500J of heat and 2200J of work is done by the system on its surroundings. Determine the change in internal energy of the system

Answers

Answer:

-700

formula is heat gained - work done

The change in internal energy if A system gains 1500J of heat and 2200J of work is done by the system on its surroundings, is 700 joules.

What is Energy?

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc.

Additionally, there is heat and work, which is energy being transferred from one body to another. Energy is always assigned based on its nature once it has been transmitted. Thus, heat transmitted may manifest as thermal energy while work performed may result in mechanical energy.

Given:

A system gains 1500J of heat and 2200J of work is done by the system on its surroundings,

Calculate the change in internal energy as shown below,

The change in internal energy = heat gained - work done

The change in internal energy = 1500 - 2200

The change in internal energy = -700 J

Thus, the change in internal energy is 700 joules.

To know more about Energy:

https://brainly.com/question/8630757

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Q 26.12: Assume current flows in a cylindrical conductor in such a way that the current density increases linearly with radius, from zero at the center to 1.0 A/m2 at the surface of the conductor. If the conductor has a cross sectional area of 1.0 m2, what can you say about the current in this conductor

Answers

Answer:

The current is 0.67 A.

Explanation:

Density, J = 1 A/m^2

Area, A = 1 m^2

Let the radius is r. And outer is R.

Use the formula of current density

[tex]I = \int J dA = \int J 2\pi r dr\\\\I = \int_{0}^{R}\frac{2\pi r^2}{R} dr\\\\I = \frac{2 \pi R^2}{3}.... (1)Now A = \pi R^2\\\\1 =\pi R^2\\\\R^2 = \frac{1}{\pi}\\\\So, \\\\I = \frac{2\pi}{3}\times \frac{1}{\pi}\\\\I = 0.67 A[/tex]

Which physical phenomenon is illustrated by the fact that the prism has different refractive indices for different colors

Answers

Answer:

The incoming white light is composed of light of different colors,

Since these different colors have different refractive indices they are refracted at different angles from one another.

The output light is then separated by color creating a color spectrum.

Since n is greater for shorter wavelengths  (violet colors) these wavelengths are refracted thru the larger angles.

While an object near the earths surface is in free fall, its
A) velocity increases
B) acceleration increases

Answers

Answer:

a

Explanation:

The rate of change of an object's location with relation to a reference point is its velocity, which is dependent on time. when an object is dropped from space at rest (t = 0) under the influence of gravity, the velocity of the object changes and increases with time while the acceleration decreases.

190 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabolic power, slightly more than they would use at rest. An air conditioner with a COP of 5.0 is being used to keep the room at a constant temperature. What minimum electric power must be used to operate the air conditioner?

Answers

Answer:

W = 4.75 KW

Explanation:

First, we will calculate the heat to be removed:

Q = (No. of students)(Metabolic Power of Each Student)

Q = (190)(125 W)

Q = 23750 W = 23.75 KW

Now the formula of COP is:

[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]

W = 4.75 KW

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.

Answers

Answer:

[tex]\lambda=6.83\times 10^{-5}\ m[/tex]

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]

So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.

What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.

Answers

Answer:

sup qwertyasdfghjk

Explanation:

Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k

Answers

Answer:

parametric representation: x = u, y = v - u ,  z = - v

Explanation:

Given vectors :

i - j ,  j - k

represent the vector equation of the plane as:

r ( u, v ) = r₀ + ua + vb

where:  r₀ = position vector

            u and v = real numbers

             a and b = nonparallel vectors

expressing the nonparallel vectors as :

a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )

hence we can express vector equation of the plane as

r(u,v) = ( x₀ + u, y₀ - u + v,  z₀ - v )

Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0

( x, y , z ) = ( x₀ + u,  y₀ - u + v,   z₀ - v )

x = 0 + u ,

y = 0 - u + v

z = 0 - v

parametric representation: x = u, y = v - u ,  z = - v

Your cell phone typically consumes about 300 mW of power when you text a friend. If the phone is operated using a lithium-ion battery with a voltage of 3.5 V, what is the current (in A) flowing through the cell-phone circuitry under these circumstances

Answers

Answer:

I = 0.0857 A

Explanation:

Given that,

Power consumed by the cellphone, P = 300 mW

The voltage of the battery, V = 3.5 V

Let I is the current flowing through the cell-phone. We know that,

P = VI

Where

I is the current

So,

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]

So, the current flowing the cell-phone is 0.0857 A.

A particle charge of 2.7 µC is at the center of a Gaussian cube 55 cm on edge. What is the net electric flux through the surface?

Answers

Answer:

3.05×10⁵ Nm²C⁻¹

Explanation:

According to Gauss' law,

∅' = q/e₀............... Equation 1

Where ∅' = net flux through the surface, q = net charge, e₀ = electric permittivity of the space

From the question,

Given: q = 2.7 μC = 2.7×10⁻⁶ C,

Constant: e₀ = 8.85×10⁻¹² C²/N.m²

Substituting these values into equation 1

∅' = (2.7×10⁻⁶)/(8.85×10⁻¹²)

∅' = 3.05×10⁵ Nm²C⁻¹

А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?​

Answers

Answer:

I am not able to answer this question please don't mind...

Explanation:

please marks me as brainliests...

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vertical height does the block reach above its starting point? Use the coefficients μk=0.20 andμs=0.50.

Answers

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta[/tex]

[tex]\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta[/tex]

[tex]\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h[/tex]

[tex]\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h[/tex]

[tex]h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}[/tex] (1)

Where:

[tex]h[/tex] - Maximum height of the wood block, in meters.

[tex]v[/tex] - Initial speed of the block, in meters per second.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]m[/tex] - Mass, in kilograms.

[tex]s[/tex] - Distance travelled by the wood block along the wooden ramp, in meters.

[tex]\theta[/tex] - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that [tex]v = 10\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.20[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the height reached by the block above its starting point is:

[tex]h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]h = 4.249\,m[/tex]

The wood block reaches a height of 4.249 meters above its starting point.

The cable lifting an elevator is wrapped around a 1.1m diameter cylinder that is turned by the elevator's motor. The elevator is moving upward at a speed of 2.6 ms. It then slows to a stop, while the cylinder turns one complete revolution. How long does it take for the elevator to stop? Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

t = 2.7 s

Explanation:

This is a kinematics problem.

How the elevator reduces its speed to zero by a distance equal to the length of the cylinder

        y = 2π r = 2 π d / 2

        y = π d

        y = π 1.1

        y = 3,456 m

now we can look for the acceleration of the system

        v² = v₀² - 2 a y

        0 = v₀² - 2 a y

        a = v₀² / 2y

        a = 2.6² / 2 3.456

        a = 0.978 m / s²

now let's calculate the time

        v = v₀ - a t

        0 = v₀ - at

         t = v₀ / a

         t = 2.6 /0.978

         t = 2.658 s

ask for the result with two significant figures

         t = 2.7 s

A 1,200kg roller coaster car starts rolling up a slope at a speed of 15m/s. What is the highest point it could reach

Answers

Answer: 11.36 m

Explanation:

Given

Mass of roller coaster is m=1200 kg

Initial speed of roller coaster is v=15 m/s

Energy at bottom and at the top is same i.e.

[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]

Thus, the highest point reach by the roller coaster is 11.36 m

Answer:

11.36m

Explanation:

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How long will it take to get to the top of its trajectory? A. 3 seconds B. 4 seconds C. 2 seconds D. 6 seconds

Answers

Answer:

A (3 seconds)

Explanation:

Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.

Lets identify our givens.

Givens:

Horizontal speed= 30m/s

Vertical Speed= 30 m/s

Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0

The ball is launched from the ground so y0=0

Final vertical velocity= 0

This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use

vy=vy0+ayt

Plug in our givens

0=30-10t

solve for t

t=3 seconds

A room has dimensions of 15 ft by 15 ft by 20 ft contains air with a density of 0.0724 pounds-mass per cubic feet. The weight of air in the room in pounds-force is

Answers

Answer:

the weight of the air in pound-force (lb-f) is 325.8 lbf

Explanation:

Given;

dimension of the room, = 15 ft by 15 ft by 20 ft

density of air in the room, ρ = 0.0724 lbm/ft³

The volume of air in the room is calculated as;

Volume = 15 ft x 15 ft x 20 ft = 4,500 ft³

The mass of the air is calculated as;

mass = density x volume

mass = 0.0724 lbm/ft³  x  4,500 ft³

mass = 325.8 lb-m

The weight of the air is calculated as;

Weight = mass x gravity

Weight = 325.8 lb-m x 32.174 ft/s²

Weight = 10482.29 lbm.ft/s²

The weight of the air in pound-force (lb-f) is calculated as;

1 lbf = 32.174 lbm.ft/s²

[tex]Weight =10,482.29\ lbm.ft/s^2\times \frac{1 \ lbf}{32.174 \ lbm.ft/s^2} \\\\Weight = 325.8 \ lbf[/tex]

Therefore, the weight of the air in pound-force (lb-f) is 325.8 lbf

Part B
What is the approximate amount of thrust you need to apply to the lander to keep its velocity roughly constant? Explain why, using Newton's first
law of motion.

Answers

Answer:

Force is zero.

Explanation:

According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.

F = m a

if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.  

Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.

Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.

For a research project, a student needs a solenoid that produces an interior magnetic field of 0.0100 T. She decides to use a current of 1.00 A and a wire 0.500 mm in diameter. She winds the solenoid in layers on an insulating form 1.00 cm in diameter and 20.0 cm long.
Determine the number of layers of wire needed. (Round your answer up to the nearest integer.)
Determine the total length of the wire. (Use the integer number of layers and the average layer diameter.)

Answers

Answer:

[tex]n=3.8[/tex]

Explanation:

From the question we are told that:

Magnetic Field [tex]B=0.01T[/tex]

Current [tex]I=1.00[/tex]

Wire Diameter [tex]d_w=0.5*10^3m[/tex]

Layers Diameter [tex]d_l=1*10^2m[/tex]

Length [tex]l=0.2m[/tex]

Generally the equation for number of layers is mathematically given by

[tex]n=\frac{Bd_w}{\mu_o I}[/tex]

Where

[tex]Vacuum\ permeability=\mu_0[/tex]

[tex]n= \frac{0.01*0.5*10^3m}{4 \pi *10^{-7}*1 }[/tex]

[tex]n=3.8[/tex]

A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential energy are zero. How is the gymnast’s mechanical energy stored for that moment? Question 12 options: rest energy chemical energy elastic energy thermal energy

Answers

Answer:

elastic energy

Explanation:

When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.

During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.

The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:

elastic energy

Soap bubble coloring example:
(reflection, interference, refraction, diffraction)​

Answers

Explanation:

Interference is the example of soap bubble colouring

EXTRA INFO:

(LOOK AT THE IMAGE)

An incoming light ray is partly reflected by the top surface of the soap film and partly reflected by the bottom surface. The wave reflected from the bottom surface has traveled further (an extra distance equal to twice the thickness of the film) so emerges out of step with the top wave. When the two waves meet, they add together, and some colors are removed by destructive interference. Where the film is thickest, the bubble appears more blueish; where it's thinner, it will look more violet or magenta.

[tex]\huge\bold\color{black}{ANSWER}[/tex]

Soap bubble coloring example: INTERFERENCE

as the ball rises the vertical component of it's velocity_____. explain​

Answers

Answer:

Decreases

Explanation:

because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again

A bullet with mass 5.35 g is fired horizontally into a 2.174-kg block attached to a horizontal spring. The spring has a constant 6.17 102 N/m and reaches a maximum compression of 6.34 cm.
(a) Find the initial speed of the bullet-block system.
(b) Find the speed of the bullet.

Answers

Answer:

a)[tex]V=1.067\: m/s[/tex]

b)[tex]v=434.65\: m/s [/tex]  

Explanation:

a)

Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.

[tex]\frac{1}{2}MV^{2}=\frac{1}{2}k\Delta x^{2}[/tex]

Where:

M is the bullet-block mass (0.00535 kg + 2.174 kg = 2.17935 kg)V is the speed of the systemk is the spring constant (6.17*10² N/m)Δx is the compression of the spring (0.0634 m)

Then, let's find the initial speed of the bullet-block system.

[tex]V^{2}=\frac{k\Delta x^{2}}{M}[/tex]

[tex]V=\sqrt{\frac{6.17*10^{2}*0.0634^{2}}{2.17935}}[/tex]

[tex]V=1.067\: m/s[/tex]

b)

Using the conservation of momentum we can find the velocity of the bullet.

[tex]mv=MV[/tex]

[tex]v=\frac{MV}{m}[/tex]

[tex]v=\frac{2.17935*1.067}{0.00535}[/tex]

[tex]v=434.65\: m/s [/tex]  

I hope it helps you!

             

 

Which one is the better material to use for an inexpensive compass? hard iron, soft iron, or any conductor ​

Answers

Answer:

Soft iron

Explanation:

A resident of a lunar colony needs to have her blood pressure checked in one of her legs. Assume that we express the systemic blood pressure as we do on earth and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still 120/80 (which may not actually be true).

Required:
If a lunar colonizer has her blood pressure taken at a point on her ankle that is 1.5 m below her heart, what will be her systemic blood-pressure reading, expressed in the standard way, if she has normal blood pressure? The acceleration due to gravity on the moon is 1.67 m/s^2

Answers

Answer:

The pressure is 2505 Pa.  

Explanation:

Height, h = 1.5 m

density of blood, d = 1000 kg/cubic meter

Gravity, g = 1.67 m/s^2

let the pressure is P.  

The pressure due to the fluid is given by

P = h d g

P = 1.5 x 1000 x 1.67

P = 2505 Pa

A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 21 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 3.3 s?

Answers

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is  ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

if Petrol diesel etc catches fire one should never try to extinguish in using water why?​

Answers

Answer:

because both petrol and diesel are oil

Explanation:

oil floats on water that's why if we will try to extinguish with water so the fire will float on water

hope u like my answer

please mark methe brainest

The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of 5.7. The work output of the engine is 940 J in each cycle. Compute the temperatures of the two reservoirs between which this engine
operates.

Answers

Answer:

The hot temperature is 157.5 K

The cold temperature is 48.8 K

Explanation:

Step 1: Data given

The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.

The volume increases by a factor of 5.7

The work output of the engine is 940 J in each cycle.

During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)

Step 2:For a carnot engine:

V2/V1 = V4/V3

Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))

⇒with Work = the work done in the cycle = 940J

⇒with n = the number of moles = 1.50 moles

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T1 = the hot temperature

⇒With T2⇒ the cold temperature

where R = 8.31 J/mol K Gas Constant

940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))

940 = 1.5 * 8.314 ln(2) * (T1-T2)

(T1-T2) = 940 / (1.5*8.314*ln(2))

(T1-T2) = 108.7K

For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)

For a monatomic ideal gas, Cv = 3/2R

When we combine both, we'll have:

T2 = T1*(1/5.7)^(R/3/2R)

T2 = T1*(1/5.7)^(2/3)

T2= T1 * 0.31

Since we know that (T1-T2) = 108.7K

we have:

T1 - 0.31T1= 108.7K

0.69T1 = 108.7K

T1 = 157.5K

T2 = 157.5*0.31 = 48.8K

Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volume and pressure of container B, respectively. If Container A has half as many molecules of the ideal gas in it as Container B does, then which of the following mathematical statements is correct regarding the absolute temperatures TA and TB in Container A and Container B. respectively?
A. TA = TB/2.
B. TA = 4TB.
C. TA = TB/4.
D. TA = 2TB.
E. TA = TB

Answers

Answer:

A. TA = TB/2.

Explanation:

Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:

[tex]V_A = \frac{1}{2}V_B[/tex]

Now, from Charle's Law:

[tex]\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}[/tex]

Hence, the correct option is:

A. TA = TB/2.

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