As shown in Fig. 4, an ideal gas of monatomic molecules expands from its initial state A to a state B through an isobaric process and then further expands to a volume C. Find the work done by the gas, increase in internal energy, and the energy transferred by heat to the gas over the whole process​

As Shown In Fig. 4, An Ideal Gas Of Monatomic Molecules Expands From Its Initial State A To A State B

Answers

Answer 1
The picture is very blurry

Related Questions

13. What type of lens bends light outwards and away from a point?
concave

Answers

Answer:

No,it isn't concave. The correct answer is convex lens.

Explanation:

A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.

Convex lens is the answer.

See the attached diagram.

A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight but catches it as it is falling back to earth. If the friend catches the ball 3.0 s after it is thrown, at what time did it pass him on its upward flight

Answers

Answer:

[tex]t=1.9 sec[/tex]

Explanation:

From the question we are told that:

Height [tex]h=28m[/tex]

Time [tex]t=3s[/tex]

Generally the Newton's equation for Initial velocity upward is mathematically given by

 [tex]s=ut+\frtac{1}{2}at^2[/tex]

 [tex]28=3u-\frac{1}{2}*9.8*3^2[/tex]

 [tex]u=24.03m/s[/tex]

Generally the velocity at  elevation and depression occurs  as ball arrives and passes through S=28

 [tex]v=\sqrt{24.03-2*9.8*28}[/tex]

 [tex]v=5.35m/s and -5.35m/s[/tex]

Generally the Newton's equation for time to reach initial velocity  is mathematically given by

 [tex]v=u+at[/tex]

 [tex]5.35=24.03-9.8t[/tex]

 [tex]t=\frac{28.03-5.35}{9.8}[/tex]

 [tex]t=1.9 sec[/tex]

A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12.0 m/s. What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping

Answers

Explanation:

The centripetal force [tex]F_c[/tex] on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.

[tex]y:\:\:\:\:N - mg = 0[/tex]

[tex]x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N[/tex]

or

[tex]m \dfrac{v^2}{r} = \mu mg[/tex]

Solving for [tex]\mu[/tex],

[tex]\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28[/tex]

g you hang an object of mass m on a spring with spring constant k and find that it has a period of T. If you change the spring to one that has a spring constant of 2 k, the new period is

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Answer:

a)   T = 2π [tex]\sqrt{\frac{m}{k} }[/tex],  b)  T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

Explanation:

a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity

          w² = k / m

angular velocity and period are related

          w = 2π /T

     

we substitute

          4π²/ T² = k / m

           T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

b) We change the spring for another with k ’= 2 k, let's find the period

           T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]

           T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]

           T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

20. Using the picture, how many neutrons does lithium have?

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Answer:

No. of Neutrons = 3

Explanation:

The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.

Mass Number = No. of Protons + No. of Neutrons = 6

Atomic Number = Number of Electrons = No. of Protons = 3

Therefore,

Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons

Mass Number - Atomic Number = No. of Neutrons

No. of Neutrons = 6 - 3

No. of Neutrons = 3

Answer:

3

Explanation:

Honestly, it is the only one in the picture and as an answer.

A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound=330 m/s.
What is the velocity v of the police car ?

Answers

Vs = 34m/s
I don’t have an explanation my apologies.

When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.

What is the Doppler formula?

The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.

The frequency increase by the Doppler effect is represented by the formula

f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f

Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀  is 0.

Substituting the value into the equation will give us the velocity of the police car.

[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)

When the car is receding, the frequency of the receiving signal f = 4500 Hz.

[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)

Solving both equation, we get the velocity of a police car.

v = 33 m/s

Therefore, the velocity v of the police car is 33 m/s.

Learn more about Doppler equation.

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what is the velocity of a 1.3 kg puppy with a forward momentum of 6 kg m/s​

Answers

Answer:

by using p = mv equation we can find v,

6 = 1.3 v

4.615 = v

An object with mass m = 0.56 kg is attached to a string of length r = 0.72 m and is rotating with an angular velocity ω = 1.155 rad/s. What is the centripetal force acting in the object?

Answers

Answer:

The centripetal force is 0.54 N.

Explanation:

mass, m = 0.56 kg

radius, r = 0.72 m

angular speed, w = 1.155 rad/s

The centripetal force is given by

[tex]F = m r w^2\\\\F =0.56\times 0.72\times 1.155\times 1.155\\\\F = 0.54 N[/tex]

A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train then moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt. The acceleration during the first 5.6 km of travel is closest to which of the following?

a. 0.19 m/s^2
b. 0.14 m/s^2
c. 0.16 m/s^2
d. 0.20 m/s^2
e. 0.17 m/s^2

Answers

Answer:

The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

Explanation:

Given the data in the question;

since the train starts from rest,

Initial velocity; u = 0 m/s

final velocity; v = 42 m/s

distance covered S = 5.6 km = ( 5.6 × 1000 )m = 5600 m

acceleration a = ?

From the third equation of motion;

v² = u² + 2as

we substitute in our values

( 42 )² = ( 0 )² + [ 2 × a × 5600 ]

1764 = 0 + [ 11200 × a ]

1764 = 11200 × a

a = 1764 / 11200

a = 0.1575 ≈ 0.16 m/s²          { two decimal place }

Therefore, The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

Water at 200 C has a bulk modulus of 2.2109 Pa, and the speed of sound in water at this temperature is 1480m/s. For 1000Hz sound waves in water at 200 C, what displacement amplitude is produced if the pressure amplitude is 310-2 Pa?​

Answers

What does the mystery stand for?

A helicopter is ascending vertically witha speed of 5.40 m/s. At a height of 105 m above the earth a package is dropped from the helicopter. How much time does is take for the package to reach the ground

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Answer: 5.21 s

Explanation:

Given

Helicopter ascends vertically with [tex]u=5.4\ m/s[/tex]

Height of helicopter [tex]h=105\ m[/tex]

When the package leaves the helicopter, it will have the same vertical velocity

Using equation of motion

[tex]\Rightarrow h=ut+\dfrac{1}{2}at^2\\\\\Rightarrow 105=-5.4t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-5.4t-105=0\\\\\Rightarrow t=\dfrac{5.4\pm \sqrt{5.4^2+4\times 4.9\times 105}}{2\times 4.9}\\\\\Rightarrow t=\dfrac{5.4\pm 45.68}{9.8}\\\\\Rightarrow t=5.21\ s\quad \text{Neglect negative value}[/tex]

So, package will take 5.21 s to reach the ground

What is the law of conservation of energy ​

Answers

energy can not be created nor destroyed
Hope this helps!!

A 55 kg person is in a head-on collision. The car's speed at impact is 12 m/s. Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answers

1.244 m per second the person driving will go

what is microeconomics ​

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Answer:

Microeconomics is a part of economics and the study of decisions made by people and businesses regarding the allocation of resources, and prices at which they trade goods and services.

Microeconomics helps business planning i.e. helps the business community to plan their costs, production, etc. in anticipation of demand in order to maximize profits. Microeconomics is useful in explaining how market mechanism determines the price in a free market economy.

Wind is caused by ___. the earth's tilt the Coriolis effect temperature differences humidity

Answers

I am guessing wind is caused by climate change in the atmosphere

Explanation:

wind is cause by climate change in the atmosphere that depends weather is is breezy really cold or rain and cold

Answer:

caused by the uneven heating of the Earth by the sun and the  own rotation.

LC-circuit of the radio receiver consists of variable capacitor (Cmin= 1 pF, Cmax=10 pF) and inductor
with inductance 1 µH. Determine the wavelength range of this radio receiver.

Answers

Answer:

the radio can tune wavelengths between 1.88 and 5.97 m

Explanation:

The signal that can be received is the one that is in resonance as the impedance of the LC circuit.

         X = X_c - X_L

         X = 1 / wC - w L

at the point of resonance the two impedance are equal so their sum is zero

         X_c = X_L

         1 / wC = w L

         w² = 1 / CL

         w = [tex]\sqrt{\frac{1}{CL} }[/tex]

let's look for the extreme values

C = 1  10⁻¹² F

         w = [tex]\sqrt{\frac{1}{ 1 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]

         w = [tex]\sqrt{1 \ 10^{18}}[/tex]

         w = 10⁹ rad / s

C = 10 10⁻¹² F

         w = [tex]\sqrt{\frac{1}{10 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]Ra 1/10 10-12 1 10-6

         w = [tex]\sqrt{0.1 \ 10^{18}}[/tex]Ra 0.1 1018

         w = 0.316 10⁹ rad / s

Now the angular velocity and the frequency are related

           w = 2π f

           f = w / 2π

the light velocity  is

           c = λ f

           λ = c / f

we substitute

          λ = c 2π/w

               

we calculate the two values

 C = 1 pF

          λ₁ = 3 10⁸ 2π / 10⁹

          λ₁= 18.849 10⁻¹ m

          λ₁ = 1.88 m

C = 10 pF

           λ₂ = 3 10⁸ 2π / 0.316 10⁹

           λ₂ = 59.65 10⁻¹ m

           λ₂ = 5.97 m

so the radio can tune wavelengths between 1.88 and 5.97 m

A golf ball is dropped from rest from a height of 8.40 m. It hits the pavement, then bounces back up, rising just 5.60 m before falling back down again. A boy then catches the ball when it is 1.40 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

Answers

Answer:

t1= 8.40/10 =.84 s

t2 = 5.60/10 = .56s

t3= 1.4/10 = .14s

total time = 1.54 sec

Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan

Answers

Answer:

F = 1010 Lb

the tension on the cable is greater than its resistance, which is why the plan is not viable

Explanation:

For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.

          v = v₀ + a t

how the car comes out of rest v₀ = 0

          a = v / t

let's reduce to the english system

          v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s

let's calculate

          a = 66/10

          a = 6.6 ft / s²

now let's write Newton's second law

X axis

         Fₓ = ma

with trigonometry

         cos 20 = Fₓ / F

         Fₓ = F cos 20

we substitute

          F cos 20 = m a

          F = m a / cos20

          W = mg

          F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]

let's calculate

          F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20

          F = 1010 Lb

Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.

ou charge a piece of aluminum foil (mass = 4.99 g) by touching it to a charged rod. The charged rod gives the aluminum foil a charge of 13 µC. Your professor brings a charged plate over and tells you to put the aluminum foil on top of the plate. To your surprise the aluminum foil hovers motionless in the air above it! Calculate the value of the electric field from the charged plate (assume it is a uniform field and the aluminum foil is a point charge).

Answers

Answer:

The appropriate answer is "3761.69 N/C".

Explanation:

Given that:

Mass,

m = 4.99 g

or,

   = [tex]4.99\times 10^{-3} \ kg[/tex]

Charge,

q = 13 µC

or,

  = [tex]13\times 10^{-6} \ C[/tex]

As we know,

⇒ [tex]F=mg=Eq[/tex]

then,

⇒ [tex]E=\frac{mg}{q}[/tex]

By putting the values, we get

        [tex]=\frac{4.99\times 10^{-3}\times 9.8}{13\times 10^{-6}}[/tex]

        [tex]=3761.69 \ N/C[/tex]

An 8.50 kg point mass and a 14.5 kg point mass are held in place 50.0 cm apart. A particle of mass (m) is released from a point between the two masses 12.0 cm from the 8.50 kg mass along the line connecting the two fixed masses.Find the magnitude of the acceleration of the particle.

Answers

Answer:

[tex]a=2.8*10^{-9}m/s[/tex]

Explanation:

From the question we are told that:

First Mass [tex]m=8.50kg[/tex]

2nd Mass [tex]m=14.5kg[/tex]

Distance

[tex]d_1=50=>0.50m\\\\d_2=>12cm=>0.12m[/tex]

Generally the Newtons equation for Gravitational force is mathematically given by

[tex]F_n=\frac{Gm_nm}{(r_n)^2}[/tex]

Therefore

Initial force on m

[tex]F_1=\frac{Gm_1m}{(r_1)^2}[/tex]

Final force on m

[tex]F_2=\frac{Gm_2m}{(r_2)^2}\\\\F=\frac{Gm_1m}{(r_1)^2}-\frac{Gm_2m}{(r_2)^2}[/tex]

Acceleration of m

[tex]a=\frac{F}{m}\\\\a=\frac{Gm_1}{r_1^2}-\frac{Gm_2}{r_2^2}[/tex]

[tex]a=6,67*10^{-11}{\frac{8.5}{0.12}}-\frac{14.5}{0.50}[/tex]

[tex]a=2.8*10^{-9}m/s[/tex]

1000 grams of water is heated from 0 degree to 200 degree . The specific heat of water is 4186 j/kg.°C. Estimate the change in entropy of the water.​

Answers

Answer:2

Explanation:

Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above​

Answers

Answer:

Explanation:no change in surface tension

An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.

Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.

The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.

As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.

Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

Learn more here: https://brainly.com/question/15785205

Electromagnetic waves from the sun carry what to the earth

Answers

Answer:

Solar radiation

Explanation:

Visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays.

==>  Energy

==>  Radio noise, heat, visible light, ultraviolet radiation, X-rays, gamma rays

==>  They carry all these kinds of energy wherever they go.  Not only to the Earth.

can anyone answer this fast pls

Answers

i believe the answer would be 4.5. because it wouldnt be c or d. and 2 seems too small.

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during the impact with the bat, how many impules of importance are used to find the final velocity of the bat

Answers

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

[tex]Mv+mu=Mv_1+mv_2[/tex]

(900 x 47) + (200 x -30)  = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

36300 =  (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)

[tex]9v_1 = 363 - 2v_2[/tex]

[tex]v_1=\frac{363 - 2v_2}{9}[/tex]

The mathematical expression for the conservation of kinetic energy is

[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]

[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex]    ................(ii)

[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]  

[tex]21681 = 9v_1^2+2v_2^2[/tex]

Substituting (i) in (ii)

[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]

[tex](363-2v_2)^2+18v_2^2=195129[/tex]

[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]

[tex]22v_2^2-145v_2-63360=0[/tex]

Solving the equation, we get

[tex]v_2=96 \ m/s, -30 \ m/s[/tex]

The negative velocity is neglected.

Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get

[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]

     = 19

Thus, only impulse of importance is used to find final velocity.

In the following circuit (Fig.3), calculate the intensity I through the resistance 3 using the principle of superposition.

Answers

Answer:

time

Explanation:

A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

Answers

Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

Human vision cuts off on the red side of the spectrum at about 675 nm. What is the energy of a photon (in J) of this wavelength?

Answers

Answer:

The energy of a photon is 2.94x10⁻¹⁹ J.

Explanation:

The energy of the photon is given by:

[tex] E = \frac{hc}{\lambda} [/tex]  

Where:

h: is Planck's constant = 6.62x10⁻³⁴ J.s

c: is the speed of light = 3.00x10⁸ m/s

λ: is the wavelength = 675 nm

Hence, the energy is:

[tex] E = \frac{hc}{\lambda} = \frac{6.62 \ccdot 10^{-34} J.s*3.00 \cdot 10^{8} m/s}{675 \cdot 10^{-9} m} = 2.94 \cdot 10^{-19} J [/tex]

Therefore, the energy of a photon is 2.94x10⁻¹⁹ J.

I hope it helps you!

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30.08 as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp.

Answers

Answer:

2.55 m/s

Explanation:

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.

Solution:

The work done by friction is given as:

[tex]W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J[/tex]

The work done by gravity is:

[tex]W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s[/tex]

A wave has angular velocity 12 rad/sec and maximum displacement X cm (X= last two digit [1] of your student’s ID). Calculate the maximum acceleration of the wave.

Answers

Answer:

[tex]a=1440\ m/s^2[/tex]

Explanation:

Given that,

The angular velocity of a wave, [tex]\omega=12\ rad/s[/tex]

The maximum displacement of the wave, A = 10 cm (let)

The maximum acceleration of the wave is given by :

[tex]a=-A\omega^2[/tex]

Put all the values,

[tex]a=10\times 12^2\\\\a=1440\ m/s^2[/tex]

So, the maximum acceleration of the wave is equal to [tex]1440\ m/s^2[/tex].

Other Questions
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