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Lorico [155]
1 year ago
10

An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the

expansion is isothermal, what are the final pressure pfi and the work Wi done by the gas? If, instead, the expansion is adiabatic, what are the final pressure pfa and the work Wa done by the gas? State your answers in terms of the given variables.
Physics
1 answer:
yawa3891 [41]1 year ago
3 0

Answer:

Isothermal :   P2 = ( P1V1 / V2 ) ,  work-done pdv = nRT * In( \frac{V2}{v1} )

Adiabatic : : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }  , work-done =

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

pdv = nRT * In( \frac{V2}{v1} )

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

P1V1^y = P2V2^y

where y = 5/3

hence : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }

Work-done

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Where    T2 = T1V1^(2/3)/V2^(2/3)

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A man climbs a ladder. Which two quantities can be used to calculate the energy stored of the man at the top of the ladder.
Dvinal [7]

Answer:The answer must be The weight of the man and the vertical distance moved.

Explanation: you calculate it by the force you applied times the distance you moved

8 0
1 year ago
When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.
tensa zangetsu [6.8K]
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2


Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
5 0
2 years ago
In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field
Kay [80]

Explanation:

It is given that,

Area of square coil, A=4\ cm^2=0.0004\ m^2

Side of the square, L = 0.02 m

Number of turns, N = 10000

Uniform magnetic field, B = 1.5 T

Speed, v = 100 m/s

An emf is induced in the coil which is given by :

E=NBLv

E=10000\times 1.5\times 0.02\times 100

E = 30000 V

Breakdown voltage of air, V=4000\ V/cm=400000\ V/m

Let d is the gap between the two wires connected to the ends of the coil and still get a spark. So,

Electric field, E'=\dfrac{V}{d}

\dfrac{30000}{d}=400000

d = 0.075 m

Hence, this is the required solution.

6 0
2 years ago
Car B is following Car A and has a greater speed than Car A. the two cars are moving in a straight line and in the same directio
ElenaW [278]

Answer:

Collision force will be same in both the cases.

Explanation:

A perfectly inelastic collision is said to take place  when a system loses the amount of its Kinetic Energy at its maximum. In a perfectly inelastic collision,  the colliding particles stick to each other. In such a collision, kinetic energy is lost by combining the two bodies with each other.

In situation 1:

Speed of Car A, v_{A} = 10 mph

Speed of Car B, v_{B} = 20 mph

Relative speed of car A and car B, v = v_{b} - v_{a} = 10 m/s

Now, in the situation 2:

Speed of car A, v_{A} = 30 mph

Speed of car B, v_{B} = 40 mph

Relative speed of car A and car B, v = v_{b} - v_{a} = 10 m/s

Therefore, Car A and Car B both have the same relative speed, v = 10 m/s

7 0
1 year ago
Automobile A starts from O and accelerates at the constant rate of 0.75 m/s2. A short time later it is passed by bus B which is
boyakko [2]

Answer:

they meet from point o at distance 50.46 m and time taken is 11.6 seconds

Explanation:

given data

acceleration = 0.75 m/s²

speed B = 6 m/s

time B = 20 s

to find out

when and where the vehicles passed each other

solution

we consider here distance = x , when they meet after o point

and time = t for meet point z

we find first Bus B distance for 20 s ec

distance B = velocity × time

distance B = 6 × 20

distance B = 120 m

so

B take time to meet is calculate by distance formula

distance = velocity × time

120 - x = 6 × t

x = 120 - 6t   .................1

and

distance of A when they meet by distance formula

distance = ut + 1/2 × at²

here u is initial speed = 0 and t is time

x = 0 + 1/2 × 0.75 × t²

x = 0.375 × t²      .............2

so from equation 1 and 2

0.375 × t²  = 120 - 6t

t = 11.6

so time is 11.6 second

and

distance from point o from equation 2

x = 0.375 (11.6)²

x = 50.46

so distance from point o is 50.46 m

6 0
1 year ago
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