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3 years ago

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A stationary source S generates circular outgoing waves on a lake. The wave speed is 5.0 m/s and the crest-to-crest distance is

2.0 m. A person in a motorboat heads directly toward S at 3.0 m/s. To this person, the frequency of these waves is:

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Answer:4 Hz

Explanation:

Speed of wave $v=5 m/s$

crest to crest distance $\lambda =2 m$

velocity of observer $v_0=3 m/s$

actual frequency $f=\frac{velocity}{\lambda }$

$f=\frac{5}{2}=2.5 Hz$

Apparent frequency $f'=f(\frac{v+v_0}{v})$

$f'=2.5\times \frac{5+3}{5}$

$f'=4 Hz$

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A piston having 7.23 g of steam at 110°C increases its temperature by 35°C. At the same time it expands from a volume of 2.00 L

My name is Ann [436]

Answer : The value of $q,w,\Delta U\text{ and }\Delta U$ is 505 J, -599 J, -94 J and -693 J respectively.

Explanation : Given,

Mass of steam = 7.23 g

Initial temperature = $110^oC$

Final temperature = $(110+35)^oC=145^oC$

Initial volume = 2 L

Final volume = 8 L

External pressure = 0.985 bar

Heat capacity of steam = 1.996 J/g.K

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

$\Delta U=q+w$

First we have to calculate the heat absorbed by the system.

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat absorbed by the system = ?

m = mass of steam = 7.23 g

$C_p$ = heat capacity of steam = $1.966J/g.K$

$T_1$ = initial temperature = $110^oC=273+110=383K$

$T_2$ = final temperature = $145^oC=273+145=418K$

Now put all the given value in the above formula, we get:

$Q=7.23g\times 1.966J/g.K\times (418-383)K$

$Q=505J$

Now we have to calculate the work done.

Formula used :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done = ?

$p_{ext}$ = external pressure = 0.985 bar = 0.985 atm (1 bar = 1 atm)

$V_1$ = initial volume of gas = 2.00 L

$V_2$ = final volume of gas = 8.00 L

Now put all the given values in the above formula, we get :

$w=-p_{ext}(V_2-V_1)$

$w=-(0.985atm)\times (8.00-2.00)L$

$w=-5.91L.atm=-5.91\times 101.3J=-599J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

$\Delta U=505J+(-599J)$

$\Delta U=-94J$

Now we have to calculate the change in enthalpy of the system.

Formula used :

$\Delta H=\Delta U+P\Delta V$

$\Delta H=\Delta U+w$

$\Delta H=(-94J)+(-599J)$

$\Delta H=-693J$

Therefore, the value of $q,w,\Delta U\text{ and }\Delta U$ is 505 J, -599 J, -94 J and -693 J respectively.

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