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2 years ago

10

A world-class tennis player can serve a tennis ball at 130 mi/h mi/h (about 58 m/s m/s ). The length of a tennis court is 78 ft

ft . Approximately how long does an opponent have to react from the instant the ball is served until the ball reaches the opponent?.

1 answer:

riadik2000 [5.3K]2 years ago

8 0

Answer:

The time taken is $t = 0.4103 \ s$

Explanation:

From the question we are told that

The speed is $v = 130 \ mi/h = 58 \ m/s$

The length of the tennis court $l = 78 \ ft = \frac{78}{3.281}= 23.8 \ m$

Generally the time taken is mathematically represented as

$t = \frac{ 23.8}{ 58}$

=> $t = 0.4103 \ s$

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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

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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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