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Sergio039 [100]

3 years ago

9

A baseball player hits a ball with a bat. At one moment the force exerted by the bat on the ball is 18,000 N. At that same momen

t, how much force is exerted on the bat by the ball?
a) 18,000 N

b) less than 18,000 N

c) more than 18,000 N

1 answer:

11111nata11111 [884]3 years ago

5 0

Answer:

Less than 18000N

Explanation:

Given

$Force\ Exerted = 18000N$

This question will be answered using Newton's third law.

Understanding this law, it implies that reaction force is equal and opposite to the force exerted.

This implies that;

If the force exerted on the ball is 18000N

the force exerted is -18000N

So, the option that answers the question is less than 18000N because -18000N < 18000N

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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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