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Olin [163]
3 years ago
6

The Gibson family have bought tickets to the Christmas Market. There are 4 mothers, 2 grand-mothers and 4 daughters. What is the

minimum number of tickets they need?
Physics
1 answer:
koban [17]3 years ago
5 0
The minimum number of tickets that could admit all of them is six (6).

This thing is impossible to explain in words, so I shall attempt it with a diagram:

Here are the six ladies:

       ( A )      ( B )
          |           |
          |           |
       ( C )      ( D )
          |           |
          |           |
       ( E )       ( F )  

--  'E'  and  'F'  are the daughters of  'C'  and  'D' .

--  'C'  and  'D'  are the daughters of  'A'  and  'B' .

So look what we have now:

--  'A'  and  'B'  are the mothers of  'C'  and  'D' .
     There's 2 of the mothers.

--  'C'  and  'D'  are the mothers of  'E'  and  'F' .
     There's the OTHER 2 mothers. 
 
--  'A'  and  'B'  are the grandmothers of  'E'  and  'F' .
    There's the 2 grandmothers.

--  'E'  and  'F'  are the daughters of  'C'  and  'D' .
     There's 2 of the daughters.

--  'C'  and  'D'  are the daughters of  'A'  and  'B' .
     There's the OTHER 2 daughters.

You want to know what ? !
The group is even bigger than THAT.
There are also 2 GRAND-daughters in the family ...  'E'  and  'F' .

So now you have a list of 12 people ! ... 4 mothers, 2 grandmothers,
4 daughters, and 2 grand-daughters ... and they all get in to the
Christmas Market with only six tickets.    Legally !

Such a deal !

Don't forget :  Christmas this year is also the first day of Chanukah !
                     All for the same price !

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Explanation:

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But,due to gravity,the object accelerates downward at a rate of 9.8ms^{-2}.

In X-Direction,

Given that initial velocity=u_{x}=33.8ms^{-1}

Using v=u+at,

v_{x}=33.8+(0)4.25=33.8ms^{-1}

In Y-Direction,

Given that initial velocity=u_{x}=0ms^{-1}

Using v=u+at,

v_{y}=0+(9.8)4.25=41.65ms^{-1}

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Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
Angelina_Jolie [31]

Answer:

a). \frac{\dot{W}}{m}= 311 kJ/kg

b). \frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

Explanation:

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\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0

\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})

\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})

\frac{\dot{W}}{m}= -30+1.1(980-670)

\frac{\dot{W}}{m}= 311 kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}

\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}

\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978

\frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

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3 years ago
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