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Olin [163]
4 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]4 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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lilavasa [31]

Answer:

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Explanation:

it's c because the moon has to be a full moon to be a solar eclipse when the sun moon and earth line up

6 0
3 years ago
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In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (
omeli [17]

Answer:

58 cm/s

Explanation:

0.5×129=0.5×(-45)+1.5×V

V=58

7 0
3 years ago
A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.
zzz [600]

Answer:

0.0257259766982 m

Explanation:

P_2 = Atmospheric pressure = 101325 Pa

d_1 = Initial diameter = 1.5 cm

d_2 = Final diameter

\rho = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m

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8 0
3 years ago
Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same
marissa [1.9K]

Answer:

maximum static frictional force of the wall on the book (Increasing)

normal force of the wall on the book (Decreasing)

weight of the book (Not changing)

Explanation:

Now according to Newton's third law of motion

"Every action has equal but opposite reaction"

By the data given in question, Albert was pressing the book against the wall.Now, Albert started to reduce his force up against the wall.

First we have to consider all the forces applied on book in this scenario.

1. Weight of book acting downwards (y-axis)

2. Friction between book and wall acting upward (y-axis)

3. Albert's force on book against wall (x-axis)

4. Normal reaction of wall against Albert's force (x-axis)

Now, when Albert reduced his force, new scenario will be

1. Weight will be remain constant as it is W = mg

Neither mass nor acceleration due to gravity changed, so weight acting upon the book will remain same.

2. When Albert reduced force, normal reaction of wall reduced against it according to Newton's third law of motion

3. Now notice that friction is a force which acts in accordance with the applied force. For example if a box is placed at floor, no friction is applied, but when you drag the box, friction starts to act and increases until its limit comes. So, when Albert reduced his force, weight will try to pull the book and maximum static friction will increase to hinder the movement of book downwards.

Notice that maximum static friction will hinder the book from movement, since Albert reduced his force, but wight didn't pull the book, which means that maximum static friction increased to hinder downward motion.

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An object is moving along a straight line at a
Dominik [7]
The answer for this question is 5 m
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