All Answers 358110 Answers

1+12       1+6+6                             1 way

2+11        2+5+6    2=6+5               2 ways

3+10        10=6+4 or 4+6, or 5+5       3 ways

4+9          9=6+3, 3+6, 5+4, 4+5         4 ways

5+8          8=6+2, 2+6, 5+3, 3+5, 4+4   5 ways

6+7        7=6+1, 1+6, 5+2, 2+5, 4+3, 3+4  6 ways

6+5+4+3+2+1=21

Total combinations  $${{\mathtt{6}}}^{{\mathtt{3}}} = {\mathtt{216}}$$

P(sum of 13) = $${\frac{{\mathtt{21}}}{{\mathtt{216}}}} = {\frac{{\mathtt{7}}}{{\mathtt{72}}}} = {\mathtt{0.097\: \!222\: \!222\: \!222\: \!222\: \!2}}$$

Thanks Alan,

I am sitting here laughing my head off!

It is all too much for me.   ROF LOL  :)))

Thank you so much for the entertainment.  

well . 

Thank you Alan . 

Careless error on my part I  shall fix it (I did mean 62 not 22)

Yes, in this case, if you resolve it parallel and perpendicular to the direction of the force, and only use the component in the direction of the force to calculate the work done.

However, I would recommend not doing this, but stick to resolving the force instead, as it is less likely to lead to confusion in more complicated situations.

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It is just 

8*2*2*2  -  7*2*7*1

= 64        -   98

=  -34

Alan , can I resolve the displacement vector  ? 

Thank you alan 

now its clear 

Just use the on-site calculator:

$$\left({\mathtt{30}}{!}\right) {mod} \left({\mathtt{899}}\right) = {\mathtt{464}}$$

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FV=PV(1+i)^n

FV=future value = this is what you want to find

PV=present value = $40000

i = interest rate per interest period

n = number of interest period

Here the interest is being compounded yearly at 5%   for 25years

so   i=0.05,   n=25

FV=40000*(1+0.05)^25

=40000*1.05^25

$${\mathtt{40\,000}}{\mathtt{\,\times\,}}{{\mathtt{1.05}}}^{{\mathtt{25}}} = {\mathtt{135\,454.197\: \!635\: \!975\: \!392\: \!668\: \!3}}$$

so it will grow to $135,454.20

Oops!  Missed the 4 equations bit!

In general there won't be a unique solution, though it's possible that all four equations are consistent.  Can't tell without seeing the equations.

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Just type this straight into the site calc on the home page - replace the commas with decimal points.

But there is 4 equations Alan.  

$$\\P(4 girls) = 6C4 * 0.5^4*0.5^2 = 15*0.5^6\\
P(5 girls) = 6C5 * 0.5^5*0.5^1 = 6*0.5^6\\
P(6 girls) = 6C6 * 0.5^6*0.5^0 = 1*0.5^6\\\\
p(more\; girls\; than \;boys)=(15+6+1)*0.5^6=\frac{22}{64}=\frac{11}{32}$$

Hmm.  I've re-read the question more carefully.  

It says to treat the runner as a point mass, m and so implicitly assumes that the force she applies to the mass comes from outside the mass.  ie her leg muscles are outside of herself, though their mass is included within!  In this case her (massless) muscles do indeed provide an upward force on her (massive) point body, and we can think of the force as being upward!

So, choose your preferred weirdness: pushing the Earth down or massless legs pushing a point body up!

I think the latter is a dangerous way of looking at this as it is may too easily lead to confusion in more complicated situations where there could be several internal forces that actually contribute nothing to the motion of the object in question (a point has no internal structure).

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The question is worded sloppily in my view.  The runner does work using a downward force to move the Earth down by a certain distance, so the two negatives cancel each other out and the runner does some positive work!

Of course, that's a slightly weird way of looking at the situation, but if you are going to insist on careful use of signs in calculating the work (as the question does) then that's the way it should be done.

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Resolve the (left-facing, horizontal) force parallel to the line L and perpendicular to it. Only the parallel component counts towards the work done.  The angle θ is the smaller angle between F and L so the component Fcosθ points along L in the direction from A to B.

This is the same direction in which it pointed in part A of the question.  However, the movement is now in the opposite direction from part A, so it's not surprising that the sign is just the opposite of that in part A.

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I am sorry for my false informations 315.  

I think that you know by now that my physics is very shaky.

I wouldn't give you wrong info on purpose but you always need to be wary of my physics answers.

thanks for the correction Alan.  

so what do you think the answer of the example !! 

But how you determine that the component of force is in the direction of movement  ?  

what are the last 2 digits of 7^2800  or what is $$7^{2800} \mod 100$$ ?

$$7^{2800} \mod 100 \quad | \quad 2800 = 2^4*175\\
= 7^{ 16 * 175 } \mod 100 \\
= ( 7^{16} )^{175} \mod 100 \quad | \quad 7^{16}= (7^4)^4\\
= ( (7^4)^4 )^{175} \mod 100 \quad | \quad 7^{4}= (7^2)^2\\
= ( ( (7^2)^2 )^4 )^{175} \mod 100 \quad | \quad 7^2 = 49\\
= ( ( (49)^2 )^4 )^{175} \mod 100 \quad | \quad 49^2 = 2401\\
= ( ( 2401 )^4 )^{175} \mod 100 \quad | \quad 2401 \mod 100 = 1 \quad\textcolor[rgb]{1,0,0}{ x^{a*b} \mod d = (x^a\mod d)^b \mod d }\\
= ( ( 1 )^4 )^{175} \mod 100 \quad | \quad 1^4 = 1 \\
= ( 1 )^{175} \mod 100 \quad | \quad 1^{175} = 1 \\
= 1 \mod 100 \\\\
7^{2800} \mod 100 = 1 \mod 100 \quad | \quad 1 \mod 100 = 01$$

the last 2 digits of 7^2800 = 01

I don't think she is applying an upward force.  She is applying a downward force on the Earth (through the stairs) and so is moving the Earth (and hence the stairs) downward, relative to herself!

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Select the word "Scientific" in the input area, from the pop-up menu select "Equations" then choose "2 variables, 2 equations"

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If Jack can lift a mass of 62kg on Earth, then he is lifting a weight of 62*9.8 N (= 607.6N).

Gravitational acceleration on the alien planet is 9.8*62/22 m/s² (= 27.6 m/s²)

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I guess you mean that the weight is 80000N.

Pressure is force/unit area, so pressure on the ground = 80000/(4*0.25) N/m² or 80000 N/m²

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