SOLUTION: Hello I'm really not sure if "Travel and Distance" is the correct section; but I hope I haven't thrown your off by choosing that. My instructor spent most of the class trying t

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Question 419726: Hello
I'm really not sure if "Travel and Distance" is the correct section; but I hope I haven't thrown your off by choosing that. My instructor spent most of the class trying to explain the concept of these type problems this evening; but no one understood it. As for me, I couldn't get a clue. Anyway, here is my problem (it's three-fold):
The force of gravity between any two objects varies jointly with the masses of the objects A and B, and inversely with the square of the distance D between the object.
a) Write the variation equation
b) If the force of gravity bween a 105 kg. man and his 80 kg wife when they are separated by a distance of 5 meters is 2.24112X10^-8 (the -8 is an exponent)newtons, find the variation constant.
c) Find the force of gravity between the man and his wife when they are 2 meters apart.
The rest of my assignment I can solve except for this one. Please help me as soon as possible. Next week Tuesday, we have a quiz.
Thank you in advance,
Lynn J. Tilmon


Found 2 solutions by ankor@dixie-net.com, duckness73:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
a) Write the variation equation
g = %28km%29%2Fd%5E2
Where
g = force of gravity
k = constant of variation
m = mass, (the more mass the more gravitational force between them)
d = distance (the more distance, the less the gravitational force between them)
:
b) If the force of gravity between a 105 kg. man and his 80 kg wife when they are separated by a distance of 5 meters is 2.24112X10^-8 (the -8 is an exponent)newtons, find the variation constant.
%28105%2A80%2Ak%29%2F5%5E2 = 2.24112(10^-8); (we use the product of the masses)
%288400k%29%2F25 = 2.24112(10^-8)
336k = 2.24112(10^-8)
k = %282.24112%2810%5E-8%29%29%2F336
k = 6.67*(10^-11) Newtons, this is the gravitational constant (G)
:
c) Find the force of gravity between the man and his wife when they are 2 meters apart.
g = %28105%2A80%2A6.67%2810%5E-11%29%29%2F2%5E2
g = 5.6028%2810%5E-7%29%2F4
g = 1.4007(10^-7) Newtons

Answer by duckness73(47) About Me  (Show Source):
You can put this solution on YOUR website!
When something "varies jointly" or "varies directly" with something else, it means that one is equal to the other multiplied by some constant. When something "varies inversely" with something else, it means that one is equal to the reciprocal of the other multiplied by some constant. So, taking what we know from the problem, we know that:
Force between two objects (call it F) varies jointly with the masses of the two objects (call these masses M1 and M2). Since they "vary jointly", force is equal to the product of the masses multipled by some constant (call it c), that is:
F = c * M1 * M2
But wait ...! We aren't done. Force also varies inversely with the square of the distance between the two. Let's call the distance D, and it's square is D%5E2. Since the Force varies inversely, Force is equal to the reciprocal multiplied by some constant, that is:
F+=+c+%2A+%281%2FD%5E2%29
So, putting the two together, we have
F+=+%28c+%2A+M1+%2A+M2%29%2F%28D%5E2%29
a) This is the variation equation: F+=+%28c+%2A+M1+%2A+M2%29%2F%28D%5E2%29

In the "b" part of the question, you are asked to find the value for the constant, c. M1 (husband's mass) is 105kg. M2 (wife's mass) is 80kg. The distance between them is 5 meters. And the force between them is 2.24112X10%5E-8. So plugging this into the variation equation, we can find the value for c:
2.24112X10%5E-8+=+%28c+%2A+105+%2A+80%29%2F25
%28%282.24112X10%5E-8%29+%2A+25%29%2F%28105+%2A+80%29+=+c
6.67+%2A+10%5E-11+=+c (approximately)
b) so the variation constant c = 6.67+%2A+10%5E-11
For the "c" part of the equation, you are asked to find the force when the husband and wife are 2 meters apart. Note that the husband's and wife's mass are still 105 and 80 kg respectively, and the constant c remains the same. So you have
F = (c * 105 * 80)/4 = 1.4+%2A+10%5E-9 approximately

(Feel free to check the arithmetic. And good luck on your quiz.)