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Four 6.0 kg spheres are located at the corners of a square of side 0.70 m. Calculate the magnitude and direction of the gravita...Asked by myra
Four 7.5 kg spheres are located at the corners of a square of side 0.60 m. Calculate the magnitude and direction of the gravitational force on one sphere due to the other three.
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Answered by
Tiffany
R=.6M
mass=7.5 kg
Going clockwise label the masses in a square. M1 top left corner, M2 top right corner, M3 bottom right corner, M4 bottom left corner.
Start with M1 and M2 use Newton's universal gravity formula;
F=G((M1*M2)/r^2)
Where G always equals 6.67*10^-11
Back to M1 and M2
equation:
Fg=6.67*10^-11((7.5kg*7.5kg)/0.6m^2)
Fg=1.04*10^-8 @ 0°
Same exact thing for M1 and M4 but instead of 0° it's 270°
For M1 and M3 it's a diagonal radius so to solve for the Diagonal radius make a right triangle and use the equation a^2+b^2=c^2 where c^2 is what we are finding.
Equation:
.6^2+.6^2=c^2
.72=c^2
Square root.72 To get the radius of .84
Now solve M1 and M3
Equation:
Fg=6.67*10^-11((7.5kg*7.5kg)/0.84m^2)
Fg=5.31*10^-9 @ 315°
Now you have to add the vectors to get the vector of C^2 by making a right triangle connecting M1,M2,M3 making M2 where the 90° is
From M1 to M2 we already know it's 1.04*10^-8 but she still need to find the cos to find the other part of the line.
Equation:
Cos(45=x/5.31*10^-9
X=3.75*10^-9
Y (aka M2 to M3) equals the same because 45 makes sin and cos the same
After finding that add
(1.04*10^-8)+(3.75*10^-9)
Getting; 1.41*10^-8 Remember both lines M1 to M2 and M2 to M3 are equal so they are both 1.41*10^-8
With that said to finalize this problem find C^2 now
Equation:
(1.41*10^-8)^2 + (1.41*10^-8)^2 =C^2
4.00*10^-16=CA
square root 4.00*10^-16 and you should get 2.0*10^-8
ANSWER: 2.0*10^-8 @ 45°
mass=7.5 kg
Going clockwise label the masses in a square. M1 top left corner, M2 top right corner, M3 bottom right corner, M4 bottom left corner.
Start with M1 and M2 use Newton's universal gravity formula;
F=G((M1*M2)/r^2)
Where G always equals 6.67*10^-11
Back to M1 and M2
equation:
Fg=6.67*10^-11((7.5kg*7.5kg)/0.6m^2)
Fg=1.04*10^-8 @ 0°
Same exact thing for M1 and M4 but instead of 0° it's 270°
For M1 and M3 it's a diagonal radius so to solve for the Diagonal radius make a right triangle and use the equation a^2+b^2=c^2 where c^2 is what we are finding.
Equation:
.6^2+.6^2=c^2
.72=c^2
Square root.72 To get the radius of .84
Now solve M1 and M3
Equation:
Fg=6.67*10^-11((7.5kg*7.5kg)/0.84m^2)
Fg=5.31*10^-9 @ 315°
Now you have to add the vectors to get the vector of C^2 by making a right triangle connecting M1,M2,M3 making M2 where the 90° is
From M1 to M2 we already know it's 1.04*10^-8 but she still need to find the cos to find the other part of the line.
Equation:
Cos(45=x/5.31*10^-9
X=3.75*10^-9
Y (aka M2 to M3) equals the same because 45 makes sin and cos the same
After finding that add
(1.04*10^-8)+(3.75*10^-9)
Getting; 1.41*10^-8 Remember both lines M1 to M2 and M2 to M3 are equal so they are both 1.41*10^-8
With that said to finalize this problem find C^2 now
Equation:
(1.41*10^-8)^2 + (1.41*10^-8)^2 =C^2
4.00*10^-16=CA
square root 4.00*10^-16 and you should get 2.0*10^-8
ANSWER: 2.0*10^-8 @ 45°
Answered by
Julia
I had been looking online for the answer to this question for 3 hours and this is the only process that I found was right THANK YOU
Answered by
Davin
You're a blessing Tiffany :D You save me ^_^
Answered by
mady
This saved me! None of the other processes online were right but this one!
Answered by
Eric
Thanks Tiffany! I wish my teacher could actually teach me this instead of puffing the devil' so lettuce...
Answered by
mario
Bless! Thank you.
Answered by
Alicia
I have been looking at the note from my class and could not figure out how to solve this equation. THANK YOU for breaking down the process in an easy way to understand!!
Answered by
Emily
Thank you so much! This helped a lot.
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