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# Please show all steps- Thank you- Constants Part A Four identical mass.docx

Please show all steps. Thank you.
Constants Part A Four identical masses of mass 5.00 kg each are placed at the corners of a square whose side lengths are 3.00 m What is the magnitude of the net gravitational force on one of the masses, due to the other three? Express your answer with the appropriate units aal
Solution
The expression for the gravitational force of attraction between the two masses in Newtons is -
F = G m?m?/rÂ²
where,
G = 6.674e-11 mÂ³/kgsÂ²
m? and m? are the masses of the two objects in kg r is the distance in meters between their centers

#1, at +x = 3?2 meter distance
F1 = G(5)Â² / (3?2)Â² = 1.39*G
angle is 0Âº

#2 at distance 1 meter
F2 = G(5)Â² / (3)Â² = 2.78*G
angle is 45Âº

#3 at distance 1 meter
same as F2 but angle is â€“45Âº

Y components of 2 and 3 cancel, so all we have to do is add the x components
F = F1 + F2cos45 + F3cos45
F = F1 + 1.414 F2
F = 1.39*G + 1.414*2.78*G = 5.32*G = 5.32*6.674*10^-11 = 3.55 x 10^-10 N
.

Please show all steps. Thank you.
Constants Part A Four identical masses of mass 5.00 kg each are placed at the corners of a square whose side lengths are 3.00 m What is the magnitude of the net gravitational force on one of the masses, due to the other three? Express your answer with the appropriate units aal
Solution
The expression for the gravitational force of attraction between the two masses in Newtons is -
F = G m?m?/rÂ²
where,
G = 6.674e-11 mÂ³/kgsÂ²
m? and m? are the masses of the two objects in kg r is the distance in meters between their centers

#1, at +x = 3?2 meter distance
F1 = G(5)Â² / (3?2)Â² = 1.39*G
angle is 0Âº

#2 at distance 1 meter
F2 = G(5)Â² / (3)Â² = 2.78*G
angle is 45Âº

#3 at distance 1 meter
same as F2 but angle is â€“45Âº

Y components of 2 and 3 cancel, so all we have to do is add the x components
F = F1 + F2cos45 + F3cos45
F = F1 + 1.414 F2
F = 1.39*G + 1.414*2.78*G = 5.32*G = 5.32*6.674*10^-11 = 3.55 x 10^-10 N
.

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### Please show all steps- Thank you- Constants Part A Four identical mass.docx

1. 1. Please show all steps. Thank you. Constants Part A Four identical masses of mass 5.00 kg each are placed at the corners of a square whose side lengths are 3.00 m What is the magnitude of the net gravitational force on one of the masses, due to the other three? Express your answer with the appropriate units aal Solution The expression for the gravitational force of attraction between the two masses in Newtons is - F = G m?m?/rÂ² where, G = 6.674e-11 mÂ³/kgsÂ² m? and m? are the masses of the two objects in kg r is the distance in meters between their centers #1, at +x = 3?2 meter distance F1 = G(5)Â² / (3?2)Â² = 1.39*G angle is 0Âº #2 at distance 1 meter F2 = G(5)Â² / (3)Â² = 2.78*G angle is 45Âº #3 at distance 1 meter same as F2 but angle is â€“45Âº Y components of 2 and 3 cancel, so all we have to do is add the x components F = F1 + F2cos45 + F3cos45 F = F1 + 1.414 F2 F = 1.39*G + 1.414*2.78*G = 5.32*G = 5.32*6.674*10^-11 = 3.55 x 10^-10 N