Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
Forces of 10.4 N north, 19.7 N east, and 15.3 N south are simultaneously applied to a 3.73 kg mass as it rests on an air table....Asked by Ann
Forces of 11.1N north, 19.9N east, and 14.0N south are simultaneously applied to a 3.62kg mass as it rests on an air table. What is the magnitude of its acceleration?
b) What is the direction of the acceleration in degrees? (Take east to be 0 degrees and counterclockwise to be positive. Enter an angle between -180 degrees and +180 degrees.)
b) What is the direction of the acceleration in degrees? (Take east to be 0 degrees and counterclockwise to be positive. Enter an angle between -180 degrees and +180 degrees.)
Answers
Answered by
Elena
F(x) =F1(x)+F2(x)+F3(x) =0+19.9 +0 =19.9 N
F(y) =F1(y)+F2(y)+F3(y) =11.1+0-14=-3 N
F=sqrt{F(x)^2 + F(y)^2}
a=F/m
tan alpha=F(y)/F(x)
unknown angle =appha +90 degrees
F(y) =F1(y)+F2(y)+F3(y) =11.1+0-14=-3 N
F=sqrt{F(x)^2 + F(y)^2}
a=F/m
tan alpha=F(y)/F(x)
unknown angle =appha +90 degrees
There are no AI answers yet. The ability to request AI answers is coming soon!