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when one boy is sitting 1.20 meter from the center of he seesaw the girl must sit on the other side 1.50 meter from the center...Asked by shanel
when one boy is sitting 1.20 meter from the center of he seesaw the girl must sit on the other side 1.50 meter from the center to maintain an even balance . however the boy carries an additional mass pf 14 kg and sit 1.80 meter from the center the girl must move to 3 meter from the center to balance . neglecting the weight of the seesaw find the weight of the boy and the girl?
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Answered by
Reiny
Basic law of playground-math:
the product of the distance and mass on both sides of the teeter-totter must be the same.
so let the boys mass be b kg
and that of the girls be g kg
1.2b = 1.5g
2.4b = 3g
1.8(b+14) = 3g
1.8b + 25.2 = 2.4b
25.2 = .6b
b = 25.2/.6 = 42
sub into 1.2b = 1.5g
1.2(42) = 1.5g
g = 50.4/1.5 = 33.6
state the conclusion by using my definitions
check:
is 1.2(42) = 1.5(33.6) ? yes!
is 1.8(42+14) = 3(33.6) ? Yes!
the product of the distance and mass on both sides of the teeter-totter must be the same.
so let the boys mass be b kg
and that of the girls be g kg
1.2b = 1.5g
2.4b = 3g
1.8(b+14) = 3g
1.8b + 25.2 = 2.4b
25.2 = .6b
b = 25.2/.6 = 42
sub into 1.2b = 1.5g
1.2(42) = 1.5g
g = 50.4/1.5 = 33.6
state the conclusion by using my definitions
check:
is 1.2(42) = 1.5(33.6) ? yes!
is 1.8(42+14) = 3(33.6) ? Yes!
Answered by
shanel
thank you!. im glad to find this site. you respond quickly.
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