A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is hanging from the board. The tension in the right chain is 250 N.
Which of the following describes where the person is hanging?

let Tension in the left chain is F

here, Fnet = 0

⇒F - 500N + 250 N - 125 N = 0

⇒F - 375 N = 0

⇒F = 375 N

let r from the left end of the board is the person sitting.

now torque at left end, τ = 0.

⇒375 × (0) - 500N × (r) + 250N × (4m) - 125 N × (2m) = 0

⇒0 - 500r + 1000 - 250 = 0

⇒750 = 500r

⇒r = 750/500 = 1.5 m

Explanation:

Can you please give this answer a brainliest answer please

the answer would be c

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2