Ask a New Question

Question

A 78-kg parachutist, in the first few seconds
of her jump, accelerates at a rate of 7.4 m/s2
[down]. Calculate the force she experiences
due to air resistance.
12 years ago

Answers

Elena
ma=mg-F(res)
F(res) = m(g-a) = 78(9.8-7.4) = 187.2 N
12 years ago

Related Questions

A parachutist bails out and freely falls 50m. Then the parachute opens and thereafter she decelerate... A parachutist bails out and freely falls 64.7 m. Then the parachute opens and thereafter she decele... A parachutist bails out and freely falls 59.7 m. Then the parachute opens and thereafter she deceler... a parachutist falling at the rate of 15 ft./sec. drops a stone. if he was 904 ft. above the ground w... A​ parachutist's elevation changes by −66 ft in 6 seconds. What is the change in the​ parachutist's... The velocity of a parachutist as a function of time is given by v = vf + (v0 - vf)e-t/2.5, where t =... After a 69.5 kg parachutist opens her parachute, she experiences a constant drag force of 307 N. Wha... A parachutist jumped from and airplane at an elevation of 11,880 feet above ground. In the first 5 s... A parachutist descends at 36 feet in 3 seconds. Express the rate of the parachutist’s change in h...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use