I hope this woll work.Best of luck.Sorry of the answer ia wrong
Answer:
I believe it would be Direction B
Explanation:
After passing A, gravity would pull the ball downwards
Linear momentum (mass x speed) has to be conserved.
-- Momentum before the jump:
(boy's mass) x (boy's speed) = (25 kg) x (4.0 m/s) = 100 kg-m/s
(cart's mass) x (cart's speed) = (15 kg) x (zero) = zero
Total momentum before the jump: (100 kg-m/s) + (zero) = (100 kg-m/s)
-- Momentum after the jump:
(mass of boy+cart) x (speed of boy+cart) = (40 kg) x (speed)
-- Momentum after the jump = momentum before the jump
(40 kg) x (speed) = 100 kg-m/s
Divide each side by 40 kg:
Speed = (100 kg-m/s) / (40 kg)
<em>Speed = 2.5 m/s</em> (d)
Answer:
its terminal velocity is 19.70 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Explanation:
Firstly,
given that
m = 580g = 0.58kg
Area A = 0.11 * 0.22 = 0.0242m
g = 9.8
idensity constant p = 1.21 kg/m^3
the terminal velocity of the sphere Vt is ;
Vt = √ ( 2mg / pCA)
we substitute
Vt = √ ( (2*0.58*9.8) / (1.21*1*0.0242)
Vt = √ (11.368 / 0.029282)
Vt = √ ( 388.22)
Vt = 19.70 m/s
its terminal velocity is 19.70 m/s
What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
The Velocity of the person is;
V2 = √ 2ax
V2 = √ ( 2 * 9.8 * 4 )
V2 = √ (78.4)
V2 = 8.85 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s