A basketball player is running at a constant speed of 2.5 m/s when he tosses a basketball upward with a speed of 6.0 m/s. How fa
1 answer:
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Response:
a) , b)
Clarification:
a) The absolute pressure at a depth of 27.5 meters is:
b) The force applied by the water is:
Answer:
5.7 m
Explanation:
AD = length of the ladder = L = 8 m
AB = the position of the ladder's center of mass = (0.5) L = (0.5) 8 = 4 m
AC = distance of the climber from the bottom of the ladder = x
W = weight of the ladder = 240 N
= weight of the climber = 710 N
F = force exerted by the wall on the ladder
N = normal force acting on the ladder from the ground =?
By applying force equilibrium in the vertical direction
N = + W
N = 710 + 240
N = 950 N
μ = Coefficient of static friction = 0.55
f = static friction force on the ladder
Static friction force can be expressed as
f = μ N
f = (0.55) (950)
f = 522.5 N
The equation for force along the horizontal axis reads
F = f
F = 522.5 N
using torque equilibrium around point A
F Sin50 (AD) = W Cos50 (AB) + ( Cos50 (AC))
(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)
x = 5.7 m
