Answer:
a1
b2
d3
c5
d4
Explanation:
Hope this helps
Check if right by the way
Answer:
fr ’= ½ F
Explanation:
For this exercise we use the translational equilibrium equation, on the axis parallel to the wall
fr - W = 0
fr = W
for the adult man they indicate that the friction force is equal to F
F = M g
we write the equilibrium equation for the child
fr ’= w’
fr ’= m g
in the statement they tell us that the mass of the adult is 2 times the mass of the child
M = 2m
we substitute
fr ’= M / 2 g
fr ’= ½ Mg
we substitute
fr ’= ½ F
therefore the force of friction in the child is half of the friction in the adult
Answer:
0.68 kg-m²
Explanation:
F = Force applied by the muscle = 2615 N
r = effective perpendicular lever arm = 2.85 cm = 0.0285 m
α = Angular acceleration of the forearm = 110.0 rad/s²
I = moment of inertia of the boxer's forearm = ?
Torque is given as
τ = I α eq-1
Torque is also given as
τ = r F eq-2
using eq-1 and eq-2
r F = I α
(0.0285)(2615) = (110.0) I
I = 0.68 kg-m²