Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
.................2
put here value and we get
<h3>v = 0.155 m </h3>
so
Magnification will be here as
m =
m =
<h3>m = 1</h3>
Kinetic Energy is calculated by KE= 1/2mv2
M g = 5.00 (9.81) = 49.05
sin 21 = x149.05
x = 17.6N
Where are the statements? You forgot to attach them lol
Answer:
a. 1715 N b. 2401 N
Explanation:
Let F = force due to calf muscle, F' = force due to tibia and N = force due to ground = weight of man = mg where m = mass of man = 70 kg and g = acceleration due to gravity = 9.8 m/s².
a. Magnitude of the forces exerted on the foot by the muscle
Since the force due to the calf muscle is 6.0 cm behind the ankle joint and the normal force due to the ground is 15.0 cm in front of the ankle joint and the force due to the tibia is at the ankle joint, taking moments about the ankle joint,
F × 6 cm + F' × 0 cm = N × 15 cm
6F = 15N = 15mg
F = 15mg/6
= 15 × 70 kg × 9.8 m/s²/6
= 1715 N
b. Magnitude of the forces exerted on the foot by the tibia
Taking moments about the calf muscle force, we have
F × 0 cm + F' × 6 cm = N × (15 cm + 6 cm)
6F' = 21N = 21mg
F' = 21mg/6
= 21 × 70 kg × 9.8 m/s²/6
= 2401 N