Un atleta de masa m = 70 kg y altura L = 1.80 m está haciendo flexiones sobre una superficie horizontal sin rozamiento. Su centro de masas se encuentra a una distancia LCM = 1.20 m del punto de apoyo O (ver figura). La distancia entre las manos y el punto B es LAB = 0.2 m. Calcular las reacciones del suelo sobre las manos y los pies del atleta cuando su cuerpo forma un ángulo θ = 20° con la horizontal.

step-by-step explanation:

first you will have to convert 3/4 into a decimal

which is .75

then you will have to divide 1.5 by .75

which is 2

no you have to divide

there are infinitely many solutions.

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west

i think the answer is c sorry if it's wrong