B, C. Also literally a quick search yielded these results, roughly half the time to type this out.
Answer:
By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.
here weight of the child =21kgx9.8m/s2 = 205.8N
the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.
torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.
net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.
b) Ta = 2.5x151 = 377.5N-m
Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.
c)Ta = 2x151 = 302N-m
Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.
Convert the given in SI units.
(44 ft/sec)(1 m/ 3.28 ft) = 13.41 m/sec
The distance traveled and the initial velocity can be related through the equation,
d = (Vf)² - (Vi)²/ 2a
where d is the distance, Vf is the final velocity, Vi is the initial velocity, a is the acceleration due to gravity. Substituting the known values from the given above,
d = ((0 m/s)² - (13.41 m/s)²)/ 2(-9.8 m/s²)
The value of d from the equation,
d = 9.17 meters
Convert this to feet,
d = (9.17 m)(3.28 ft / 1 m) = 30 ft
Answer: 30 ft
Expand each vector into their component forms:
Similarly,
Then assuming the resultant vector 
is the sum of these three vectors, we have
and so has magnitude
and direction 
such that
