Mass m = 68 kg
center of gravity from his palms x = 0.7 m
center of gravity from his feet x ' = 1 m
forces exerted by the floor on his palms and feet are F and F ' respectively.
with respect to palms :---------------------
( F*0 ) - (W * x ) + [ F ' * (x+x') ] = 0
-mg*0.7 + F ' * 1.7 = 0 where W = weight = mg
F ' * 1.7 = mg * 0.7
F ' = mg * 0.7 / 1.7
= 68 *9.8 * ( 0.7 / 1.7 )
= 274.4 N
with respect to feet :--------------------
( F ' * 0 ) -( W* x ' ) + [F * ( x + x') ] = 0
-mg*1 + [ F * 1.7 ]= 0
F = mg / 1.7
= 392 N
Henry Moseley, its a generator which uses energy to generate electricity
Answer:
Explanation:
A )
The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³
B )
Tension in the ball will be equal to net force acting on the ball
Net force on the ball = buoyant force - weight .
4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )
= 40.65 x 10⁻⁶ N .
C )Tension in the 3 rd ball will be equal to net force acting on the ball
Net force on the ball = weight - buoyant force
= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000 )
= 121.6 x 10⁻⁶ N .
