For #5 It's helpful to draw a free body diagram so you know which way the forces are acting on the block.
the weight mg is acting downwards, and you need to find the vertical and horizontal components of mg using sin and cosine. so do 15x9.8xsin40 which is the force. Assuming no friction, this is the only force acting on the block, as the forces on the vertical plane cancel out i.e the normal force and weight of the block.
after, just do F=ma And since you know F and m, solve for a.
Answer:
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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
Answer:
The weight of the body, W = 793.8 m/s²
Explanation:
Given,
The volume of the body, v = 45,000 cm³
The density of the body, ρ = 1.8 g/cm³
The mass of the body is given by the formula,
m = ρ x v
= 1.8 g/cm³ x 45,000 cm³
= 81,000 g
Hence, the mass of the body is m = 81 kg
The weight of the body,
W = m x g
= 81 kg x 9.8 m/s²
= 793.8 m/s²
Hence, the weight of the body, W = 793.8 m/s²
