Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.
Ek = mv^2 / 2 — multiply both sides by 2
2Ek = mv^2 — divide both sides by m
2Ek / m = V^2 — switch sides
V^2 = 2Ek / m — plug in values
V^2 = 2*30J / 34kg
V^2 = 60J/34kg
V^2 = 1.76 m/s — sqrt of both sides
V = sqrt(1.76)
V = 1.32m/s (roughly)
This imply that the force applied by the gravity of the earth on the person, is equal to the force applied by the person on the earth.
According to the Newton's third law, for every force there is an equal reaction force, in the case of person standing on the earth, the gravity of the earth pulls the person by the same magnitude as the force applied by the person on the earth (which is equal to the weight of the person)..
Answer:
see from this analysis, the apparent weight of the body is lower due to the push created by the air brujuleas
Explanation:
We will propose this exercise using Archimedes' principle, which establishes that the thrust on a body is equal to the volume of the desalted liquid.
B = ρ g V
The weight of a submerged body is the net force between the weight and the thrust
F_net = W - B
we can write the weight as a function of the density
ρ_body = m / V
m = ρ_body V
W = mg
W = ρ _body g V
we substitute
F_net= ( ρ_body - ρ _fluid) g V
In general this force is directed downwards, we can call this value the apparent weight of the body. This is the weight of the submerged body.
W_aparente = ( ρ_body - ρ _fluid) g V
If some air bubbles formed in this body, the net force of these bubbles is
F_net ’= #_bubbles ( ρ_fluido - ρ_air) g V’
this force is directed upwards
whereby the measured force is
F = W_aparente - F_air
As we can see from this analysis, the apparent weight of the body is lower due to the push created by the air brujuleas
