For a body falling freely from rest (disregarding air resistance), the distance the body falls varies directly as the square o
f the time. If an object is dropped from the top of a tower 490 ft high and hits the ground in 7 sec, how far did it fall in the first 5 sec?
2 answers:
You might be interested in
Answer:
it is separated by 80 cm distance
Explanation:
As per Coulombs law we know that force between two point charges is given by
here we know that
force between two charges is given as
now we have
so it is separated by 80 cm distance
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²