In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocit
1 answer:
You might be interested in
Answer:
v’= 279.66 m / s
Explanation:
We work this exercise using the conservation of the moment. For this we define the system formed by the two blocks, therefore the forces during the collision are internal of the action and reaction type.
Initial instant. Before the crash
p₀ = m v₀ + 0
Final moment. After the crash
p_f = m v + M v ’
how the tidal wave is preserved
p₀ = p_f
m v₀ = m v + M v ’
v =
let's calculate
v ’=
v ’=
v ’= 279.66 m / s
To solve the problem we will first start considering the Pressure given the hydrostatic definition of the product between the density, the gravity and the depth. We will define the area where the liquid acts and later we will use the definition of the force as a product between the pressure and the area to calculate the force given in the two depths. The gauge pressure at the depth x will be
This pressure acts on the strip of area
The force acting on that strip is given by,
To evaluate the force, we will then consider the integral of the pressure as a function of the Area, or the integral of the previously found terms.
Evaluating at the initial depth of 1.8m and the final depth of 4.4 we have then that,
Therefore the Net force will be