Question

a horizontal 4-cm-diameter water jet with a velocity of 18 m/s. He impinges the jet normally upon a vertical plate of mass 750 kg. The plate rides on a nearly frictionless track and is initially stationary. When the jet strikes the plate, the plate begins to move in the direction of the jet. The water always splatters in the plane of the retreating plate. Determine (a)the acceleration of the plate when the jet first strikes it (time

Answer 1

Answer:

0.5429 m/s^2

Explanation:

velocity of waterjet = 18 m/s

diameter of water jet ( d ) = 4 cm = 0.04 m

mass of vertical plate(m) = 750 kg

Determine the acceleration of plate when the jet first strikes ( i.e. t = 0 )

first we will determine the impact force

F = β*A*V^2 ----- ( 1 )

where ; β = 1000 kg/m^3  ,  A = π/4 * d^2 , V = 18 m/s

input values into equation 1

F = 407.15 N

finally determine the acceleration at t = 0

F = m*a

a = F / m =  407.15 / 750 = 0.5429 m/s^2