A uniform cubical crate is 0.740 m on each side and weighs 600 N. It rests on the floor with one edge against a very small, fixe
1 answer:
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To solve this problem we will start by defining the relationship of the pitot tube and the value indicated in the pressure. We will take advantage of this relationship to obtain a ratio between the defined pressures and proceed to calculate the total and static pressure of the system through the correlation of pressure in isotropic systems. Finally we will perform this same procedure for pressure. Pitot tube measure the total pressure minus static pressure, then
Express the total pressure and static pressure ratio
Hence the static pressure is 400.4kPa
Express stagnation to static temperature ratio,
Therefore the stagnation temperature is 316.224K
Answer:
v₂ = 0.0529 [m/s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, where momentum is conserved before or after launching the baseball. Momentum is defined as the product of mass by Velocity.
where:
P = lineal momentum [kg*m/s]
m = mass = 0.1 [kg]
v = velocity of the ball = 45 [m/s]
Now when the catcher receives the ball it will carry the same momentum as when it was thrown.