<span>the answer would actually be student 1 :9.87 m/s2, not 9.78</span>
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
At the stagnation point, the velocity is zero.
The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Answer: 2.2 psi
Answer:
A. the internal energy stays the same
Explanation:
From the first law of thermodynamics, "energy can neither be created nor destroyed but can be transformed from one form to another.
Based on this first law of thermodynamic, the new internal energy of the gas is the same as the internal energy of the original system.
Therefore, when the partition separating the two halves of the box is removed and the system reaches equilibrium again, the internal energy stays the same.
Answer:
The range is maximum when the angle of projection is 45 degree.
Explanation:
The formula for the horizontal range of the projectile is given by
The range should be maximum if the value of Sin2θ is maximum.
The maximum value of Sin2θ is 1.
It means 2θ = 90
θ = 45
Thus, the range is maximum when the angle of projection is 45 degree.
If the angle of projection is 0 degree
R = 0
It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.
If the angle of projection is 30 degree.
R = 0.088u^2
If the angle of projection is 45 degree.
R = u^2 / g
