When computing using dimensional analysis:
1 answer:
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Answer:
v₀ = √(2gH/(sin²θ)) = (sin θ)√(2gH)
v₀ = √(gR/(sin2θ))
Explanation:
An image of the artillery officer, the hill and path of motionof the projectile is attached to this solution.
Given, R, H, g and θ (theta)
Using the equations of motion, we can get the initial velocity v₀
First of, we need to resolve this motion into the vertical and horizontal axis.
The horizontal component of the initial velocity, v₀ₓ = v₀ cos θ
Vertical component of the initial velocity, v₀ᵧ = v₀ sin θ
When the projectile reaches maximum height, Velocity at max height, vₕ = 0m/s
From equations of motion,
vₕ = v₀ᵧ - gt
0 = v₀ sinθ - gt
t = v₀ sinθ/g
This is the time taken to reach maximum height. The time take to comolete the toyal flight, T = 2t = (2v₀ sinθ)/g
The maximum height to be reached, H can be calculated from the equations of motion too
H = vₕt - 0.5gt² = 0 - 0.5g((v₀ sinθ)/g)²
H = (0.5g v₀² sin²θ)/g²
H = (v₀² sin²θ)/2g
The range, or horizontal distance to be covered by the projectile, R, will be calculated using the horizontal component of the initial Velocity, v₀ₓ = v₀ cos θ, this horizontal velocity is constant all through the motion, so, no acceleration in the horizontal direction.
R = v₀ₓT = (v₀ cos θ)((2v₀ sinθ)/g)
R = (v₀²(2cosθsinθ)/g)
2cosθsinθ = sin2θ
R = v₀²(sin2θ)/g
So, writing v₀ in terms of all the other parameters,
v₀ = √(2gH/(sin²θ)) = (sinθ)√(2gH
v₀ = √(gR/(sin2θ))
Answer:
236.9g
Explanation:
Given parameters:
Volume of gas = 81.3L
Pressure of gas = 204kPa
temperature of gas = 95°C
Unknown:
Mass of carbondioxide gas = ?
Solution:
To solve this problem, the ideal gas law will be well suited. The ideal gas law is a fusion of Boyle's law, Charles's law and Avogadro's law.
Mathematically, it is expressed as;
PV = nRT
the unknown here is n which is the number of moles;
P is the pressure, V is the volume, R is the gas constant and T is the temperature.
convert pressure into atm
101.325KPa = 1atm
204 kPa = = 2atm
Convert temperature to Kelvin; 95 + 273 = 368K
2 x 81.3 = n x 0.082 x 368
n = = 5.38moles
Since the unknown is mass;
Mass = number of moles x molar mass
Molar mass of carbon dioxide = 12 + 2(16) = 44g/mol
Mass = 5.38 x 44 = 236.9g