Absolute zero is A. the temperature at which the motion of particles theoretically ceases. B. defined as 0°C. C. the triple poin
2 answers:
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<em>What volume do 5 moles of a gas occupy at 28 ° C and 3 atm of pressure?</em>
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<h3>Further explanation</h3>In general, the gas equation can be written
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
n= 5 moles
T=28 +273=301 K
P=3 atm
The volume of the gas :
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))
