First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
The specific answer for that will be 71.38 kg
0.495 m/s
Explanation
the formula for the terminal velocity is given by:
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20%5Ctext%7Bwhere%7D%20%5C%5C%20%20%5Cend%7Bgathered%7D)
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient
then
Step 1
Let's find the mass
now, replace
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2%280.002kg%29%289.81%5Ctext%7B%20%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%7D%7B%282%5Ccdot10%5E3%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E3%7D%29%280.0001m%5E2%290.8%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.03924%5Cfrac%7B%5Coperatorname%7Bkg%7Dm%7D%7Bs%5E2%7D%7D%7B0.16%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E%7B%7D%7D%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B0.2452%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%7D%20%5C%5C%20v%3D0.495%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
hence, the answer is 0.495 m/s
Answer:
Mass released = 8.6 g
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
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Explanation:
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