The time spent in the air by the ball at the given momentum is 6.43 s.
The given parameters;
- <em>momentum of the ball, P = 0.9 kgm/s</em>
- <em>weight of the ball, W = 0.14 N</em>
The impulse experienced by the ball is calculated as follows;
where;
is impulse
is change in momentum
The time of motion of the ball is calculated as follows;
Thus, the time spent in the air by the ball at the given momentum is 6.43 s.
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The angular momentum of an object is equal to the product of its moment of inertia and angular velocity.
L = Iω
I = 1/2 MR²
I = 1/2 x 13 x (0.2)
I = 1.3
ω = 2π/t
ω = 2π/0.3
ω = 20.9
L = 1.3 x 20.9
= 27.2 kgm²/s
Answer:
ill get back to this question once i get the answer
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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Answer:
Θ=0.01525 rad
or
Θ=0.87°
Explanation:
Given data
wavelength λ=2.5 µm =2.5×10⁻⁶m
Diameter d=0.20 mm =0.20×10⁻³m
To find
Angle Θ in radians and degree
Solution
Circular apertures have first dark fringe at
Θ=(1.22λ)/d
Substitute the given values
So
Θ=[1.22(2.5×10⁻⁶m)]/0.20×10⁻³m
Θ=0.01525 rad
or
Θ=0.87°
