A cylinder has a radius r and a length l. if the radius were to double and the length were to stay the same, by what factor woul
2 answers:
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Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Answer:
the charge per unit area on the plastic sheet is - 3.23 x 10⁻⁷ C/m²
Explanation:
given information:
styrofoam mass, m = 16 g = 0.016 kg
net charge, q = - 8.6 μC
to calculate the charge per unit area on the plastic sheet, we can use the following equation:
where
the force between the electric field
m = mass
g = gravitational force
where
q = charge
E = electric field
and
E = σ/2ε₀
where
ε₀ = permitivity
thus
mg = qσ/2ε₀
σ = (2mg ε₀)/q
= 2 (0.016) (9.8) (8.85 x 10⁻¹²)/( - 8.6 x 10⁻⁶)
= - 3.23 x 10⁻⁷ C/m²