PLEASE HELP: Why are shooting stars bright?
1 answer:
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Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = / vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
Answer:
a. The volume V₁ and V₂
b. The case that involves the greatest momentum change = Case B
c. The case that involves the greatest impulse = Case B
d. b. The case that involves the greatest force = Case B
Explanation:
Here we have
Case A: V₁ = 150-mL, v₁ = 8 m/s
Case B: V₂ = 600-mL, v₁ = 8 m/s
a. The variable that is different for the two cases is the volume V₁ and V₂
b. The momentum change is by the following relation;
ΔM₁ = Mass, m × Δv₁
The mass of the balloon are;
Δv₁ = Change in velocity = Final velocity - Initial velocity
Mass = Density × Volume
Density of water = 0.997 g/mL
Case A, mass = 150 × 0.997 = 149.55 g
Case B, mass = 600 × 0.997 = 598.2 g
The momentum change is;
Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s
Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s
Therefore Case B has the greatest momentum change
The case that has the gretest momentum change = Case B
c. The momentum change = impulse therefore Case B involves the greatest impulse
d. Here we have;
Impulse = Momentum change = × Δt = mΔV
∴ = m·ΔV/Δt
∴ For Case A = 149.55×8/Δt = 1196.4/Δt N
For Case B = 598.2×8/Δt = 4785.6/Δt
Where Δt is the same for Case A and Case B, for Case B >>
for Case B
Therefore, Case B involves the greatest force.