Answer:
No he should not attempt the pass
Explanation:
Let t be the time it takes for the car to pass the truck. The driver should ONLY attempt to pass when the distance covered by himself plus the distance covered by the oncoming car is less than or equal 400 m (a near miss)
At acceleration of 1m/s2 and a clear distance of 10 + 20 + 10 = 40 m, we can use the following equation of motion to estimate the time t in seconds
Within this time frame, the first car would have traveled a total distance of the clear distance (40m) plus the distance run by the truck, which is
8.94 * 25 = 223.6m
So the total distance traveled by the first car is 223.6 + 40 = 263.6m
The distance traveled by the 2nd car within 8.94 s at rate of 25m/s is
8.94 * 25 = 223.6 m
So the total distance covered by both cars within this time frame
223.6 + 263.6 = 487.2m > 400 m
So no, he should not attempt the pass as we will not clear it in time.
Answer:
Intensity
Explanation:
The intensity of a sound wave is equal to the ratio between to the power emitted by the source divided by the area of the spherical surface through which the wave propagates:
where
P is the power
is the area of the spherical surface
r is the distance from the source
As we see from the formula, the intensity is inversely proportional to the square of the distance from the source:
so, intensity is the correct answer.
Sound waves will either be reflected (echo effect) or absorbed (dissipated) depending upon the material make-up of the barrier
Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s
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