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
Answer:
Acceleration is percieved, not constant velocity.
Explanation:
You are most aware when the vehicle is accelerating. At constant velocity you would not be aware of the motion. Only if the system is accelerated the dynamics must be solved considering a pseudo-force (of inertial origin) acting.
It's because of this that:
(A) False. The acceleration can be detected from the inside of a closed car.
(B) False. You would be aware of the motion, but not because humans can sense speed but acceleration.
(C) False. Constant velocity cannot be felt in a closed car.
(D) False. Again, you can't feel constant speed.
Il existe troi types de rayons produits lors de la désintégration des éléments radioactifs:
-- "particules alpha" . . . noyaux d'hélium, composés chacun de 2 protons et 2 neutrons
-- "rayons bêta" ou "particules bêta" . . . flux d'électrons
-- "rayons gamma" . . . rayonnement électromagnétique avec les longueurs d'onde les plus courtes connues et l'énergie la plus élevée
By definition we know that the force is the vector product of the vector of the current by the length with the magnetic field vector. The current in this case goes in a positive "Y" direction. If we assume that the magnetic field goes in the positive "K" direction, then the result will be in the positive "X" direction. Attached solution.
Answer:
<em>The second option has a lower power output. P=30 W</em>
Explanation:
<u>Mechanical Power
</u>
It is a physical magnitude that measures the rate a work W is done over time t.
Since W=F.d
The first option means the worker will lift the box by a distance of 1.2 meters in 3 seconds by applying 250 N of force. That produces a power of
The second option requires the worker applies 75 N of force and travel a distance of 4 meters for 10 seconds, thus the power is
The second option has a lower power output
