The movement of fluid as a result of differential heating or convection. Earth convection currents refer to the motion of molten rock in the mantle as radioactive decay heats up magma, causing it to rise and driving the global scale flow of magma.
Answer:They come in different kinds, called elements, but each atom shares certain characteristics in common. All atoms have a dense central core called the atomic nucleus. Forming the nucleus are two kinds of particles: protons, which have a positive electrical charge, and neutrons, which have no charge
Explanation:
Answer:
The final velocity of the object after 2 seconds is 30 m/s
Explanation:
Given;
constant downward acceleration, a = 10 m/s²
initial velocity of the object falling down, v = 10 m/s
time of fall, t = 2 s
The final velocity of the object is given by;
v = u + at
where;
v is the final velocity
v = 10 + (10)(2)
v = 10 + 20
v = 30 m/s
Therefore, the final velocity of the object after 2 seconds is 30 m/s
<span>¿Qué estás pidiendo en esta situación. Hay entornos de diferencia de los animales. Algunos viven en entornos de tundra, otra veraniega en vivo, ambientes cálidos.</span>
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