Now I can actually edit my answer directly: I'm fairly sure I've got this wrong, and my mind has gone blank for how to do it, if someone could delete this that would be great and I'll think about it and see if I can figure it out!
Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
Answer:
b. they get blown in from colder or warmer areas.
Answer:
Re = 1 10⁴
Explanation:
Reynolds number is
Re = ρ v D /μ
The units of each term are
ρ = [kg / m³]
v = [m / s]
D = [m]
μ = [Pa s]
The pressure
Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]
μ = [Pa s] = [kg / m s²] [s] = [kg / m s]
We substitute the units in the equation
Re = [kg / m³] [m / s] [m] / [kg / m s]
Re = [kg / m s] / [m s / kg]
RE = [ ]
Reynolds number is a scalar
Let's evaluate for the given point
Where the data for methane are:
viscosity μ = 11.2 10⁻⁶ Pa s
the density ρ = 0.656 kg / m³
D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m
Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶
Re = 1.19 10⁴
I don't actually understand what your question is, but I'll dance around the subject
for a while, and hope that you get something out of it.
-- The effect of gravity is: There's a <em>pair</em> of forces, <em>in both directions</em>, between
every two masses.
-- The strength of the force depends on the <em>product</em> of the masses, so it doesn't matter whether there's a big one and a small one, or whether they're nearly equal.
It's the product that counts. Bigger product ==> stronger force, in direct proportion.
-- The strength of the forces also depends on the distance between the objects' centers. More distance => weaker force. Actually, (more distance)² ==> weaker force.
-- The forces are <em>equal in both directions</em>. Your weight on Earth is exactly equal to
the Earth's weight on you. You can prove that. Turn your bathroom scale face down
and stand on it. Now it's measuring the force that attracts the Earth toward you.
If you put a little mirror down under the numbers, you'll see that it's the same as
the force that attracts you toward the Earth when the scale is right-side-up.
-- When you (or a ball) are up on the roof and step off, the force of gravity that pulls
you (or the ball) toward the Earth causes you (or the ball) to accelerate (fall) toward the Earth.
Also, the force that attracts the Earth toward you (or the ball) causes the Earth to accelerate (fall) toward you (or the ball).
The forces are equal. But since the Earth has more mass than you have, you accelerate toward the Earth faster than the Earth accelerates toward you.
-- This works exactly the same for every pair of masses in the universe. Gravity
is everywhere. You can't turn it off, and you can't shield anything from it.
-- Sometimes you'll hear about some mysterious way to "defy gravity". It's not possible to 'defy' gravity, but since we know that it's there, we can work with it.
If we want to move something in the opposite direction from where gravity is pulling it, all we need to do is provide a force in that direction that's stronger than the force of gravity.
I know that sounds complicated, so here are a few examples of how we do it:
-- use arm-muscle force to pick a book UP off the table
-- use leg-muscle force to move your whole body UP the stairs
-- use buoyant force to LIFT a helium balloon or a hot-air balloon
-- use the force of air resistance to LIFT an airplane.
-- The weight of 1 kilogram of mass on or near the Earth is 9.8 newtons. (That's
about 2.205 pounds). The same kilogram of mass has different weights on other planets. Wherever it is, we only know one of the masses ... the kilogram. In order
to figure out what it weighs there, we need to know the mass of the planet, and
the distance between the kilogram and the center of the planet.
I hope I told you something that you were actually looking for.
