Answer:
=
Explanation:
given data:
Pressure P = 3 * 10^ 5 Pa
speed v = 5 m / s
Area A = A
from the information given in equation final Area is 1/3 of initial area i.e.
A ' = A / 3
we know that density of water = 1000 kg / m^ 3
from continuity equation
Av = A ' v'
so we have
speed v ' = 3*A*v / A
v ' = 3*A*5/ A
v = 15 m / s
from bernoulli's equation we can calculate final pressure
Required pressure P ' = P + ( 1/ 2) \rho [ v^ {2} v'^{ 2}]
= ![P ' = P + ( 1/ 2) \rho_{water} [ v^ {2} - v'^{ 2}]](https://tex.z-dn.net/?f=P%20%27%20%3D%20P%20%2B%20%28%201%2F%202%29%20%5Crho_%7Bwater%7D%20%5B%20v%5E%20%7B2%7D%20-%20v%27%5E%7B%202%7D%5D)
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Answer:
R = 6.3456 10⁴ mile
Explanation:
For this exercise we will use Newton's second law where force is gravitational force
F = m a
The satellite is in a circular orbit therefore the acceleration is centripetal
a = v² / r
Where the distance is taken from the center of the Earth
G m M / r² = m v² / r
G M / r = v²
The speed module is constant, let's use the uniform motion relationships, with the length of the circle is
d = 2π r
v = d / t
The time for a full turn is called period (T)
Let's replace
G M / r = (2π r / T)²
r³ = G M T²²2 / 4π²
r = ∛ (G M T² / 4π²)
We have the magnitudes in several types of units
T = 88.59 h (3600 s / 1h) = 3.189 10⁵ s
Re = 6.37 10⁶ m
Let's calculate
r = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (3,189 10⁵)²/4π²)
r = ∛ (1.027487 10²⁴)
r = 1.0847 10⁸ m
This is the distance from the center of the Earth, the distance you want the surface is
R = r - Re
R = 108.47 10⁶ - 6.37 10⁶
R = 102.1 10⁶ m
Let's reduce to miles
R = 102.1 10⁶ m (1 mile / 1609 m)
R = 6.3456 10⁴ mile
Answer:
[See Below]
Explanation:
✦ Formula = 
✧ 
So 
grams is equal to 
kilograms.
~<em>Hope this helps Mate. If you need anything feel free to message me. </em>
