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:
Increasing
Explanation:
It’s increasing because your are adding more weight
Answer:
The frequency of the piano string is either 1053 HZ or 1059 HZ.
Explanation:
Here we know that frequency of beats is equal to the difference between the frequencies between two waves .
Given that frequency of tuning fork is 1056 HZ .
Let the frequency of the piano be ' f ' .
Given that number of beats = 3.
We know that | 1056 - f | = 3 ;
⇒ 1056- f = ±3,
Upon solving this we get
f = 1056-3 and 1056 + 3
⇒ f = 1053 or 1059 .
A mode that uses sliding friction would be sledding. :)
Answer:68.15m/s
Explanation:
<u><em>Given: </em></u>
v₁=15m/s
a=6.5m/s²
v₁=?
x=340m
<u><em>Formula:</em></u>
v₁²=v₁²+2a (x)
<u>Set up:</u>
=
<h2><u><em>
Solution:</em></u></h2><h2><u><em>
68.15m/s</em></u></h2>
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