V = 340 m/s
f = 256 Hz
lambda (wavelength)
v = f*lambda
340 = 256 * lambda
340/256 = lambda
lambda = 1.328 m
It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.
To find the answer, we need to know about the third law of Kepler.
<h3>What's the Kepler's third law?</h3>
- It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
- Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
- The time period of geosynchronous orbit is 24 hours or 1440 minutes.
- As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
- If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
- a1= (T1/T2)⅔×a2
= (1440/90)⅔×6780
= 43,090 km
- Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km
Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.
Learn more about the Kepler's third law here:
brainly.com/question/16705471
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Answer:
speed wind Vw = 54.04 km / h θ = 87.9º
Explanation:
We have a speed vector composition exercise
In the half hour the airplane has traveled X = 108 km to the west, but is located at coordinated 119 km west and 27 km south
Let's add the vectors in each coordinate axis
X axis (East-West)
-Xvion - Xw = -119
Xw = -Xavion + 119
Xw = 119 -108
Xwi = 1 km
Calculate the speed for time of t = 0.5 h
Vwx = Xw / t
Vwx= 1 /0.5
Vwx = - 2 km / h
Y Axis (North-South)
Y plane - Yi = -27
Y plane = 0
Yw = 27 km
Vwy = 27 /0.5
Vwy = 54 km / h
Let's use the Pythagorean theorem and trigonometry to compose the answer
Vw = √ (Vwx² + Vwy²)
Vw = R 2² + 54²
Vw = 54.04 km / h
tan θ = Vwy / Vwx
tan θ = 54/2 = 27
θ = Tan⁻¹ 1 27
θ = 87.9º
The speed direction is 87. 9th measure In the third quadrant of the X axis in the direction 90-87.9 = 2.1º west from the south
The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
And if we substitute t=2.7 s, we find the distance covered:
