How does Einstein's famous equation, E = mc2, relate to the production of the Sun's energy? Be sure to include what each factor
1 answer:
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Answer:
<em>a) 17.05 mph</em>
<em>b) 54.7° northeast direction</em>
<em>c) 10.71 mph</em>
<em>The direction is -22.58° relative to the east.</em>
<em></em>
<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>
Explanation:
The question is a little confusing but, I guess the correct question should be;
You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.
a) What is your airspeed?
b) What angle (direction) are you flying?
c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?
NB: The difference in the question and my suggestion is highlighted boldly.
Your speed = 14 mph
direction is 45° northeast
Th wind speed = 4 mph
direction is north
We resolve the your speed and the wind speed into the horizontal and vertical components
For vertical the component component
= 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph
For the horizontal speed component
= 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph
Resultant speed =
==> = <em>17.05 mph This is your airspeed</em>
b) To get your direction, we use
tan ∅ = /
tan ∅ = 13.89/9.89 = 1.413
∅ = (1.413) = <em>54.7° northeast direction</em>
c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .
In this case,
= 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph
Resultant speed =
==> = <em>10.71 mph This is your airspeed</em>
Your direction will be,
tan ∅ = /
tan ∅ = -4.11/9.89 = -0.416
∅ = (-0.416) =<em> -22.58° this is the angle you'll travel relative to the east.</em>
<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>
Answer:
The ratio is
Explanation:
From the question we are told that
The radius of Phobos orbit is R_2 = 9380 km
The radius of Deimos orbit is
Generally from Kepler's third law
Here M is the mass of Mars which is constant
G is the gravitational constant
So we see that
=>
Here is the period of Deimos
and is the period of Phobos
So
=>
=>