Answer:
the drag coefficient = 0.2703126
Explanation:
see the attached file
You apparently skipped some words and didn't copy them into the question. Fortunately, those words were part of the "red herring" section ... the part that only distracts you, and is completely unnecessary in order to answer the question.
The actual question here is: "What fraction of an hour is 7.2 seconds ?"
(7.2 sec) x (1 hour/3,600 sec) = 7.2 / 3600 = <em>0.002 hour </em>.
Answer:
V = 308.1 m/s θ = -64º
Explanation:
a and b) We will use the projectile launch equations where we are asked to find the speed when the tank reaches the ground
Let's start by breaking down the speed
Vox = Vo cos θ
Voy = Vo sint θ
Vox = 140 cos 15 = 135.2 m / s
Voy = 140 sin 15 = 36.2 m / s
Let's look for vertical speed when it hits the ground
Vy² = Voy² - 2gy
Vy = √(36.2² - 2 9.8 3.98 103) = √ (1310-78008)
Vy = 276.9 m / s
We have both components.
V² = Vx² + Vy²
V = √ (135.2² + 276.9²)
V = 308.1 m / s
tan θ = Vy / Vx
tan θ = -276.9 / 135.2 = 2,048
θ = -64º
The negative sign means that it is measured from the x-axis clockwise
c and d) We repeat the same calculation for tank B, the only difference is the angle T = -15º
Vox = 140 cos (-15)
Voy = 140 sin (-15)
Vox = 135.2 m / s
Voy = 140 sin (-15)
Voy = -36.2 m / s
We calculate the vertical speed
Vy² = Voy² - 2 g Y
Vy = √ ((-36.2)² - 2 9.8 3980)
Vy = 276.9 m / s
V = √(135.3²2 + 276.9²)
V = 308.1 m/s
tan θ = -276.9 / 135.2
θ = -64
You can see that the speeds and angles are the same in both cases, the difference between these two situations is in the horizontal distance that runs each story
A delightful problem !
I'm pretty sure that what we need here is the speeds, not the velocities,
and that's the way I'm going to do it.
Regular speed is (distance covered) divided by (time to cover the distance) .
Angular speed is very much the same.
It's
(angle turned) divided by (time to turn the angle) .
<u>Earth's orbit around the sun</u>:
..... Once per year.
..... Roughly 360° in 365 days ..... <em>almost exactly 1° per day</em>.
Let's see what it is more accurately:
(360°) / (<span>365.25636<span> days) = 0.985609° per day.
============================================
<u>Earth's rotation on its axis</u>:
..... Once per "day".
..... Roughly 360° in 24 hours ..... <em>almost exactly 15° per hour</em>.
This one is slightly trickier to do more accurately, because a day is
not necessarily 24 hours. It depends on what you call 1 day.
-- If you say the day is the period of time between when the sun is
highest in the sky, then that averages out to 24 hours in the course
of a year.
-- If you say that the day is the period of time it takes for a star
to reach the same point in the sky tomorrow night, then that's </span></span>
23 hours, 56 minutes, 4.09 seconds .
Using this to calculate the angular speed of rotation, you get
(360°) / (23h 56m 4.09s) = 15.041° per hour
Answer:
Time domain and frequency domain