Which of these statements correctly describes a key difference between aerobic activity and anaerobic activity?
1 answer:
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Refer to the diagram shown below.
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)
After 5 minutes (300 seconds):
The man travels north by (2 ft/s)*(300 s) = 600 ft
The woman, located at q, 500 east of p, begins walking south at 4 ft/s.
The distance separating them is
d₁ = √(600² + 500²) = 781.025 ft
After 20 minutes:
The man has traveled for 20 minutes (1200 s).
The woman has traveled for 15 minutes (900 s).
The man has moved (2 ft/s)*(1200 s) = 2400 ft north of p.
The woman has moved (4 ft/s)*(900 s) = 3600 ft south of q.
The distance separating them is
d₂ = √(6000² + 500²) = 6020.8 ft
The separation from d₁ to d₂ occurs in 15 minutes (900s).
Therefore the rate of separation is
Rate = (d₂ - d₁ ft)/(900 s) = (6020.8 - 781.025)/900 = 5.822 ft/s
or
Rate = (5.822 ft/s)*(60 s/min) = 349.32 ft/min
Answer: 349.32 ft/min (or 5.82 ft/s)
Answer:
g(h) = g ( 1 - 2(h/R) )
<em>*At first order on h/R*</em>
Explanation:
Hi!
We can derive this expression for distances h small compared to the earth's radius R.
In order to do this, we must expand the newton's law of universal gravitation around r=R
Remember that this law is:
In the present case m1 will be the mass of the earth.
Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):
Therefore, we can see that
With a the acceleration due to the earth's mass.
Now, the taylor series is going to be (at first order in h/R):
a(R) is actually the constant acceleration at sea level
and
Therefore:
Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power: