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3 years ago

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Mary wishes to lose 5 pounds before her vacation in approximately 5 weeks. Her average consumption is 2,100 kilocalories per day

. How many kilocalories should she consume daily to begin to lose this weight, assuming her activity pattern remains unchanged?

1 answer:

nordsb [41]3 years ago

7 0

Answer:

1,600 to 1,700 kilocalories per day

Explanation:

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Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu

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Answer:

(a) Determine the astronaut's orbital speed

The velocity of the astronaut is 1436m/s.

(b) Determine the period of the orbit.

The period of the orbit is 10413 seconds.

Explanation:

<em>(a) Determine the astronaut's orbital speed</em>

The orbital speed of astronaut can be found by means of the Universal law of gravity:

$F = G\frac{M \cdot m}{r^{2}}$ (1)

Then, replacing Newton's second law in equation 3 it is gotten:

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (2)

However, a is the centripetal acceleration since the astronaut describes a circular motion around the Moon:

$a = \frac{v^{2}}{r}$ (3)

Replacing equation 3 in equation 2 it is gotten:

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (4)

Where v is the orbital speed, G is the gravitational constant, M is the mass of the Moon, and r is the orbital radius.

Notice that the orbital radius will be given by the sum of the radius of the Moon and the height of the astronaut above the surface.

But it is necessary to express the height of the astronaut above the surface in units of meters before it can be used.

$r = 680km \cdot \frac{1000m}{1km}$ ⇒ $680000m$

$r = 1.70x10^{6}m+680000m$

$r = 2380000m$

However it is neccesary to find the mass of Moon in order to use equation 4.

Then Newton's second law ( $F = ma$ ) will be replaced in equation (1):

$ma = G\frac{Mm}{r^{2}}$

Then, M will be isolated

$M = \frac{r^{2}a}{G}$ (5)

Where r is the orbital radius of the astronaut and a is the acceleration due to gravity.

$M = \frac{(2380000m)^{2}(0.867 m/s^{2})}{6.67x10^{-11}N.m^{2}/kg^{2}}$ (5)

$M = 7.36x10^{22}Kg$

$v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(7.36x10^{22}Kg)}{2380000m}}$

$v = 1436m/s$

Hence, the velocity of the astronaut is 1436m/s.

<em>(b) Determine the period of the orbit. </em>

The period of the orbit can be determined by the next equation:

$v = \frac{2\pi r}{T}$ (6)

Where v is the orbital velocity, r is the orbital radius and T is the period of the orbit.

Then, T can be isolated from equation 6.

$T = \frac{2\pi r}{v}$ (7)

$T = \frac{2\pi (2380000m)}{1436m/s}$

$T = 10413s$

Hence, the period of the orbit is 10413 seconds.

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3 years ago