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3 years ago

How does the resting energy expenditure change as a person ages

Physics

1 answer:

Lady bird [3.3K]3 years ago

6 0

Answer:

Resting energy expenditure (REE) decreases from young to old age by 1% to 2% per decade [1]. This is partly explained by age-related decreases in fat free mass (FFM) [2]. FFM accounts for 50%–70% of the variance in REE [3,4,5].Explanation:

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Amy counts the wave crest traveling down a stretched string five wave crests pass Amy in two seconds what is the frequency of th

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<span>One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle, theta, with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.390. A. what is the maximum value...</span>

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4 years ago

During the execution of a play, a football player carries the ball for a distance of 33 m in the direction 58° north of east. To

podryga [215]

Answer:28 m

Explanation:

Given

Direction is $58^{\circ}$ North of east i.e. $58 ^{\circ}$ with x axis

Also ball moved by 33 m

therefore its east component is 33cos58=17.48 m

Northward component $=33sin58=27.98 m\approx 28 m$

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3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

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3 years ago

A snowboarder is sliding back and forth on a half pipe at one point she leaves the top of the half pipe and slides to the other

r-ruslan [8.4K]

Answer:

he kinetic energy increases on the descent, being maximum at the lowest point of the trajectory.

Explanation:

In these semicircular sections the skaters slide from one side to the other, in the downward path their kinetic energy increases and their potential energy decreases; When it leaves the ramp and is in the air, the kinetic energy decreases rapidly, up to the point of maximum height where the kinetic energy is zero.

Consequently, the kinetic energy increases on the descent, being maximum at the lowest point of the trajectory.

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3 years ago

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