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Harrizon [31]
2 years ago
7

Cellular respiration occurs in the body 24 hours per day. During exercise, the rate of cellular respiration increases. Why does

increased cellular respiration cause people to breathe faster?
More oxygen is needed to produce energy, and more carbon dioxide waste must be removed from the body.


More nutrients are needed to produce more energy, and more oxygen must be removed from the body.


More heat is generated by the body, and breathing cools the body quickly.


More water is lost from the body, and breathing hydrates the body.
Physics
2 answers:
lions [1.4K]2 years ago
8 0

Answer: more oxygen is needed to produce energy

Colt1911 [192]2 years ago
6 0

Answer: I think is C

Explanation:

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4 0
2 years ago
(5, 3) and (7, 3) are two coordinate points for a single object on a position-versus-time graph. Assume time is measured in seco
Maru [420]
Since the y axis stayed consistent, we can assume it did not move at all.
(So your answer would be A)
6 0
3 years ago
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A car traveling at 28 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?
Maru [420]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) minus (speed at the beginning)

change in speed = (zero) minus (28 m/s) = -28 m/s

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Acceleration = -2.15 m/s²

5 0
3 years ago
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timama [110]

Answer:

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3 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
3 years ago
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