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sineoko [7]
3 years ago
14

What objects can be seen from earth because they producde there own light? 

Physics
1 answer:
Lena [83]3 years ago
7 0

- neon signs
- fireworks
- rockets during the ascent
- meteors
- fireflies
- stars
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Explain how oxygen is obtained by most complex organisms and then used during cellular respiration.(Please help quick)​
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Discuss the importance of oxygen to all living organisms, and describe how three different organisms—human, flatworm, and fern—obtain the oxygen they need.

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(Question 1):Oxygen is the key for generating energy in cellular respiration. The cellular respiration is a complex biological process that breaks down sugars, fat and proteins and transforms these materials into energy for the functioning of the organism. Oxygen is needed at the end of this process, when electrons liberated in the respiration are transported through the inner membranes of the cells, and the oxygen "attracts" these electrons (like bees to honey) and makes possible the production of great quantities of chemical energy in the said membranes.(Question 2):Earth's natural bodies of water have a low oxygen concentration--much lower than the level of oxygen in air at the earth's surface. Smaller organisms can obtain sufficient oxygen through the skin (e.g. flatworms), but larger organisms require special structures to collect enough oxygen to sustain life. This oxygen comes from molecules of oxygen gas dissolved in the water. The oxygen atom present in the water molecule is n…

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What type of motion occurs when an object spins around an axis without altering its linear position?
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7 0
3 years ago
Read 2 more answers
During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What m
Tanzania [10]

Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.

Explanation:

Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.

Now, according to the law of conservation of energy the formula used is as follows.

mgh = \frac{1}{2} mv^{2}_{y}\\v_{y} = \sqrt{2gh}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.2}\\= 4.85 m/s

As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.

v = \sqrt{v^{2}_{x} + v^{2}_{y}}\\= \sqrt{(6.8)^{2} + (4.85 m/s)^{2}}\\= 11.65 m/s

Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.

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