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Nimfa-mama [501]
3 years ago
14

Mary is campaigning for improved air quality in her city. She is very concerned about carbon and methane outputs, but not phosph

orous. Why is that? A. She thinks phosphorous makes a positive contribution to air quality. B. Phosphorous is most commonly found as phosphate rock. C. She doesn’t know a lot about phosphorous. D. She doesn’t consider phosphorous to be a greenhouse gas.
Biology
1 answer:
Galina-37 [17]3 years ago
4 0

The answer is; D

Carbon dioxide and methane are potent greenhouse gases. They cause heating of the lower atmosphere, by trapping sunlight, that results to climate change. Due to the higher melting and boiling temperatures of phosphorus, it is rare that it occurs as gas in room temperatures. Phosphorus, therefore, does not pose a great threat of climate change.


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1) Students in a science class created Eco columns to mimic a small ecosystem. Grass seeds were planted in 50 grams of potting s
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B) The grass height was the dependent variable.

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7 0
3 years ago
Read 2 more answers
Giardia does it lead to elevated levels of IgE and eosinophilia?
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Answer:

No, Giardia is a protozoan that does not cause eosinophilia.

Explanation:

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Many factors might cause. One of them is parasite infections, in which helminths trigger the IgE generation, producing eosinophilia.

In the presence of the parasite antigen, eosinophils have a shorter medullar generation time, and they express a higher number of receptors for IgE and IgG. Their function is to damage the parasite, directly or indirectly, and to decrease the damages caused by their presence.

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4 0
3 years ago
Unlike record players which rotate at a constant angular velocity, the CD-ROM driver in a CD player rotates the CD at different
zimovet [89]

Answer:

A. 21 rad/s

B. 200.5 rpm

C. 26.46 rad/s2

D. 0.299 Hz

Explanation:

Parameters given are:

Tangential velocity,v = 1.26m/s

Diameter (outer edge) = 0.120m

Outer edge radius, ro = 0.12/2

= 0.06m

A. Calculate Angular velocity

Angular velocity, w = v/r

= v/ro

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In rpm,

2pi/60 rad/s = 1 rpm

1 rad/s = 60/2pi rpm

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= 200.5 rpm

B. Calculate the inner edge radius, ri given w = 500 rpm

Converting rpm to rad/s,

= 2pi/60 * 500 rad/s

= 52.34 rad/s

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= 1.26/ 52.36

= 0.024m

C. Calculating centripetal acceleration, a from the outer edge, ro

a = v2/r = w2r

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= 2pi/21

= 0.299 Hz (or per second)

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