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Zepler [3.9K]
3 years ago
7

Compare the benefits of wildfires to grasslands, northern forests, and deciduous forests.

Physics
2 answers:
Dimas [21]3 years ago
6 0
Wildfires benefit grasslands, northern forests, and deciduous forests. Grasslands are benefited by improved soil quality and control of tree cover. Invertebrate species diversity is maintained through wildfire as well. Northern forests, like grasslands, experience increased production and nutritional quality of food as a result of wildfires. Deciduous forests experience an increase in the nutritional quality of food as well, but the effects are more temporary. The amount of shrubs in deciduous forests is reduced as a result of wildfires, allowing more herbaceous plants such as mosses and lichens to grow.
zhannawk [14.2K]3 years ago
6 0

Wildfires benefit grasslands, northern forests, and deciduous forests. Grasslands are benefited by improved soil quality and control of tree cover. Invertebrate species diversity is maintained through wildfire as well. Northern forests, like grasslands, experience increased production and nutritional quality of food as a result of wildfires. Deciduous forests experience an increase in the nutritional quality of food as well, but the effects are more temporary. The amount of shrubs in deciduous forests is reduced as a result of wildfires, allowing more herbaceous plants such as mosses and lichens to grow.

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water has an index of refraction of 1.33. What is the critical angle for light leaving a pool of water into air? 0 37 90 49
vesna_86 [32]
When light travels from a medium with higher refractive index into a medium with lower refractive index, there is a maximum angle (called critical angle) for which all the light is reflected, so there is no refraction.

The value of the critical angle is given by:
\theta = \arcsin ( \frac{n_2}{n_1} )
when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
\theta = \arcsin ( \frac{1.00}{1.33} )=49^{\circ}
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Iodine 131 half life is 8.0 days. Ten percent of the original sample o his isotope remains after (a) 22.7 days (b) 24.9 days (c)
Artyom0805 [142]

Answer:

option (c) is correct

Explanation:

Half life of a substance is the time in which the element becomes half of is initial value.

half life, T = 8 days

Amount remaining, N = 10 % of original value

Let the original value is No.

N = 10% of No

N = 0.1 No

Let the time taken is t and the decay constant is λ.

The relation between the decay constant and the half life is given by

\lambda =\frac{0.6931}{T}=\frac{0.6931}{8}=0.08664 per day

Us the equation of radioactivity

N=N_{0}e^{-\lambda t}

0.1N_{0}=N_{0}e^{-0.08664 t}

e^{0.08664 t}=10

Taking natural log on both the sides, we get

0.08664 t = 2.303

t = 26.6 days

4 0
3 years ago
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