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Zepler [3.9K]
2 years ago
7

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

Physics
2 answers:
Dimas [21]2 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]2 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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A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me
never [62]

1) Current in the wire: 0.0875 A

The current in the wire is given by:

I=\frac{Q}{t}

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

I=\frac{420 C}{4800 s}=0.0875 A

2) Drift velocity of the electrons: 1.78\cdot 10^{-6} m/s

The drift velocity of the electrons in the wire is given by:

u = \frac{I}{nAq}

where

I = 0.0875 A is the current

n=5.8\cdot 10^{28} is the number of free electrons per cubic meter

A is the cross-sectional area

q=1.6\cdot 10^{-19} C is the charge of one electron

The radius of the wire is

r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m

So the cross-sectional area is

A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2

So, the drift velocity is

u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s

4 0
2 years ago
How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?
noname [10]

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

U=mgh

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

U=(50 kg)(9.8 m/s^2)(15 m)=7350 J

3 0
2 years ago
A man weighing 750 n and a woman weighing 500 n have the same momentum. what is the ratio of the man's kinetic energy km to that
miss Akunina [59]
Because weight W = M g, the ratio of weights equals the ratio of masses.

(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

but p's are equal, so

K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662
4 0
2 years ago
To compare the hearing capacities of 5 of his friends, Ravi designs a simple experiment. He places a CD player at the end of a l
aliya0001 [1]

Answer:

A) having each person listen to just 1 song instead of 5.

Explanation:

Ravi could accurately use the outcome to compare the hearing capacities of his friends by this experiment if appropriate precautions are observed.

Playing 5 well known songs simultaneously could result to the interference of sounds when each participant moves away from the player. Which could naturally cause the variation in volume of the songs played with respect to the increasing distance.

Therefore, the reliability of the result from this experiment can be increased if each person listen to just 1 song instead of 5 at a predetermined volume. Each participant would be able to focus on hearing a song during the experiment.

8 0
2 years ago
the brightest , hottest, and most massive stars are the brilliant blue stars designated as spectral class O. if a class O star w
4vir4ik [10]

The speed is 0.956 m / s.

<u>Explanation</u>:

The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.

                   K.E = 1/2 \times m \times v^{2}

where K.E represents the kinetic energy,

           m represents the mass,

            v represents the velocity.

                  K.E = 1/2 \times m \times v^{2}

    1.10 \times 10^42 = 1/2 \times 3.26 \times 10^31 \times v^{2}

                     v^{2} = (1.10 \times 10^42 \times 2) / (3.26 \times 10^31)

                     v = 0.956 m / s.

6 0
2 years ago
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