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Klio2033 [76]
2 years ago
8

How are forest fires beneficial to conifers like Jack Pines? a. Jack pine seeds are protected by forest fires. b. Pine cones are

carried on the winds produced by forest fires. c. Forest fires release the seeds stored in Jack pine cones. d. All of the above.
Physics
2 answers:
Gre4nikov [31]2 years ago
5 0
C. Forrest Fires release the seeds stored in Jack Pines
In-s [12.5K]2 years ago
5 0

Answer: c. Forest fires release the seeds stored in Jack pine cones.

Fire ecology is a scientific discipline that deals with the natural processes that involves fire in an ecosystem and in the ecological effects. Fire in the ecosystem either generates naturally or intentionally by humans. The fire generated naturally is useful in the forests ecosystem, especially forests with coniferous trees. The trees exhibit special adaptation, which involves the release of seeds from the cones by the aid of fire. The naturally generated fires melts the resin covering the cones and helps in the release of seeds from these cones of coniferous trees such as pine.

Hence, forest fires release the seeds stored in Jack pine cones is the correct option.

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Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with e
My name is Ann [436]

To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.

However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that

PV =NkT

Where,

N = Number of molecules

k = Boltzmann constant

V = Volume

T = Temperature

P = Pressure

Our values are given as,

N = 1

k = 1.38*10^{-23}J/K

T = 27\°C = 27\°C + 273 = 300K

P = 1atm = 101325Pa

Rearrange the equation to find V we have,

V = \frac{NkT}{P}

V = \frac{1(1.38*10^{-23})(300K)}{101325Pa}

V = 4.0858*10^{-26}m^3

We know that length of a cube is given by

V = L^3

Therefore the Length would be given as,

L = V^{1/3}

L = (4.0858*10^{-26})^{1/3}

L = 3.445*10^{-9}m

Therefore each length of the cube is 3.44nm

7 0
3 years ago
PLEASE HEELP!!!
Ann [662]

Answer:

The "pressure" of the electricity is electric potential. Electric potential is the amount of energy available to push each unit of charge through an electric circuit. The unit of electric potential is the volt. ... A volt is the force needed to move one amp through a conductor that has 1 ohm of resistance

8 0
2 years ago
At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri
viva [34]

Answer:

v_{f2} =6.5%v_{i1}

Explanation:

Mass of the ball: m_{1} =0.12kg]

Initial velocity of the ball:   v_{i1}

final velocity of the ball: v_{f1} which is -30/100 of v_{i1} =-0.3v_{i1}

Mass of the bottle: m_{2} =2.4kg

Initial velocity of the bottle: v_{i2}=0m/s

final velocity of the bottle: v_{f2} is unknown (to find)

<em>by using conservation momentum, which stated that the initial momentum is equal to the final momentum.</em>

<em />m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}<em />

<em>so since the bottle is at rest firstly, therefore </em>v_{i2} =0<em />

<em />m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}<em />

<em />m_{1} v_{i1}  =m_{1} v_{f1} +m_{2} v_{f2}<em>         </em><em>equation 1</em>

so now substitute v_{f1} into equation 1

m_{1} v_{i1}  =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}

<em />m_{1} v_{i1}  = -0.3m_{1}v_{i1}  +m_{2} v_{f2}<em />

<em>collect the like terms</em>

m_{1} v_{i1}   +0.3m_{1}v_{i1}  =m_{2} v_{f2}

1.3m_{1} v_{i1}   =m_{2} v_{f2}

divide both  side by m_{2}

v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }

Now substitute

v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2}    =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}

v_{f2} =6.5%v_{i1}

<em />

6 0
2 years ago
Read 2 more answers
Indicate on the chart whether you would classify each model as representing an element, compound, or mixture.
Ne4ueva [31]

Answer:

1 compound

2 mixture

3 elements

4 0
3 years ago
Read 2 more answers
If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J

So, the required work done is 16 J.

7 0
2 years ago
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