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Klio2033 [76]
3 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]3 years ago
5 0
C. Forrest Fires release the seeds stored in Jack Pines
In-s [12.5K]3 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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Multiply 0.00032 cm by 4.02 cm and express the answer in scientific notation
FromTheMoon [43]
0.00032cm*4.02=1.2864 × 10^-3 in scientific notation.
4 0
3 years ago
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David is investigating the properties of soil using the sample shown.
xenn [34]

Although the sample is not shown in this question, we can conclude that it would be reasonably easy for David to provide evidence of the color, consistency, temperature, and texture of the soil.

<h3 />

<h3>What are these properties an example of?</h3>

These are all examples of the physical properties of a sample. Since we cannot see the sample that David is using, it would be safest to assume that he would have no trouble providing evidence as to the physical properties of the soil, the:

  • Color
  • Consistency
  • Temperature
  • Texture

are all examples of this.

Therefore, we can confirm that David can provide evidence of the color, consistency, temperature, and texture of the soil.

To learn more about physical properties visit:

brainly.com/question/24632287?referrer=searchResults

7 0
3 years ago
A collapsible plastic bag (figure below) contains a glucose solution. If the average gauge pressure in the vein is 1.29 A collap
expeople1 [14]

Answer:

0.125 m

Explanation:

Pressure in fluids is given as the product of density, height and acceleration due to gravity and expressed as

P=hdg

Where h is the height, d is density, g is acceleration due to gravity and P is pressure.

Making h the subject of formula then

h=P/dg

Given specific gravity of a substance, its density is equal to specific gravity multiplied by density of water. Taking density of pure water as 1000 kg/m³ then the density of reference fluid will be 1.05*1000=1050 kg/m³

Substituting pressure with 1.29*10³ pa as given then taking g as 9.81 m/s² then

H=1.29*10³÷(9.81*1050)=0.1252366389981068880151448958788408329692m

Rounded off, the height is approximately 0.125 m

8 0
3 years ago
Help in solving this question please<br>(question is on the picture )<br>​
MakcuM [25]

Answer:

Jesus

Explanation:

he is the way

6 0
2 years ago
Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.87 Earth years in its
dolphi86 [110]

Answer: 4.487(10)^{11}m

Explanation:

This problem can be solved using the Third Kepler’s Law of Planetary motion:

<em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.  </em>

<em />

This law states a relation between the orbital period T of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size a of its orbit:

T^{2}=\frac{4\pi^{2}}{GM}a^{3} (1)  

Where:

T=3.87Earth-years=122044320s is the period of the orbit of the exoplanet (considering 1Earth-year=365days)

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}  

M=3.59(10)^{30}kg is the mass of the star

a is orbital radius of the orbit the exoplanet describes around its star.

Now, if we want to find the radius, we have to rewrite (1) as:

a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}} (2)  

a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}} (3)  

Finally:

a=4.487(10)^{11}m This is the radius of the exoplanet's orbit

3 0
3 years ago
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