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3 years ago

How does a wildfire impact a population of oak trees?

Physics

2 answers:

miss Akunina [59]3 years ago

8 0

Answer:

it impacts the tree form.

Wildfire o Northwest California oak woodlands would naturally be subjected to ... o Fire does not necessarily have a positive impact on growth or recruitment of.

Explanation:

Many species of mature oaks are relatively tolerant to fire. Oaks might replace pines, or drought-tolerant shrubland might take the place ... “Areas with low tree density could maintain droughts for some time because of the cyclical effect of ... diseased plants and removing them from the flora population.

Ivanshal [37]3 years ago

8 0

Answer:

it burns the tree into another form

Explanation:

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Liquid droplets that fall from the sky are? A) rain B)snow C) sleet D) hail

alukav5142 [94]

The answer is a)rain

6 0

3 years ago

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If I travel 300 m east, then 400 m west, what is my distance & displacement?

ad-work [718]

Answer:100m west

Explanation:

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3 years ago

Describe the motion of a swing that requires 6 seconds to complete one cycle. What is its period and the frequency? Round to the

shutvik [7]

Period = 6 seconds and $frequency = 0.167Hz$ .

Explanation:

We have , the motion of a swing that requires 6 seconds to complete one cycle. Period is the amount of time needed to complete one oscillation . And in question it's given that 6 seconds is needed to complete one cycle. Hence ,Period of the motion of a swing is 6 seconds . Frequency is the number of vibrations produced per second and is calculated with the formula of $\frac{1}{t}$ . SI unit of frequency is Hertz or Hz. We know that time period is 6 seconds so frequency = $\frac{1}{t}$

⇒ $frequency = \frac{1}{time}$

⇒ $frequency = \frac{1}{6}$

⇒ $frequency = 0.167Hz$

Therefore , Period = 6 seconds and $frequency = 0.167Hz$ .

7 0

3 years ago

A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p

Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0

2 years ago