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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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Which memory disorder is a common illness on television and films

Alenkinab [10]

amnesia is the most common illness used in tv an films

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3 years ago

Read 2 more answers

PE=30J, m=?, g=10m/s2, h=10m

OleMash [197]

Based on the given, this is probably a gravitational potential energy problem (PEgrav). The formula for PEgrav is:

PEgrav = mgh

Where:
m = mass (kg)
g = acceleration due to gravity
h = height (m)

With this formula you can derive the formula for your unknown, which is mass. First put in what you know and then solve for what you do not know.

$PEgrav=mgh$
$30J=m(10)(10[tex] \frac{30}{100} =m$ )[/tex]

Do operations that you can with what is given first.

$30J=m(100m)$

Transpose the 100 to the other side of the equation. Do not forget that when you transpose, you do the opposite operation.

$\frac{30}{100} =m$

m = 0.30kg

5 0

2 years ago

Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from

qwelly [4]

Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0

2 years ago

A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20

statuscvo [17]

Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

Assuming no friction, total mechanical energy must keep constant, so the following is always true:

$\Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0 (1)$

Choosing the ground level as our zero reference level, Uf =0.
Since the child starts from rest, K₀ = 0.
From (1), ΔU becomes:
$\Delta U = 0- m*g*h = -m*g*h (2)$
In the same way, ΔK becomes:
$\Delta K = \frac{1}{2}*m*v_{1}^{2} (3)$
Replacing (2) and (3) in (1), and simplifying, we get:

$\frac{1}{2}*v_{1}^{2} = g*h (4)$

In order to find v₁, we need first to find h, the height of the slide.
From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

$h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)$

Replacing (5) in (4) and solving for v₁, we get:

$v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s (6)$

As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
We can do this in more than one way, but a very simple one is using kinematic equations.
If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

$v_{1}^{2} - v_{o}^{2} = 2*a* x_{1} (7)$

Since v₀ = 0 (the child starts from rest) we can solve for a:

$a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2 (8)$

Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

$t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s (9)$

Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

$x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2} (10)$

Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

$x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2} = 17.2 m (11)$

From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

$v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)$

Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

$\frac{1}{2}*v_{2}^{2} = g*h_{2} (13)$

Replacing from (12) in (13), we can solve for h₂:

$h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m (14)$

Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

$x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)$

4 0

2 years ago

If a fuse or circuit breaker directly opens a circuit when the current limit is reached, do overload heaters open and break the

OLEGan [10]

Answer:

Explanation:

No, overload heaters do not open to break the circuit when their current limit is reached. These are safety device to prevent over heating.

Thermal overload relays in a motor are intended to protect the conductors (windings). Such safety devices are made to maintain the current flow in a circuit at a safety level so as to avoid overheating of the circuitry conductors. And the excess of current is often called as over current.

3 0

2 years ago