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2 years ago

Which of the following is an abiotic limiting factor for a plant population in an ecosystem?

Physics

1 answer:

Ludmilka [50]2 years ago

8 0

Correct answer is Availability of soil minerals.

There are two types of limiting factors that affect plant population in a ecosystem. They are:

Biotic
Abiotic

Biotic factors includes food, diseases etc. Such as Invasive weed species, seed dispersal by wind, disease-causing fungal spores are examples of biotic limiting factor.

Abiotic factors include sunlight, temperature and chemical environment. The availability of soil mineral is an example of abiotic limiting factor. Growth of plant population depends on availability of soil minerals.

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True or false? A protostellar cloud spins faster as it contracts

Aleks04 [339]

Answer:

No. The protostellar cloud spins faster in the collapsing stage (stage 1) and becomes much slower in the contraction stage (stage 2)

Explanation:

Once the cloud is so dense that the heat which is being produced in its center cannot easily escape, pressure rapidly rises, and catches up with the weight, or whatever external force is causing the cloud to collapse, and the cloud becomes stable, as a protostellar cloud.

The protostellar cloud will become more dense over thousands of years. This stage of decreasing size is known as a contraction, rather than a collapse. In the contraction stage the cloud has become much slower, and because weight and pressure are more or less in balance. In the first stage of formation, the decrease of size is very rapid, and compressive forces completely overwhelm the pressure of the gas, and we say that the cloud is collapsing.

3 0

3 years ago

Which statements are true about galaxies, stars, and the universe? (more than one true answer by the way)

Elina [12.6K]

Answer:

statements <em><u>2, 3, 4, and 7</u></em> are true

Explanation:

8 0

3 years ago

If the only forces acting on a 2.0kg mass are F1 = (3i-8j)N and F2 = (5i+3j)N, what is the magnitude of the acceleration of the

belka [17]

Answer: 4.7m/s²

Explanation:

According to newton's first law,

Force = mass × acceleration

Since we are given more the one force, we will take the resultant of the two vectors.

Mass = 2.0kg

F1+F2 = (3i-8j)+(5i+3j)

Adding component wise, we have;

F1+F2 = 3i+5i-8j+3j

F1+F2 = 8i-5j

Resultant of the sum of the forces will be;

R² = (8i)²+(-5j)²

Since i.i = j.j = 1

R² = 8²+5²

R² = 64+25

R² = 89

R = √89

R = 9.4N

Our resultant force = 9.4N

Substituting in the formula

F = ma

9.4 = 2a

a = 9.4/2

a = 4.7m/s²

Therefore, magnitude of the acceleration of the particle is 4.7m/s²

3 0

3 years ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

7 0

3 years ago

Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200

andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

ΔK.E = ΔP.E

K_2 - K_1 = P_1- P_2

Where, K_2: Kinetic energy of hammer head as it hits the I-beam

K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

P_1: Initial gravitational potential energy of hammer head

- The expression simplifies to:

K_2 = P_1

Where, 0.5*m*v2^2 = m*g*s12

v2 = √(2*g*s12) = √(2*9.81*3)

v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

Wnet = ( P_3 - P_1 ) + W_friction

Wnet = m*g*s13 + F*s23

n*s23 = m*g*s13 + F*s23

Where, n: average force the hammerhead exerts on the I-beam.

s13 = s12 + s23

Hence,

n = m*g*( s12/s23 + 1) + F

n = 200*9.81*(3/0.074 + 1) + 60

n = 81562 N

6 0

3 years ago