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2 years ago

Which of the following is a result of a limiting factor in an ecosystem?

Physics

2 answers:

Morgarella [4.7K]2 years ago

4 0

Answer:

a. A pond has only produced a small amount of algae, so the fish population is also small.

Explanation:

serious [3.7K]2 years ago

3 0

Answer:A - a pond has only produced a small amount of algae, so the fish population is also small. i’m pretty sure that’s correct.

Explanation: The amount of alage determines the size,or carrying capasity of the fish population,so it is a limiting factor.

Hope I could help!Brainlist?plz

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A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

The rate of a reaction is the speed at which products form or reactants disappear.<br> T/F

Natali [406]

Answer: TRUE (:

Explanation:

3 0

2 years ago

A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.

PolarNik [594]

Answer:

11.95m/s

Explanation:

A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.

a) Find the speed of the object. Answer in units of m

K. E =½mv²

150= ½mv²

Multiply both sides by 2

mv² = 300

Divide both sides by v²

m = 300/v² .................. Equation 1

Momentum is the product of mass and velocity

Momentum = mv

25.1 = mv

Divide both sides by v

m = 25.1/v ................ Equation 2

Equate equations 1 and 2

300/v² = 25.1/v

Cross multiply

25.1v² = 300v

Multiply v with both sides

25.1v = 300

Divide both sides by 25.1

v = 300/25.1

V = 11.95m/s

I hope this was helpful, please mark as brainliest

8 0

2 years ago

What does the zeroth law of thermodynamics allow us to define?

NeTakaya

It defines that if two thermodynamic systems are each in equilibrium with a third system, then they are in equilibrium with each other.

3 0

3 years ago

Read 2 more answers

a wall of glass 2cm thickhas inside temperature of 30°C,outside temperature of15°C.how much heat is flowing through the glass(k=

kenny6666 [7]

Answer:46.5

Explanation:is the topical formula of F= 30x+15

5 0

3 years ago