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2 years ago

How does competition affect population size? Use the terms carrying capacity and limiting factor with your example.

2 answers:

6 0

Limiting factors are resources or other factors in the environment that can lower the population growth rate. Competition for resources like food and space cause the growth rate to stop increasing, so the population levels off. This flat upper line on a growth curve is the carrying capacity.

cestrela7 [59]2 years ago

3 0

Answer:

they can eat all of the food and kill off the population

Explanation:

if the competition eats all of the food there's no food for the population and they will die off.

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A father pushes his child on a swing. He pushes with a force of 20 N or a distance of 1 m, with the force always maintained para

Firlakuza [10]

Answer:

20 Joules

Explanation:

Work is done whenever a force moves a body through a certain distance in the direction of the force. So, work done is the product of force and distance moved.

Therefore, we have;

Work done = Force x distance

i.e Wd = Fs

Given that: F = 20 N and s = 1 m, then;

Wd = 20 N x 1 m

= 20 Nm

The work done by the father is 20 Joules(Nm).

7 0

2 years ago

An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t

elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

$\mu = \frac{f_0}{f_e}$

Here,

$\mu$ = Magnification

$f_e$ = Focal length eyepieces

$f_0$ = Focal length of the Objective

Rearranging to find the focal length of the objective

$f_0 = \mu f_e$

Replacing with our values

$f_0 = 7.5* 3.7cm$

$f_0 = 27.75cm$

Therefore the focal length of th eobjective lenses is 27.75cm

5 0

2 years ago

A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th

Mila [183]

Answer:

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

$\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0$

$\frac{3}{(r + a)^2} = \frac{2}{r^2}$

$\frac{r+a}{r} = \sqrt{\frac{3}{2}}$

$1 + \frac{a}{r} =\sqrt{\frac{3}{2}}$

$\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}$

so we will have

$r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

so the x coordinate of this position is given as

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

6 0

3 years ago

Nobel gases share which characteristics?

Leto [7]

Full outer shell of electrons, they are colorless and odorless ,their melting and boiling points are close together which gives them very narrow liquid range

8 0

3 years ago

Read 2 more answers

The specific heat capacity of ethanol is 2440J/kg °C. How many joules of energy will be required to heat 150g ethanol to 35°C if

m_a_m_a [10]

Answer:

Required energy = 4758 J

Explanation:

Specific heat capacity of a material is the amount of energy required to raise the temperature of one kilogram (kg) of that material through one degree Celsius (°C).

Given data :

Specific heat capacity = c = 2440 J/kg.°C

Mass = m = 150 g = 0.15 kg

Initial temperature = 22°C

Final temperature = 35°C

Change in Temperature = ΔT = 13°C

Energy = E = ?

Using the following formula and substituting the values, we get:

E = m × c × ΔT

E = 0.15 × 2440 × 13

E = 4758 J

8 0

3 years ago