A finite resource in the environment that keeps a population from steadily increasing is known as

1 answer:

3 0

The choices can be found elsewhere and as follows:

 (1) dynamic equilibrium
(2) a limiting factor
(3) a reproductive enzyme
(4) ecological succession

I think the correct answer from the choices is option 2. A finite resource in the environment that keeps a population from steadily increasing is known as a limiting factor.

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Two 4.0 kg masses are 1.0 m apart on a frictionless table. Each has 1.0 μC of charge.What is the magnitude of the electric force

xeze [42]

Coulomb's law:

Force = (8.99×10⁹ N m² / C²) · (charge₁) · (charge₂) / distance²

= (8.99×10⁹ N m² / C²) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C) / (1.0 m)²

= (8.99×10⁹ x 1×10⁻¹² / 1.0) N

= 8.99×10⁻³ N

= 0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses'; don't need it.
 Mass has no effect on the electric force between them.

-- 'frictionless table'; don't need it.
 Friction has no effect on the force between them,
only on how they move in response to the force.

7 0

2 years ago

30 POINTS! (Question is located in picture under graph)

harkovskaia [24]

Hi i think its E because the the numbers on both sides are to spread out and are not even one side is 4 and the other is 3

8 0

3 years ago

Read 2 more answers

A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain

Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

For maximum speed

$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16$

At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$