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3 years ago

Which of the following is an example of a population?

Physics

2 answers:

svlad2 [7]3 years ago

7 0

D the gray wolves in the forest.is the correct answer.

Bogdan [553]3 years ago

7 0

<span>D is the answer because it describes a single species in a specific environment.

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You might be interested in

You drop a 20 Ton object and a

Alborosie

Explanation:

it dosent depend on the weights of the items. I'll reach the ground at same time taking as no air friction or restrictions.

i.e

v = u + gt

whte v is final velocity of the object

u is initial velocity of the object

g is acceleration due to gravity and

t is time. thanks

please if found helpful rate brainliest

7 0

1 year ago

Are colavent bonds more likely to be found in compounds contaning both materals and nonmetals or compounds contaning only nonmet

atroni [7]

Answer:

bonds between 2 alike atoms or most non-metals get a covalent bond when reacting to the ... It tells what kinds of atomas are in a compound and how many.

Hope this helped

-Gavin

7 0

3 years ago

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The power of a machine is 6000 W. This machine is scheduled for design improvements. Engineers have reduced the

Georgia [21]

Explanation:

Work = power × time

W = (6000 W) (7.5 s)

W = 45,000 J

7 0

3 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

3 years ago

The wavelength of a wave is the horizontal distance from a ________ to an adjacent ________.

goldenfox [79]

<span>A. crest, crest

hope im right!(: </span>

5 0

3 years ago

Read 2 more answers