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Georgia [21]
3 years ago
11

Which kingdom does the owl belong to

Physics
1 answer:
Hoochie [10]3 years ago
8 0

<em>Owls, </em><em>birds</em><em> of the </em><em>order</em><em> Strigiformes, include about 200 </em><em>species</em><em> of mostly </em><em>solitary</em><em> and </em><em>nocturnal</em><em> </em><em>birds of prey</em><em> typified by an upright stance, a large, broad head, binocular vision and binaural hearing, and feathers adapted for silent flight. Exceptions include the diurnal </em><em>northern hawk-owl</em><em> and the gregarious </em><em>burrowing owl</em><em>.</em><em>Owls hunt mostly small </em><em>mammals</em><em>, </em><em>insects</em><em>, and other birds, although a few species specialize in hunting </em><em>fish</em><em>. They are found in all regions of the Earth except </em><em>Antarctica</em><em> and some remote islands.</em>

<em>Owls are divided into two </em><em>families</em><em>: the </em><em>true owls</em><em>, Strigidae; and the </em><em>barn-owls</em><em>, Tytonidae.</em>

i think you can find an answer from this ;)

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kati45 [8]

Answer:

The value is c  = 8 \  km

Explanation:

From the question we are told that

The distance of friends house from your point is a =  10 \  km

The distance of your friends street from your street is b =  6 \  km \  in the \ direction \  towards \  the  \  north

The diagram illustrating this question is shown on the first uploaded image

From the diagram we can apply by Pythagoras theorem as follows

a^2 =  b^2 +  c^2

=>     c  =  \sqrt{^2 -  b^2}

=>     c  =  \sqrt{ 10^2 -  6^2}

=>     c  = 8 \  km

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4 years ago
A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin
pychu [463]

a) See graph in attachment

b) The suvat equation to use is v_f^2 - v_i^2 = 2as

c) The acceleration is a=\frac{v_f^2-v_i^2}{2s}

d) The acceleration is 1.25 m/s^2

e) The time needed is 8 s

Explanation:

a)

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is x_i = 0, the origin

- The  final position of the boat is x_f = 200 m

- The initial velocity of the boat is v_i = 20.0 m/s

- The final velocity of the boat is v_f = 30.0 m/s

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

b)

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

v_i = 20.0 m/s, the initial velocity

v_f = 30.0 m/s, the final velocity

s = x_f - x_i = 200 m, the  displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

v_f^2 - v_i^2 = 2as

where

a is the acceleration

c)

Now we have to solve the equation

v_f^2 - v_i^2 = 2as

In order to find the acceleration.

This can be done by dividing both terms by 2s: this way, we find

\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}

And so the acceleration is

a=\frac{v_f^2-v_i^2}{2s}

d)

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

v_i = 20.0 m/s, the initial velocity

v_f = 30.0 m/s, the final velocity

s = x_f - x_i = 200 m, the  displacement of the boat

And substituting into the equation,

a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2

e)

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

v_f = v_i + at

where:

v_i is the initial velocity

v_f is the final velocity

a is the acceleration

t is the time

Here we have:

v_i = 20.0 m/s

v_f = 30.0 m/s

a=1.25 m/s^2

Solving for t, we find:

t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s

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Hence the metalcore is found in the given X area.

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Answer:

https://physicsabout.com/acceleration-and-velcoity/

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