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3 years ago

Which kingdom does the owl belong to

1 answer:

8 0

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

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

i think you can find an answer from this ;)

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In what way do pigments act differently from light?

Karo-lina-s [1.5K]

A. Because they reflect their color and absorb all the others

6 0

3 years ago

Read 2 more answers

A force of 350N is applied to a body. If the work done is 40kJ, what is the distance through which the body moved?

Studentka2010 [4]

The distance covered by the body is 114.3 m

Explanation:

The work done by a force exerted on an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the object in this problem, we have

F = 350 N is the force applied

$W=40 kJ = 40,000 J$ is the work done

$\theta=0^{\circ}$ if we assume that the force is applied parallel to the motion of the object

Solving for d, we find the distance covered by the object:

$d=\frac{W}{F cos \theta}=\frac{40,000}{(350)(cos 0)}=114.3 m$

Learn more about work:

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brainly.com/question/6443626

#LearnwithBrainly

7 0

2 years ago

Because she had serious traffic accident on Friday the 13th of last month, Felicia is conceived that all Friday the 13th will br

elena-s [515]

Answer:

Illusory correlation

Explanation:

Illusory correlation is a psychological phenomenon of sensing a correlation or relationship between a particular quantity and other variables however unrelated the two variables may be. A person may establish such connections, which are indeed false connections, as a result of over emphasis on novel incidents that are easily noticed due to them being salient

For example, when a waiter at a restaurant is being rude to a person after the person was made to walk under a ladder due to ongoing construction.

4 0

2 years ago

a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig

suter [353]

The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

$2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}$

Then, the final velocity will be

$v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s$

Thus, the final velocity of the truck is found as 146.969 m/s.

8 0

3 years ago

A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee

Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

$F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2} = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\$

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

$v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\$

v = 7.69 x 10³ m/s

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

$v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v = \frac{2\pi r}{T}\\\\$

where,

T = Time Period of Satellite = ?

Therefore,

$T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\$

T = 5500 s = 91.67 min = 1.53 h

7 0

2 years ago