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olchik [2.2K]
3 years ago
8

What is the kinetic energy of a 26 kg eagle flying at an altitude of 65 m at a speed of 19 m/s?

Physics
1 answer:
sweet [91]3 years ago
4 0

Answer:

4693 J

Explanation:

The kinetic energy of the eagle is given by:

K=\frac{1}{2}mv^2

where

m is the mass of the eagle

v is the speed of the eagle

In this problem, m = 26 kg and v = 19 m/s. Substituting into the formula, we find

K=\frac{1}{2}(26 kg)(19 m/s)^2=4693 J

As we can see, the altitude of the eagle is not relevant for the calculation of the kinetic energy.

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Explain how this happens?
grandymaker [24]

Answer:

When the polythene rod is rubbed with the woolen cloth, static electric charges move from the cloth and into the rod. The rod becomes negatively charged as negative charges move from the cloth and into the rod leaving the cloth positively charged as well.

5 0
3 years ago
Ultrasonic images are obtained from the inside organs of our body. This process uses which property of sound wave?
Temka [501]

This question involves the concepts of echo, ultrasonic images, ultrasonic sound waves.

The process of ultrasonic images uses the "echo" property of the sound waves.

Echo is the property of the sound wave by the virtue of which the sound wave reflects back to the source of the sound after hitting a surface or an object.

Ultrasonic images are obtained from inside organs of our body. This process involves the use of ultrasonic sound waves that have a frequency greater than 20,000 Hz. These sound waves are out of the range of audible sound by the human ear. When these ultrasonic sound waves are sent in form of pulses into the human body by the use of probes, they reflect back from the tissues of different organs to the probe. The probe then records the reflection properties of these sound waves and displays them in form of an image, known as ultrasonic images.

Learn more about echo here:

brainly.com/question/14335186?referrer=searchResults

The attached picture shows the process of ultrasonic imaging.

4 0
2 years ago
What are the three rules for writing the electron configuration of elements
astraxan [27]

<em></em>

Answer:

<u><em>The aufbau principle</em></u>

<u />

<u><em>The Pauli exclusion principle</em></u>

<u><em></em></u>

<u><em>Hund's rule of maximum multiplicity</em></u>

Explanation:

<u><em>The aufbau principle:</em></u>

<em></em>

The fundamental electronic configuration is achieved by placing the electrons one by one in the different orbitals available for the atom, which are arranged in increasing order of energy.

<u><em>The Pauli exclusion principle:</em></u>

<em></em>

Two electrons of the same atom cannot have their four equal quantum numbers.  Because each orbital is defined by the quantum numbers n, l, and m, there are only two possibilities ms = -1/2 and ms = +1/2, which physically reflects that each orbital can contain a maximum of two electrons, having opposite spins

<u><em>Hund's rule of maximum multiplicity:</em></u>

This rule says that when there are several electrons occupying degenerate orbitals, of equal energy, they will do so in different orbitals and with parallel spins, whenever this is possible.  Because electrons repel each other, the minimum energy configuration is one that has electrons as far away as possible from each other, and that is why they are distributed separately before two electrons occupy  the same orbital.

4 0
2 years ago
A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
Paraphin [41]

Answer:

\omega = 0.016\,\frac{rad}{s}

Explanation:

The rotation rate of the man is:

\omega = \frac{v}{R}

\omega = \frac{0.80\,\frac{m}{s} }{5\,m}

\omega = 0.16\,\frac{rad}{s}

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

The final angular speed is:

\omega = 0.016\,\frac{rad}{s}

3 0
3 years ago
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

3 0
3 years ago
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