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Vika [28.1K]
2 years ago
7

United States citizens have many ways to participate in national life. The most common form of participation is?

Physics
1 answer:
Travka [436]2 years ago
3 0

Answer:

You would have to give better explanation on subject.

Explanation:

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You are lost at night in a large, open field. Your GPS tells you that you are 122.0 m from your truck, in a direction 58.0o east
Drupady [299]

Answer:

The person is 187[m] farther and 70° south to east.

Explanation:

We can solve this problem by drawing a sketch of the location of the person and the truck, then we will draw the displacement vectors and finally the length of the vector and the direction of the vector will be measured in order to give the correct indication of where the person will have to move.

First we establish an origin of a coordinate system.

We can see in the attached schema that the red vector is the displacement vector from the last point to where the truck is located.

The length of the vector is 187 [m], and the direction is 70 degrees south to East.

3 0
3 years ago
The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5
nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

G\frac{mM}{R^2}=m\frac{v^2}{R}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

m=11,100 kg is the mass of the telescope

M=5.97\cdot 10^{24} kg is the mass of the Earth

R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s

6 0
3 years ago
What is the characteristic that allows an object to have kinetic energy?
Shtirlitz [24]
Kinetic energy is movement, thus movement would be a characteristic that allows an object to have kinetic energy
6 0
2 years ago
Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su
dybincka [34]
  1. La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
  2. La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua (v), en metros por segundo:

v = \sqrt{\frac{K}{\rho} } (1)

Donde:

  • K - Módulo de compresibilidad, en newtons por metro cuadrado.
  • \rho - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que \rho = 1\times 10^{3}\,\frac{kg}{m^{3}} y K = 2,16\times 10^{9}\,\frac{N}{m^{2}}, entonces la velocidad de las ondas sonoras es:

v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }

v\approx 1469,694\,\frac{m}{s}

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda (\lambda), en metros, mediante la siguiente fórmula:

\lambda = \frac{v}{f} (2)

Donde f es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que v\approx 1469,694\,\frac{m}{s} y f = 1000\,hz, entonces la longitud de onda de las ondas sonoras es:

\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}

\lambda = 1,470\,m

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

6 0
2 years ago
If a dog walks north for 10 meters and then east for 10 meters, what is the direction of its displacement?
Katyanochek1 [597]

The direction of its displacement wil be

c.northeast

In fact, the dog walks north for 10 meters and east for another 10 meters. The path of the dog can be represented with two vectors, A pointing north (of magnitude 10 meters) and B pointing east (of magnitude 10 meters). The direction of the resultant vector (due to east) will be given by

tan \theta =\frac{A}{B}=\frac{10}{10}=1

\theta=tan^{-1} (1)=45^{\circ}

and the direction will be north-east.

5 0
3 years ago
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