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USPshnik [31]
3 years ago
5

Sequence eye light enters the eye

Physics
1 answer:
Burka [1]3 years ago
8 0

Answer:

Well it is the courtnes

Explanation:

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The potential energy of a 40-kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?
nasty-shy [4]
Pe = mgh.
14000 J = (40kg)(10m/s^2)(h)
h = 35 meters
7 0
3 years ago
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Por que el movimiento de un auto que recorre una pista circular ni es M.R.U.V​
tigry1 [53]

Answer:

Salta al contenido principal

Contenido principal

Movimiento rectilíneo uniforme (MRU)

Movimiento Rectilíneo

Movimiento rectilíneo uniforme (MRU)

Imagina que eres un astronauta en la Estación Espacial Internacional. Estás arreglando unos paneles solares averiados, cuando de pronto, al presionar, tu destornillador sale disparado de tus manos. Si no lo atrapas a tiempo, el destornillador estará viajando por el espacio en línea recta y a velocidad constante, a menos que algo se interponga en su camino. Esto sucede porque la herramienta se mueve con movimiento rectilíneo uniforme, o MRU.

Foto de la Estación Espacial Internacional

Foto de la Estación Espacial Internacional

Estación Epacial Internacional orbitando nuestro planeta. Créditos: International Space Station orbiting above earth de la National Reconnaissance Office.

El MRU se define el movimiento en el cual un objeto se desplaza en línea recta, en una sola dirección, recorriendo distancias iguales en el mismo intervalo de tiempo, manteniendo en todo su movimiento una velocidad constante y sin aceleración.

Recuerda que la velocidad es un vector, entonces, al ser constante, no varía ni su magnitud, ni su dirección de movimiento.

Condiciones del MRU

Para que un cuerpo esté en MRU, es necesario que se cumpla la siguiente relación:

t−t

0

x−x

0

Constante

Donde

xxx: es la posición en el espacio y

ttt: es el tiempo.

De esta condición, llegamos a la ecuación del MRU:

x = x_0 + v(t-t_0)x=x

0

+v(t−t

0

)x, equals, x, start subscript, 0, end subscript, plus, v, left parenthesis, t, minus, t, start subscript, 0, end subscript, right parenthesis

Donde:

\Large x_0x

0

x, start subscript, 0, end subscript: posición en el instante \Large t_0t

0

t, start subscript, 0, end subscript

\Large xxx: Posición en el instante \Large ttt

Esto quiere decir que si conocemos la posición x_0x

0

x, start subscript, 0, end subscript en el instante t_0t

0

t, start subscript, 0, end subscript y sabemos cuál es la de la velocidad vvv, podremos conocer la posición xxx en cualquier instante ttt.

¡No olvides fijarte bien en las unidades que utilizas y de convertirlas si es necesario!

Veamos un ejemplo:

Imagínate que has programado un carro robótico para que tenga una velocidad constante ¿Puedes calcular a qué distancia desde el punto de partida estará luego de 30\text{ s}30 s30, start text, space, s, end text?

Tienes los siguientes datos:

v

x

0

t

0

t

=10 m/s

=0 m

=0 s

=30 s

Explanation:

espero y esto te ayude

5 0
3 years ago
Which of these organs is an endocrine structure?
Gre4nikov [31]

Answer: Actually three of them are. The ovaries, the uterus and fallopian tubes.

6 0
2 years ago
Why is venus the hottest planet in the solar system
mr Goodwill [35]

Venus is the hottest world in the solar system. Although Venus is not the planet closest to the sun, its dense atmosphere traps heat in a runaway version of the greenhouse effect that warms Earth.

8 0
3 years ago
Read 2 more answers
The figure above represents a stick of uniform density that is attached to a pivot at the right end and has equally spaced marks
zavuch27 [327]

Answer:

After 2.0s the  angular momentum is L= 2(4A+3B+2C+D)x

Explanation:

Let us call forces acting on the rod, A, B, C, and D, and the separation between them x .

Then, the  torque due to force A is

\tau_a = 4Ax,

due to the force B

\tau_b = 3Bx,

due to force C

\tau_c = 2Cx,

and the torque due to force D is

\tau_d = Dx.

Therefore, the total torque on the the stick is

\tau_{tot} =\tau_a+\tau_b+\tau_c+\tau_d

\tau_{tot} =4Ax+3Bx+2Cx+Dx

\tau_{tot} =x(4A+3B+2C+D)

Now, this torque causes angular acceleration \alpha according to the equation

I \alpha = \tau_{tot}

where I is moment of inertia of the stick and it has the value

I = \dfrac{1}{3} m(4x)^2

Therefore the angular acceleration is

\alpha = \dfrac{\tau_{tot} }{I}

\alpha =\dfrac{x(4A+3B+2C+D)}{\dfrac{1}{3}m(4x)^2 }

\boxed{\alpha =\dfrac{3(4A+3B+2C+D)}{16mx } .}

Now, the angular momentum L of the stick is

L = I\omega,

where \omega is the angular velocity.

Since \omega = \alpha t, we have

$L = \dfrac{1}{3}m (4x)^2  *\dfrac{3(4A+3B+2C+D)}{16mx }* t$

L= (4A+3B+2C+D)x t

Therefore,   t = 2.0s, the angular momentum is

\boxed{ L= 2(4A+3B+2C+D)x. }

5 0
3 years ago
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