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SIZIF [17.4K]
4 years ago
8

A 110 kg crate is pulled along a level surface by an engine. The coefficient of kinetic friction between the crate and the surfa

ce is 0.585. How much power must the engine deliver to move the crate at a constant speed of 7.59 m/s? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of W.
Physics
1 answer:
Leno4ka [110]4 years ago
3 0

Answer:

Net power will be 1150.637 watt

Explanation:

We have given mass of the crate m = 110 kg

Coefficient of friction \mu =0.585

Acceleration due to gravity g=9.8m/sec^2

Constant speed v=7.59m/sec

Force applied F=ma=110\times 7.59=834.9N

Frictional force f=\mu mg=0.585\times 110\times 9.8=630.63N

So net force =834.9-630.6=204.3N

Power is given by P=Fv=204.3\times 7.59=1550.637Watt

So net power will be 1550.637 watt

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Can a molecule have bond dipoles but not have a molecular dipole?
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Answer: Yes

Explanation: A molecule has a bond dipole but not have a molecular dipole.

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Example:  

lets take an example of  CO_{2} molecule in which two electro negative oxygen atoms are attached with Carbon atom. Oxygen being electro negative will attract the shared pair electrons towards itself and  partial negative charge will create on oxygen atom and partial positive charge on carbon atom C.

due to formation of partial positive and partial negative charges dipole moment is created between oxygen and Carbon bond on both the sides in opposite direction.

Since, the dipole moment acting in opposite direction the net dipole moment in the molecule is zero.

Hence, Yes, a molecule has bond dipoles but not have a molecular dipole.

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If the weight of an object that is submerged in a fluid is 10 N and the buoyant force on it is
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Por que el movimiento de un auto que recorre una pista circular ni es M.R.U.V​
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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
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