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spin [16.1K]
3 years ago
7

Choose whether each of the following statements is true or false. In order to move a massive crate sitting on the floor, the for

ce you apply to the crate has to the force overcome the force of the crate pushing back on you. An object can be moving in one direction while the net force acting on it is in another direction. An object can be in motion and still have zero net force acting on it. An object at rest has no forces acting on it.
Physics
1 answer:
velikii [3]3 years ago
6 0

Answer:

 

Explanation:

1. False

The force you apply on crate is equal and opposite to the force that crate applies on you by Newton's third law of motion.

The force must over come the static frictional force between the crate and the  floor.

2. True

The object can move along another direction than the direction of net force. For example, when a car slows down, the net force is opposite to the direction of motion.

3. True

An object moving at constant velocity has zero net force acting on it.

4. False

An object at rest has forces acting on it but the summation of all the forces is zero i.e. the net force is zero.

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RUDIKE [14]

Answer:

Hey

Your answer would be Radioactive Decay and Heat of formation.

When earth first formed it was very violant. Some planetary scientists believe that the moon was created from a collision involving earth and Thea (a theoretical dwarf planet). these collisions that have formed earth heated it so much that it is still hot from them.

Radioactivce decay is another major fource of internal heat for earth.

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3 years ago
Parts of the brain ​
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8 0
3 years ago
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Un trineo de 20 kg descansa en la cima de una pendiente de 80 m de longitud y 30° de inclinación. Si µ = 0.2, ¿cuál es la veloci
Mariulka [41]

Answer:

v= 26.70 m/seg

Explanation:  Ver anexo ( diagrama de cuerpo libre)

De acuerdo a la segunda ley de Newton

∑ F  =  m*a

∑ Fx  =  m* a(x)             ∑ Fy  =  m* a(y)

También sabemos que el coeficiente de roce dinámico es:

  μ  = 0.2 = F(r)/N            siendo N la fuerza normal.

Si descomponemos la fuerza P = mg  =  20Kg* 9.8m/seg²

P =  196 [N]    en sus componentes sobre los ejes x y y tenemos

Py  =  P* cos30  =  196* √3/2  =  98*√3

Px  = P* sen30   =  196*1/2  =  98

La sumatoria sobre el eje y es :

∑ F(y)  =  m*a         Py  - N  = 0          98*√3  = N       ( no hay movimiento en la dirección y)

∑ F(x)  = m*a    P(x)  -  Fr  =  m*a

Fr  =   μ *N  =  0.2* 98*√3

Fr  =  19.6*√3  [N]

98 -  19.6*√3  =  m*a

98  -  33.52  = m*a

a =  (98  -  33.52 ) / 20

a = 3.22 m/seg²

Para calcular la velocidad del trineo al pié del plano, sabemos que al pié del plano el trineo ha recorrido 80 m, y que de cinemática

v²  =  v₀²  +  2*a*d             ( se pueden chequear unidades para ver la consistencia de la ecuación  v  y  v₀    vienen dados en m/seg  entonces  v²  y  v₀²  vienen en m²/seg²,  el producto de a (m/seg²) por la distancia d (m) resulta en m²/seg²  entonces es consistente la relación

v²   =  0   +  2*3.22*80       ( la velocidad inicial es cero)

v²  = 515.2  m²/seg²

v  =  √515.2  m/seg

v= 26.70 m/seg

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