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poizon [28]
3 years ago
15

A 2kg object initially going 4m/s to the right is later going 8m/s. whats the change in velocity?

Physics
1 answer:
Nezavi [6.7K]3 years ago
6 0

The change in velocity is +4 m/s to the right (or -4 m/s to the left).

The object's mass is irrelevant.

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Angelina_Jolie [31]

Answer:

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3 years ago
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Neporo4naja [7]

Answer:

Explanation:

When a force hits something, an equal amount of force is exerted back on it.

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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

6 0
2 years ago
Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football
faltersainse [42]

Answer:

a=159.32\ m/s^2

Explanation:

It is given that,

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Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

v=r\omega

v=0.085\times 43.35

v = 3.68 m/s

The centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(3.68)^2}{0.085}

a=159.32\ m/s^2

So, the centripetal acceleration of the laces on the football is 159.32\ m/s^2. Hence, this is the required solution.

6 0
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A force of 30 N is applied to an object with a mass of 15 kg. What is the resulting acceleration?
svet-max [94.6K]

Answer:

We know that Force = mass × acceleration

By substituting the values we get,

30 N = 15 kg × a (where a is acceleration)

Or we can write it as

15 kg × a = 30 N

Transposing 15 to RHS,

a = 30 ÷ 15 m/s²

Therefore, acceleration = 2 m/s²

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