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valentinak56 [21]
3 years ago
10

Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2

.0 N, and you do 12 J of work on each. Match each box to the vertical distance through which it is lifted.
Physics
1 answer:
Alona [7]3 years ago
8 0

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

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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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3 years ago
What is the build up of electrons on a surface
Minchanka [31]

Answer:

Static energy

Explanation:

Think of it as a balloon rubbing against your hair, the two attractions of friction causes Static energy.

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A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

v=u+at

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

v = 15.5 +(-9.8)(2.64)=-10.4 m/s

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

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