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3 years ago

In a chart the independent variable goes on the ___________. Question 5 options: left right middle

Physics

2 answers:

I am Lyosha [343]3 years ago

6 0

The answer i think is left

PSYCHO15rus [73]3 years ago

4 0

Independent goes on the right

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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 total energy equation?

netineya [11]

The total energy equation would be Kinetic energy+Potential energy

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4 years ago

True or False: A 24 year old motorcycle driver would not be required to wear protective eyewear when riding in Florida.

n200080 [17]

False people should always wear gear

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3 years ago

Read 2 more answers

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

$PE_{initial}+KE_{initial} = PE_{final}+KE_{final}$

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

$KE_{final} = 3600MJ$

Therefore the final kinetic energy is 3600MJ

5 0

3 years ago

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)