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laiz [17]
3 years ago
6

Resuelva el siguiente ejercicio. Un armario de 120 kg descansa sobre una superficie plana, determine la fuerza de rozamiento si

el coeficiente de rozamiento para el cuerpo vale 0,4 y realice un gráfico indicando las fuerzas indicando las fuerzas que actúan el en armario.
Physics
1 answer:
aalyn [17]3 years ago
4 0

Responder:

470.4N

Explicación:

La fuerza de fricción se expresa mediante la fórmula;

Ff = nR

n es el coeficiente de fricción

R es la reacción

R = peso = mg

Ff = nmg

Dado

n = 0,4

masa m = 120kg

g es la aceleración debida a la gravedad = 9,8 m / s²

Sustituir en la fórmula;

Ff = 0,4 (120) (9,8)

Ff = 470,4 N

Por tanto, la fuerza de fricción es 470,4 N

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An object is thrown upward from the top of a 128​-foot building with an initial velocity of 112 feet per second. The height h of
lorasvet [3.4K]

Answer:

Time, t = 8 seconds

Explanation:

An object is thrown upward from the top of a 128​-foot building with an initial velocity of 112 feet per second. The height h as a function of time t is given by :

h=-16t^2+112t+128

We need to find the time when the object will hit the ground. When it will hit the ground, h = 0

So,

-16t^2+112t+128=0

On solving the above quadratic equation, we get the value of t = 8 seconds. So, after 8 seconds the object will hit the ground. Hence, this is the required solution.

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What 2 key vocabulary words are used when you send an object in motion?
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3 years ago
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A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c
Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

  • The plane wall is initially with constant properties is initially at a uniform temperature, To.
  • Suddenly the surface x=L is exposed to convection process such that T∞>To.
  • The other surface x=0 is maintained at To
  • Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

  1. There is only conduction in 1-D.
  2. The system bears constant properties.
  3. The volumetric heat generation is uniform

From the given data, the condition are as follows

<u>Initial Condition</u>

At t≤0

T(x,0)=T_o

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

<u>Boundary Conditions</u>

<u>At x=0</u>

<u />T(0,t)=T_o<u />

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

<u>At x=L</u>

<u />-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]<u />

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0
3 years ago
A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is conne
kipiarov [429]

Answer:

0.832 m/s

Explanation:

The work done by the spring W equals the kinetic energy of the object K

The work done by the spring W = 1/2k(x₀² - x₁²) where k = spring constant, x₀   = initial compression = 0.065 m and x₁ = final compression = 0.032 m

The kinetic energy of the object, K = 1/2mv² where m = mass of object and v = speed of object

Since W = K,

1/2k(x₀² - x₁²) = 1/2mv²

k(x₀² - x₁²) = mv²

mv² = k(x₀² - x₁²)

v² = [(k/m)(x₀² - x₁²)]

taking square root of both sides, we have

v = √[(k/m)(x₀² - x₁²)] since ω = angular frequency = √(k/m),

v = √[(k/m)√(x₀² - x₁²)]

v = ω√(x₀² - x₁²)]

Since ω = 14.7 rad/s, we substitute the other variables into the equation, so we have

v = 14.7 rad/s × √((0.065 m)² - (0.032 m)²)]

v = 14.7 rad/s × √(0.004225 m² - 0.001024 m²)]

v = 14.7 rad/s × √(0.003201 m²)

v = 14.7 rad/s × 0.056577

v = 0.832 m/s

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In the diagram, disk 1 has a moment of inertia of 3.4 kg · m2 and is rotating in the counterclockwise direction with an angular
nlexa [21]

Answer:

I = 3.6 kg•m²

Explanation:

Conservation of angular momentum

Let's assume CW is the positive direction

3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)

I(9.3 - 1.8) = 3.4(1.8 + 6.1)

I(7.5) = 3.4(7.9)

I = 3.4(7.9)/(7.5) = 3.5813333333...

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3 years ago
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