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2 years ago

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Un cuerpo de 3,5 kg se encuentra en reposo sobre un plano inclinado 37o. Está sujeto al extremo superior del plano inclinado med

iante un muelle de constante recuperadora 15 N/m. Sabiendo que el coeficiente de rozamiento vale 0,6, calcula el alargamiento del muelle. Sol: 42 cm.

1 answer:

Tems11 [23]2 years ago

8 0

Answer:

1,1 m

Explanation:

Dado que;

coeficiente de fricción = 0,6

sabemos que W = R = mgcos 37 = 3.5Kg * 10m / s ^ 2 * cos37 = 27.95 N

coeficiente de fricción = fuerza / reacción normal (R)

Fuerza = 0.6 * 27.95 N

Fuerza (F) = 16.77 N

Recuerda que F = Ke

dónde;

K = constante de fuerza (15N / m)

e = extensión (lo desconocido)

e = F / K

e = 16,77 N / 15 N / m

e = 1,1 m

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Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is $P = 104.04 \hat{i} -314.432 \hat{j}$

Explanation:

The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

the peripheral velocity that is directed downward $(-V_y)$ along the y-axis

the linear velocity $(V_x)$ that is directed along the x-axis

Now;

$V_x = \frac{d}{dt}(12t^3+2) = 36 t^2$

$V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s$

Also,

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where $\omega$ (angular velocity) = $\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)$

$-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s$

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3 years ago

The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a

dimaraw [331]

The absolute pressure is given by the equation,

$P_{abs}=P_{atm}-P_{vac}$

Here, $P_{abs}$ is absolute pressure, $P_{atm}$ is atmospheric pressure and $P_{vac}$ is vacuum pressure.

Therefore,

$P_{abs}=98 kPa-80 kPa=18kPa$

The gage pressure is given by the equation,

$P_{gage}=P_{abs}-P_{atm}$ .

Thus,

$P_{gage}=18kPa-98 kPa=-80 kPa$ .

In kn/m^2,

The absolute pressure,

$P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2$ .

In lbf/in2

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2$

In psi,

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi$ .

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi$

In mm Hg

The absolute pressure,

$P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg$

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$P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg$

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2 years ago

Which volcanic hazard can block the sunlight and temporarily cool the Earth’s surface?

BaLLatris [955]

If a volcano epulses massive amounts of dust into the atmosphere, those two things will/can happen.
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2 years ago

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nadya68 [22]

Answer:

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2 years ago

Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of

Aneli [31]

Answer:

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Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

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Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

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Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

3 0

2 years ago