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Nonamiya [84]
3 years ago
7

1.- Un balín de acero a 20 C tiene un volumen de 0.004 m3. ¿Cuál es la dilatación que sufre cuando su temperatura aumenta a 50 C

?
Physics
1 answer:
Brrunno [24]3 years ago
0 0

Answer:

\Delta V = 1.440\times 10^{-6}\,m^{3}

Explanation:

El coeficiente de expansión volumétrica se calcula a partir de la siguiente ecuación diferencial parcial (The volumetric expansion coefficient is computed by means of the following partial differential equation):

\alpha = \frac{1}{V} \cdot \left(\frac{\partial V}{\partial T} \right)

Se integra la fórmula a continuación (The formula is integrated herein):

\alpha\, dT = \frac{dV}{V}

Supóngase que el coeficiente es constante (Let suppose that coefficient is constant):

\alpha \int\limits^{T_{f}}_{T_{o}}\,dT = \int\limits^{V_{f}}_{V_{o}}\, \frac{dV}{V}

\alpha \cdot (T_{f}-T_{o}) = \ln \frac{V_{f}}{V_{o}}

El volumen final es (The final volume is):

V_{f} = V_{o}\cdot e^{\alpha \cdot (T_{f}-T_{o})}

El coeficiente de expansión volumétrica del acero es 12\times 10^{-6}\,^{\circ}C^{-1} (The volumetric expansion coefficient of steel is 12\times 10^{-6}\,^{\circ}C^{-1}):

V_{f} = (0.004\,m^{3})\cdot e^{(12\times 10^{-6}\,^{\circ}C^{-1})\cdot (50^{\circ}C-20^{\circ}C)}

V_{f} \approx 4.001\times 10^{-3}\,m^{3}

Finalmente, la dilatación experimentada por el balín es (Lastly, the dillatation experimented by the pellet is):

\Delta V = 4.001\times 10^{-3}\,m^{3} - 4.000\times 10^{-3}\,m^{3}

\Delta V = 1.440\times 10^{-6}\,m^{3}

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Explanation:

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3 years ago
11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle
Alja [10]

Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Explanation:

To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.

For M vector you obtain:

M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}

For N vector:

N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}

The resultant vector is the sum of the components of M and N:

F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}

The magnitude of the resultant vector is:

|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm

And the direction of the vector is:

\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°

hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

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3 years ago
Water flows over a section of Niagara Falls at rate of 1.1×10^6 kg/s and falls 49.4 m. How much power is generated by the fallin
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Answer:

<em>The power generated is =  5.33×10⁸ Watt. </em>

Explanation:

Power: Power can be defined as the time rate of doing work. The S.I unit of power is <em>Watt(W).</em>

<em>Mathematically,</em>

<em>Power (P) = Work done/time or Energy/time</em>

P = mgh/t............................... Equation 1

P = δgh............................. Equation 2

Where δ = fall rate, g = acceleration due to gravity, h = height.

<em>Given: </em>δ = 1.1×10⁶ kg/s, h = 49.4 m g = 9.81 m/s²

Substituting these values into equation 2

P = 1.1×10⁶×49.4×9.81

P = 533.08×10⁶

<em>P = 5.33×10⁸ Watt.</em>

<em>Thus the power generated is =  5.33×10⁸ Watt. </em>

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A bat hits a 0.150 kg baseball for
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Answer:

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Explanation:

impulse = mass ( change in velocity )

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               = 0.150 ( 32.0 - (-47.0))

                = 0.150 ( 32.0 +47.0)

                = 0.150 (79)

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Answer: B

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