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3 years ago

7

Using aluminum wire(2.82x10^-8 ohm) of diameter 1.60mm, you want to wind a resistor that will dissipate 25.0 W when 25.0 V is ap

plied across it . What length( in m) of wire is needed?

1 answer:

kiruha [24]3 years ago

4 0

Answer:

4565 meters

Explanation:

W = V²/R

25 = (25)²/R

R = 25 ohm

R = (pl)/A

$R = \frac{pl}{2\pi( \frac{d}{2} )}$

$25 = \frac{( 2.82\times {10}^{ - 8} \times l)}{2\pi (\frac{1.60 \times {10}^{ - 6} }{2}) }$

l = 4565 meters approximately

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Answer:

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A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

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To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

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What makes science different from other fields of knowledge like art or philosophy?

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As opposed to art or philosophy, science is an objective phenomenon. It deals with something which is verifiable, meaning that it can be tested, and it is based on facts, which are true and can be proven. On the other hand, art is highly subjective, which means that it can be interpreted in many different ways. Science is more concrete and testable, whereas art and philosophy lie in the domain of the abstract.

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So to solve for this problem, this is computed by the following steps:

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So plugging in our values: <span>
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lina2011 [118]

Answer:

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The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Explanation:

GIven that:

Diameter of the piston-cylinder = 12 cm

Pressure of the piston-cylinder = 100 kPa

Temperature =24 °C

Length of the piston = 20 cm

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The gas is compressed and The temperature of the gas remains constant during this process.

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Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ

b. What is the final volume and pressure in the cylinder?

In an isothermal process;

Workdone W = $\int dW$

$W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}$

Since the gas is compressed ; then $v_f< v_i$

However;

$W =- nRT \ In \dfrac{V_f}{V_i}$

$W =- P_1V_1 \ In \dfrac{V_f}{V_i}$

The initial volume for the cylinder is calculated as ;

$v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3$

Replacing over values into the above equation; we have :

$100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}$

The final pressure can be calculated by using :

$P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f$

$P_f =\dfrac{P_iV_i}{V_f}$

$P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}$

$P_f = 1.565*10^2 \ kPa$

$\mathbf{P_f = 156.5 \ kPa}$

c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?

The change in entropy of the gas is given by the formula:

$\Delta S=\dfrac{\Delta Q}{T}$

where

T = 24 °C = (24+273)K

T = 297 K

$\Delta S=\dfrac{-100 \ J}{297 \ K}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

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3 years ago