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Un material determinado tiene un espesor de 30 cm y una conductividad térmica (K) de 0,04 w/m°C. En un instante dado la distribu

aksik [14]

Answer:

Para x=0:

$\phi=1.2 W/m^{2}$

Para x=30 cm:

$\phi=-2.4 W/m^{2}$

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

$\phi=-k\frac{dT}{dx}$ (1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

$\frac{dT(x)}{dx}=300x-30$

Remplezando esta derivada en (1):

$\phi=-0.04(300x-30)$

Para x=0:

$\phi=0.04(30)$

$\phi=1.2 W/m^{2}$

Para x=30 cm:

$\phi=-0.04(300*0.3-30)$

$\phi=-2.4 W/m^{2}$

Espero que te haya ayudado!

4 0

3 years ago

Two particles have a mass of 7.8 kg and 11.4 kg , respectively. A. If they are 800 mm apart, determine the force of gravity acti

aleksley [76]

Answer:

A) About $9.273 \times 10^{-9}$ newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) $F_g=\dfrac{GM_1M_2}{r^2}$ , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

$F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}$

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

3 0

3 years ago

A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled intension with a load o

grigory [225]

Answer:

The elastic modulus of the steel is 139062.5 N/in^2

Explanation:

Elastic modulus = stress ÷ strain

Load = 89,000 N

Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2

Stress = load/area = 89,000/0.64 = 139.0625 N/in^2

Length of steel bar = 4 in

Extension = 4×10^-3 in

Strain = extension/length = 4×10^-3/4 = 1×10^-3

Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2

7 0

3 years ago

In a website browser address bar, what does “www” stand for?

Ludmilka [50]

Answer:

www stands for world wide web

Explanation:

It will really help you thank you.

3 0

3 years ago

P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si

klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V 0.184 W

2.499 mV 125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

$V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\$ *5

= 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:

V_o = A_voc*V_i*(R_o/R_l+R_o)

= 4.545*(100/100+50)

= 3.03 V

Now we can determine delivered power:

P_L = V_o^2/R_L

= 0.184 W

Apply voltage-divider principle to the circuit:

V_o = (R_o/R_o+R_s)*V_s

= 50/50+100*10^3*5

= 2.499 mV

Now we can determine delivered power:

P_l = V_o^2/R_l

= 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.

4 0

3 years ago