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3 years ago

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A 3 column table with 5 rows labeled table A. The first column is labeled voltage in volts with entries 1.0, 5.0, 10, 20, 50. Th

e second column is labeled current: calculated in amps with no entries. The last column is labeled current: experimental in amps. According to Ohm’s law, determine the calculated current for these values in Table A. voltage = 5 V, resistance = 20 Ω: A voltage = 20 V, resistance = 20 Ω: A voltage = 50 V, resistance = 20 Ω: A

2 answers:

dimaraw [331]3 years ago

7 0

Answer:

<em>0.25A</em>

<em>1.0A</em>

<em>2.5A</em>

Explanation:

Volgvan3 years ago

6 0

Answer:

0.25A

1.0A

2.5A

Explanation:

this is only for the calculated current column on edge

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Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

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$t=\frac{v}{a}$ ; $d=s*(t-t_{0} )$

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$L=\frac{F}{\pi*r*P}$ ; $d=\frac{w}{F*cos(o)}$

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$t^{2}=\frac{2*x}{g} ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} } \\$

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$h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}$

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$h=\frac{m}{(1/2)*\pi *r^{2} } ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2} }$

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$b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}$

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$a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2} }{M}$

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$v_{o}=\frac{x-\frac{1}{2}*a*t^{2} }{t} ; F=\frac{W+uNd}{d*cos(o)}$

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$h=\frac{E-\frac{1}{2}*m*v^{2} }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2} }{\frac{1}{2}m }$

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$N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2} }{\frac{1}{2}*k }$

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$x=x_{o} +\frac{v^{2-v_{o}^{2} } }{2a} ; m=\frac{P*A-F_{1}-F_{2} }{g}$

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3 years ago

Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13

svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

Steady operations exist;
The heat transfer coefficient (h) is uniform over the entire fin surfaces;
Thermal conductivity (k) is constant;
Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

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