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alexira [117]
3 years ago
8

Consider a circuit element, with terminals a and b, that has vab= -12V and iab= 3A. Over a period of 2 seconds, how much charge

moves through the element? If electrons carry the charge, which terminal do they enter? How much energy is transferred? Is it delivered to the element or taken from it?
Engineering
1 answer:
miv72 [106K]3 years ago
5 0

Answer:

a) 6 coulombs

b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current

c) = -72 joules

Energy is taken from element

Explanation:

Given data:

V ab = -12 v

I ab = 3A

period ( t ) = 2 seconds

a) determine how much charge moves through the element

q = I * t

  = 3 * 2 = 6 coulombs

b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current

c) determine how much energy is transferred

= Vab * Iab * t

= -12 * 3 * 2

= -72 joules

Energy is taken from element

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Acetylene and propane

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214Bi83 --> 214Po84 + eBismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle,
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(C) ln [Bi]

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3 years ago
Read 2 more answers
A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

4 0
3 years ago
Create a variable pounds to store weight in pounds. Convert this to kilograms and assign the result to a variable kilos. The con
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Answer:

>>pounds=13.2

>>kilos=pounds/2.2

Explanation:

Using Matlab to write the program, consider at any time when the weight in pounds is 13.2 lb, this variable of weight is created in MATLAB by typing >>pounds=13.2. To convert it from lb to Kg, we simply divide it by 2.2 hence the second command to created is kilos. For this, the output of the program will be 6 Kg.

5 0
3 years ago
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