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satela [25.4K]
3 years ago
14

If you were to plot the voltage versus the current for a given circuit, what would you expect the slope of the line to be? If no

w the resistance of the circuit were a function of temperature would you still expect to see a straight line when plotting voltage versus current if the temperature was allowed tovary?

Engineering
1 answer:
Brut [27]3 years ago
7 0

Answer:

Part 1: It would be a straight line, current will be directly proportional to the voltage.

Part 2: The current would taper off and will have negligible increase after the voltage  reaches a certain  value. Graph attached.

Explanation:

For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.

V=I*R

where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.

In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.

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Explanation:

Given that;

diameter bolts d = 1.83 in

ultimate shear strength of the bolts = 60 ksi

we know that

shear area = 2×(π/4)d²

= 2×(π/4)×(1.83)² = 5.2604 in²

so

p/3(5.2604) = 60000/3.9

p/15.7812 = 15384.6153

p = 15.7812 × 15384.6153

p = 242787.691 lb

p = 242.7877 kips

therefore the allowable load P is 242.7877 kips

3 0
3 years ago
Assume a steel pipe of inner radius r1= 20 mm and outer radius r2= 25 mm, which is exposed to natural convection at h = 50 W/m2.
Mekhanik [1.2K]

Answer:

98,614.82 W/m²

Explanation:

Q = 2\pi hL(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})

Where;

Q = the amount of heat loss from the pipe

h =  the heat transfer coefficient of the pipe = 50 W/m².K

T₁ = the ambient temperature of the pipe = 30⁰C

T₂  = the outside temperature of the pipe = 100⁰C

L= the length of pipe

r₁ = inner radius of the pipe = 20mm

r₂ = outer radius of the pipe = 25mm

To determine the amount of heat loss from the pipe per unit length

From the equation above

\frac{Q}{L} = 2\pi h(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})

\frac{Q}{L} = 2\pi* 50(\frac{100-30}{Ln\frac{25}{20}})

\frac{Q}{L} = 314.159(\frac{70}{0.223})

\frac{Q}{L} = 314.159(313.901) = 98,614.82 W/m²

3 0
3 years ago
28. What is the value of a resistor in a series circuit if you measure 0.5 amps flowing through it and 15 volts
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In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in
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Answer:

V = 125.7m/min

Explanation:

Given:

L = 400 mm ≈ 0.4m

D = 150 mm ≈ 0.15m

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F = 0.30mm ≈ 0.0003m

To calculate the cutting speed, let's use the formula :

T = \frac{pi* D * L}{V*F}

We are to find the speed, V. Let's make it the subject.

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Substituting values we have:

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3 years ago
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
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as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
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