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2 years ago

14

A circuit contains four 100 ohm resistors connected in series. If you test the circuit with a DMM you should read a total resist

ance value of
A. 250 ohms.
. B. 1000 ohms
C. 400 ohms
D. 100 ohms

1 answer:

olga nikolaevna [1]2 years ago

8 0

C is the correct answer

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A pressure cylinder has an outer diameter 200 mm, maximum external pressure 4 MPa, and maximum allowable shear stress 27.5 MPa.

ludmilkaskok [199]

Answer:

The minimum value of wall thickness t=3.63 mm.

Explanation:

Given:

D=200 mm

P=4 MPa

t= Wall thickness

maximum shear stress=27.5 MPa

We know that

hoop stress $\sigma _{h}=\frac{Pd}{2t}$

Longitudinal stress $\sigma _{l}=\frac{Pd}{4t}$

So maximum shear tress in plane $\tau _{max}=\dfrac{\sigma _h-\sigma _l}{2}$

$\tau _{max}=\dfrac{Pd}{8t}$

Now by putting the value

$27.5=\dfrac{4\times 200}{8t}$

So t=3.36 mm

The minimum value of wall thickness t=3.63 mm.

4 0

3 years ago

A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem

lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is $\Delta h=9.975\,m$

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

$\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2} \Delta v^2 =0$

( $\rho$ stands here for density, $h$ for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

$\frac{\Delta P}{\rho}+g\, \Delta h =0$

$\Delta P= -g\, \rho\, \Delta h$

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

$\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\$

We insert that into our last equation and get:

$\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m$

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

8 0

3 years ago

A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground

7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

T ≈ 17199 lbf

4 0

3 years ago

What gadgets are charge coupled devices used in?

Alinara [238K]

Answer:

They are used in imaging application gadgets such as video cameras,TV, surveillance cameras and document scanners

Explanation:

A charge couple device (CCDs) are highly capable in imagery detector.Its common application is in video and digital imaging.The quality of a charge couple device is determined by factors such as the dynamic range, dark charge level and the quantum efficiency.These devices serve the purpose of detecting optical images though some are installed with applications for data storage.

5 0

2 years ago

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value

OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

$Y = 1 - e^{-kt^n}$

here Y is percentage of completion of reaction = 50%

t is duration of reaction = 146 sec

so,

$0.50 = 1 - e^{-k^146^2.1}$

$0.50 = e^{-k306.6}$

taking natural log on both side

ln(0.5) = -k(306.6)

$k = 2.26\times 10^{-3}$

for 86 % completion

$0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}$

$e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14$

$-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)$

$t^{2.1} = 869.96$

t = 25.10 sec

5 0

2 years ago