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3 years ago

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A circuit consists of three resistors in parallel with values of 6.3k, 4.2k 12k, and 780 and a total voltage of 150 V. What is t

he current of the circuit in milli-amps? Round to
the nearest mA. Do Not include units.

1 answer:

k0ka [10]3 years ago

4 0

Answer:

32.5

Explanation:

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10. True or False? A disruptive technology<br> radically changes the way people live and<br> work.

Anettt [7]

Answer:

<u>True</u>

Explanation:

According to Investopedia.com, "Disruptive technology is an innovation that significantly alters the way that consumers, industries, or businesses operate".

So yes, a disruptive does radically change the way people live and work.

5 0

3 years ago

Read 2 more answers

Water flows through a horizontal 60 mm diameter galvanized iron pipe at a rate of 0.02 m3/s. If the pressure drop is 135 kPa per

maksim [4K]

Answer:

pipe is old one with increased roughness

Explanation:

discharge is given as

$V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}$

V = 7.07 m/s

from bernou;ii's theorem we have

$\frac{p_1}{\gamma} +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} +\frac{V_2^2}{2g} + z_2 + h_l$

as we know pipe is horizontal and with constant velocity so we have

$\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}$

$P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma$

$135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81$

solving for friction factor f

f = 0.0324

fro galvanized iron pipe we have $\epsilon = 0.15 mm$

$\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025$

reynold number is

$Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}$

Re = 378750

from moody chart

$For Re = 378750 and \frac{\epsilon}{d} = 0.0025$

$f_{new} = 0.025$

therefore new friction factor is less than old friction factoer hence pipe is not new one

now for Re = 378750 and f = 0.0324

from moody chart

we have $\frac{\epsilon}{d} =0.006$

$\epsilon = 0.006 \times 60$

$\epsilon = 0.36 mm$

thus pipe is old one with increased roughness

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3 years ago

Find an expectation value in the n th state of the harmonic oscillator.Find an expectation value in the n th state of the harmon

s344n2d4d5 [400]

The classical motion for an oscillator that starts from rest at location x₀ is

x(t) = x₀ cos(ωt)

The probability that the particle is at a particular x at a particular time t

is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average

to get the spatial density. Our natural time scale for the averaging is a half

cycle, take t = 0 → π/ ω

Thus,

ρ = $\frac{1}{\pi / w} \int\limits^\pi_0 {d(x - x_o cos(wt))} \, dt$

Limit is 0 to π/ω

We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt) so that

ρ(x) = $-\frac{w}{\pi } \int\limits^x_x {\frac{d ( x - y)}{x_ow sin(wt)} } \, dy$

Limit is x₀ to -x₀

$\frac{1}{\pi } \int\limits^x_x {\frac{d (x-y)}{x_o\sqrt{1 - cos^2(wt)} } } \, dy$

Limit is -x₀ to x₀

$= \frac{1}{\pi } \int\limits^x_x {\frac{d(x-y)}{\sqrt{x_o^2 - y^2} } } \, dy\\ \\= \frac{1}{\pi\sqrt{x_o^2 - x^2} }$

This has $\int\limits^x_x {p(x)} \, dx = 1$ as expected. Here the limit is -x₀ to x₀

The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.

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4 years ago

How do we need to prepare for the future?

Paraphin [41]

B. to achieve sustainable development

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3 years ago

A smelter emits SO2 at a rate of 10,000 kg/day. The stack is 200 m tall and the plume rise is 100 m. The windspeed is 2 m/s at a

Olegator [25]

Answer:

lol

Explanation:

lol

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4 years ago