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2 years ago

Hi gospelgamer10 lol

Engineering

2 answers:

Katyanochek1 [597]2 years ago

6 0

Answer:

I’m not that person but

Explanation:

thamks for the pointzzz

brainliest too plz??? <3<3<3

Kaylis [27]2 years ago

5 0

Answer:

yo not that person either but points here bye

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What is the measurment unit of permeability?

nikitadnepr [17]

Answer:

Henries :)

Explanation:

I looked it up

7 0

2 years ago

Read 2 more answers

The current at resonance in a series L-C-R circuit is 0.2mA. If the applied voltage is 250mV at a frequency of 100 kHz and the c

iVinArrow [24]

Answer:

The resistance of the circuit is 1250 ohms

The inductance of the circuit is 0.063 mH.

Explanation:

Given;

current at resonance, I = 0.2 mA

applied voltage, V = 250 mV

resonance frequency, f₀ = 100 kHz

capacitance of the circuit, C = 0.04 μF

At resonance, capacitive reactance ( $X_c$ ) is equal to inductive reactance ( $X_l$ ),

$Z = \sqrt{R^2 + (X_ l - X_c)^2} \\\\But \ X_l= X_c\\\\Z = R$

Where;

R is the resistance of the circuit, calculated as;

$R = \frac{V}{I} \\\\R = \frac{250 \ \times \ 10^{-3}}{0.2 \ \times \ 10^{-3}} \\\\R = 1250 \ ohms$

The inductive reactance is calculated as;

$X_l = X_c = \frac{1}{\omega C} = \frac{1}{2\pi f_o C} = \frac{1}{2\pi (100\times 10^3)(0.04\times 10^{-6} ) } = 39.789 \ ohms\\$

The inductance is calculated as;

$X_l = \omega L = 2\pi f_o L\\\\L = \frac{X_l}{2\pi f_o}\\\\L = \frac{39.789}{2\pi (100 \times 10^3)} \\\\L= 6.3 \ \times \ 10^{-5} \ H\\\\L = 0.063 \times \ 10^{-3} \ H\\\\L = 0.063 \ mH$

5 0

2 years ago

Consider laminar, fully developed flow in a channel of constant surface temperature Ts. For a given mass flow rate and channel l

Lisa [10]

Answer:

Please see attachment.

Explanation:

6 0

2 years ago

An automobile having a mass of 1100 kg initially moves along a level highway at 110 km/h relative to the highway. It then climbs

Greeley [361]

Answer:

Change in kinetic energy=-513.652 KJ

Change in potential energy=431.64KJ

Explanation:

We are given that

Mass of an automobile , m=1100 kg

Initial speed, u=110 km/h= $110\times \frac{5}{18}=30.56 m/s$

Where $1 km/h=5/18 m/s$

Height , $h_2=40 m$

$g=9.81 m/s^2$

Final speed, v=0

Change in kinetic energy, $\Delta K.E=\frac{1}{2}m(v^2-u^2)$

$\Delta K.E=\frac{1}{2}(1100)(0-(30.56)^2)=-513652.48 J$

$\Delta K.E=-\frac{513652.48}{1000}=-513.652 KJ$

Where 1 KJ=1000 J

Change in potential energy, $\Delta P.E=mgh(h_2-h_1)$

Initially height, h1=0

Using the formula

$\Delta P.E=1100\times 9.81(40-0)$

$\Delta P.E=431640J$

$\Delta P.E=431.64KJ$

6 0

2 years ago

The penalty for littering 15 lb or less is _____.<br> A. $25<br> B. $50<br> C. $100<br> D. $150

Marina CMI [18]

Answer:

D the closest

Explanation:

i looked it up and 8t says 200

5 0

2 years ago