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3 years ago

To measure the current through a resistor, a _ must be placed in _ with the resistor.

Engineering

1 answer:

fgiga [73]3 years ago

5 0

Answer:

Explanation:

To measure the current through a resistor, the Ammeter must be placed in series with the resistor so that the same current that passes through the resistor also passes through the Ammeter. Ammeter is a measuring device used to measure current.

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3 years ago

Read 2 more answers

(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1

olya-2409 [2.1K]

Answer:

The answer is " $4.35 \times 10^{-3}\ mm$ and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point

$y = yo + \frac{k}{\sqrt{x}}$

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.

Replacing the equation for each condition:

$y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k$

People can get yo = 275 MPa with both equations and k= 15.5 Mpa $mm^{\frac{1}{2}}$ .

To substitute the answer,

$310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\ mm$

In point b:

The equation is $\sigma y = \sigma 0 + k y d^{\frac{1}{2}}$

equation is:

$75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\$

by putting the above value in the formula we get the $\sigma y$ value that is= 157.5 MPa

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3 years ago

Why are funeral home adding dining facilities to the business

Gelneren [198K]

Because they think it will make them more money

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3 years ago

A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c

Eduardwww [97]

Answer:

$h=1.99998\ W/m^2.C$

$k=33.333\ W/m.C$

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

$\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0$ Eq (1)

Where:

k is thermal Conductivity

$\dot e_{gen}$ is uniform thermal generation

$T(x) = a(L^2-x^2)+b$

$\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a$

Putt in Eq (1):

$-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C$

Energy balance is given by:

$Q_{convection}=Q_{conduction}$

$h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L$ Eq (2)

$T(x) = a(L^2-x^2)+b$

Putting x=L

$T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC$

$\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2$

From Eq (2)

$h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C$

7 0

3 years ago

To measure the current through a resistor, a _______________ must be placed in _______________ with the resistor.

To measure the current through a resistor, a _ must be placed in _ with the resistor.