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3 years ago

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What is the composition, in atom percent, of an alloy that contains 44.5 lbmof Ag, 83.7 lbmof Au, and 5.3 lbmof Cu? What is the

composition in weight percent?

1 answer:

Vlad [161]3 years ago

4 0

Answer:

rr

Explanation:

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2. BCD uses 6 bits to represent a symbol. a) True b) False

Goshia [24]

Answer:

true because BCD used 6 bits to represent a symbol .

Explanation:

mark me brainlist

4 0

3 years ago

A coil with an average diameter of 5 inch will have an area of ""blank"" square meters

nadezda [96]

Answer:

19.64 square inches

Explanation:

Area will be (¶d^2)/4

= (3.142 x 5^2)/4

= 19.64 square inches

8 0

3 years ago

A vector AP is rotated about the Z-axis by 60 degrees and is subsequently rotated about X-axis by 30 degrees. Give the rotation

Musya8 [376]

Answer:

R = $\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]$ $\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]$

Explanation:

The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:

Let $\theta$ and $\alpha$ be the the angles 60⁰ and 30⁰ respectively

that is $\theta$ = 60⁰ and

$\alpha$ = 30⁰

The matrix is given by the following expression:

$\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]$ $\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]$

The angles can be evaluated and left in the surd form.

4 0

3 years ago

Technician A says that mechanical shifting controls can wear out over time. Technician B says that vacuum control rubber diaphra

diamong [38]

Based on the information, both technician A and technician B are correct.

<h3>How to depict the information?</h3>

From the information given, Technician A says that mechanical shifting controls can wear out over time.

Technician B says that vacuum control rubber diaphragms can deteriorate over time.

In this case, both technicians are correct as the information depicted is true.

Learn more about technicians on:

brainly.com/question/1548867

#SPJ12

8 0

2 years ago

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027

Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = $\frac{d}{2}$ = 3 inches

density, $\rho}$ = 493 lbm/ $ft^{3}$

S.G = 1.0027

g = 9.8 m/ $m^{2}$ = 386.22 inch/ $s^{2}$

Solution:

Using the formula for terminal velocity,

$v_{T}$ = $\sqrt{\frac{2V\rho g}{A \rho C_{d}}}$ (1)

$(Since, m = V\times \rho)$

where,

V = volume of sphere

$C_{d}$ = drag coefficient

Now,

Surface area of sphere, A = $4\pi r^{2}$

Volume of sphere, V = $\frac{4}{3} \pi r^{3}$

Using the above formulae in eqn (1):

$v_{T}$ = $\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2gr}{3C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}$

Therefore, terminal velcity is given by:

$v_{T}$ = $\frac{27.79}{\sqrt{C_d}}$ inch/sec

3 0

3 years ago