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3 years ago

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Compute the electrical resistivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in length i

n which a current of 0.25 A passes in an axial direction. A voltage of 24 V is measured across two probes that are separated by 45 mm (1.75 in.).

1 answer:

snow_tiger [21]3 years ago

8 0

No no no no no no no no no no

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How is a disc brake system different from a drum brake system? Short answer

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Answer:

Disc brake system use a slim rotor and small caliper to halt wheel movement but a drum brake system allow heat to build up inside the drum during heavy braking .

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2 years ago

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A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of

OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

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2 years ago

How do u charge ur phone? :)

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With a phone charger.

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3 years ago

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A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo

Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, $e_{min}$ = 0.51

Void ratio in the loosest state, $e_{max}$ = 0.87

Now,

Dry density, $\gamma_d=\frac{\gamma_t}{1+w}$

$=\frac{18}{1+0.05}$

= 17.14 kN/m³

Also,

$\gamma_d=\frac{G\gamma_w}{1+e}$

here, G = Specific gravity = 2.7 for sand

$17.14=\frac{2.7\times9.81}{1+e}$

or

e = 0.545

Relative density = $\frac{e_{max}-e}{e_{max}-e_{min}}$

= $\frac{0.87-0.545}{0.87-0.51}$

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

or

S = 0.2477

or

S = 0.2477 × 100% = 24.77%

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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.

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3 years ago